mastering physics hw 5 ch 14 - oscillations
TRANSCRIPT
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 113
HW 5 Ch 14 - Oscillations
Due 1159pm on Tuesday October 13 2015
To understand how points are awarded read the Grading Policy for this assignment
PhET Tutorial Masses amp Springs
Learning Goal
To understand how the motion and energetics of a weight attached to a vertical spring depend on the mass the spring constant and initial conditions
For this tutorial use the PhET simulation Masses amp Springs You can put a weight on the end of a hanging spring stretch the spring and watch the
resulting motion
Start the simulation When you click the simulation link you may be asked whether to run open or save the file Choose to run or open it
You can drag a weight to the bottom of a spring and release it You can put only one weight on any spring With the weight on the spring you can click an
drag the weight up or down and release it Adjusting the friction slider bar at top right increases or decreases the amount of thermal dissipation (due to a
resistance and heating of the spring) You can adjust the spring constant of spring 3 using the softness spring 3 slider bar The horizontal dashed line
well as the ruler can be dragged to any position which is helpful for comparing positions of the springs
Feel free to play around with the simulation When you are done and before starting Part A set the friction slider bar to the middle and the gravitational
acceleration back to Earth
Part A
Place a 50 weight on spring 1 and release it Ev entually the weight will come to rest at an equilibrium position with the spring somewhat stretchcompared to its original (unweighted) length At this point the upward force of the spring balances the force of gravity on the weight
With the weight in its equilibrium position how does the amount the spring is stretched depend on the mass of the weight
ANSWER
Correct
Since the force of gravity on the weight increases as the mass increases the upward force of the spring must increase for the two forces to
balance (and the weight to therefore be in equilibrium) The force the spring exerts on the weight increases the more the spring is stretched fromits unweighted length
Part B
Use the simulation to estimate the masses of the three colored unlabeled weights Then place them into the appropriate mass bins
Hint 1 How to approach the problem
You learned in Part A that a heavier weight stretches the spring more at the equilibrium position Compare the equilibrium positions for each of
the colored weights to those for the weights with labeled masses You might want to verify that the three springs all stretch the same amount
for a particular weight so you can compare the equilibrium positions simultaneously
g
The spring stretches less for a heavier weight
The spring stretches more for a heavier weight
The stretch does not depend on mass
Typesetting math 100
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 213
ANSWER
Correct
Part C
Recall that in the equilibrium position the upward force of the spring balances the force of gravity on the weight Use this concept to estimate the
spring constant of spring 1
Express your answer in to one significant figure
Hint 1 How to approach the problem
The spring constant is used in Hookersquos law for the force exerted by the spring where is the amount the spring is stretched
When the spring is stretched to the equilibrium position the force exerted by the spring pulling up on the weight is equal to the force of gravity
pulling down Use this relationship to solve for
ANSWER
Correct
Part D
Now for parts D-F yoursquoll investigate the energetics of the spring
Select 1 in the Show Energy of box which shows an energy bar diagram Select the g = 0 option (under the planet names) which simulates what
happens without any gravitational forces (and consequently removes gravitational potential energy from the energetics) Adjust the friction slider to
none (this prevents any thermal energy from being generated) Place a weight on spring 1 stretch it and release it Watch how the kinetic energy
and elastic potential energy vary with time (You can slow down or stop time using the buttons next to the list of planets)
When is the elastic potential energy of the spring a maximum
ANSWER
983147
N m
983147 F = minus 983147 983160 983160
983147
= 10983147 N m
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 313
Correct
The elastic potential energy depends on the magnitude of the change in the length of the spring Mathematically the elastic potential energy is
given by where is the spring constant and is the difference between the length of the spring and its unweighted length A
compression of therefore results in the same elastic potential energy as a stretch of
Part E
When is the kinetic energy of the mass a maximum
ANSWER
Correct
The total energy of the system is equal to the kinetic energy of the mass (since the spring has negligible mass) plus the elastic potential energy
Since this total energy is conserved the kinetic energy is a maximum when the elastic potential energy is a minimum which occurs when the
spring is at its unweighted length
Part F
Select Earth in the menu box so that there is now a force of gravity Now the total energy of the massspring system is the sum of the kinetic energy
the elastic potential energy and the gravitational potential energy
When is the kinetic energy a maximum (It may help to watch the simulation in slow motion - 116 time)
ANSWER
Correct
The total potential energy (the gravitational potential energy plus elastic potential energy) is a minimum at the equilibrium position even though
there is some elastic potential energy when the mass is at this location The kinetic energy is always a maximum when the total potential energy
is a minimum (since the total energy is conserved)
Part G
Now for parts G-I yoursquoll investigate what determines the frequency of oscillation For these parts turn off the friction using the slider bar
Select the stopwatch and time how long it takes for a weight to oscillate back and forth 10 times The period of oscillation is this time divided by 10
The frequency of oscillation is one divided by the period
How does the frequency of oscillation depend on the mass of the weight
ANSWER
When the spring isnrsquot stretched or compressed
When the spring is most stretched
When the spring is most compressed
Both when the spring is most compressed and when the spring is most stretched
= ( 1 2 ) 983147 U
e l a s
983160
2
983147 983160
983160 983160
When the spring is most compressed
Both when the spring is most compressed and when the spring is most stretched
When the spring is most stretched
When the spring is at its unweighted length (when it isnrsquot stretched or compressed)
When the spring is at its unweighted length (its equilibrium position without the weight attached)
When the mass is at the equilibrium position
When the spring is most stretched or most compressed
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 413
Correct
A greater mass results in a lower frequency and a longer period of oscillation
Part H
The amplitude of oscillation is the maximum distance between the oscillating weight and the equilibrium position Determine the frequency of
oscillation for several different amplitudes by pulling the weight down different amounts
How does the frequency depend on the amplitude of oscillation
ANSWER
CorrectEven though the weight has to travel farther each oscillation if the amplitude is greater the spring on average exerts a stronger force causing a
greater acceleration and a greater average speed The effects of the longer distance and faster speed cancel out so that the period of oscillation
doesnrsquot change
Part I
The spring constant of spring 3 can be adjusted with the softness spring 3 slider bar (harder means a greater spring constant or stiffer spring)
How does the frequency of oscillation depend on the spring constant
Hint 1 How to approach the problem
You can place a weight on spring 3 with the softness of spring 3 set to a very low (soft) value and place another weight with the same masson spring 1 Then release the two weights and compare their frequencies of oscillation
ANSWER
Correct
It turns out that the frequency of oscillation depends on the square root of the ratio of the spring constant to mass where
is the frequency A stiffer spring constant causes the frequency to increase Sports cars use stiff springs whereas large plush Cadillacs use softsprings for their suspension
PhET Interactive Simulations
University of Colorado
httpphetcoloradoedu
PhET Tutorial Pendulum Lab
Learning Goal
To understand the relationships of the energetics forces acceleration and velocity of an oscillating pendulum and to determine how the motion of a
pendulum depends on the mass the length of the string and the acceleration due to gravity
The frequency increases as the mass increases
The frequency is independent of the mass
The frequency decreases as the mass increases
The frequency increases as the amplitude increases
The frequency decreases as the amplitude increases
The frequency is independent of the amplitude
The frequency increases as the spring constant increases
The frequency is independent of the spring constant
The frequency decreases as the spring constant increases
983142 = 2 π ( 983147 983149 )
minus minus minus minus minus
radic 983142
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 513
For this tutorial use the PhET simulation Pendulum Lab This simulation mimics a real pendulum and allows you to adjust the initial position the mass
and the length of the pendulum
Start the simulation You can drag the pendulum to an arbitrary initial angle and release it from rest You can adjust the length and the mass of the
pendulum using the slider bars at the top of the green panel Velocity and acceleration vectors can be selected to be shown as well as the forms of
energy
Feel free to play around with the simulation When you are done click the Reset button
Part A
Select to show the energy of pendulum 1 Be sure that friction is set to none
Drag the pendulum to an angle (with respect to the vertical) of and then release itWhen the pendulum is at what form(s) of energy does it have
Check all that apply
ANSWER
Correct
The pendulum starts off with no kinetic energy since it is released from rest so it initially only has potential energy When the pendulum is at
it is just as high above the ground as when it started so it must have the same amount of potential energy as it initially had Since thetotal energy is conserved it canrsquot have other forms of energy at (if it did it would have more energy there than it initially had) so it again has
only potential energy
Part B
Drag the pendulum to an angle (with respect to the vertical) of and then release it
Where is the pendulum swinging the fastest
ANSWER
Correct
The pendulum has the least potential energy at this location since it is at the lowest point in the arc (in fact for this simulation the potential
energy reference location is here so it has no potential energy) This means that the kinetic energy is greatest here so the pendulum is moving
the fastest
Part C
3 0
∘
minus 3 0
∘
Kinetic energy
Thermal energy
Potential energy
minus 3 0
∘
minus 3 0
∘
3 0
∘
at
at
at
at
1 5
∘
minus 3 0
∘
0
∘
3 0
∘
3 0
∘
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 613
Drag the pendulum to an angle (with respect to the vertical) of and then release it Select to show the acceleration vector
With the pendulum swinging back and forth at which locations is the acceleration equal to zero
ANSWER
Correct
The pendulum is moving in a circular path so its velocity is never constant In fact for most locations the acceleration has both a radial
component (the centripetal acceleration which is directed along the rope) and a tangential component (due to the speed changing directed along
the path the only place the tangential acceleration is zero is when the angle is )
Part D
With the pendulum swinging back and forth how does the tension of the rope compare to the force of gravity when the angle is
Hint 1 How to approach the problem
Look at the direction of the acceleration when the angle is (you can slow down or pause the simulation to see this more clearly) and think
about the relationship between the net force acting on the pendulum and the acceleration (Newtonrsquos 2nd law of motion) It would probably help todraw a free-body diagram for the mass
ANSWER
Correct
Since the acceleration of the pendulum is directed up when the angle is the net force must be directed up (Newtonrsquos 2nd law) This meansthat the upward force of tension must be stronger than the downward force of gravity
Part E
Drag the pendulum to an angle (with respect to the vertical) of and then release it
With the pendulum swinging back and forth where is the tension equal to zero
ANSWER
Correct
At these locations the acceleration is solely due to gravity and directed downward Thus the net force acting on t he pendulum i s also directed
downward meaning there are no horizontal forces This requires the tension to be zero
Part F
Now for parts F-I you will investigate how the period of oscillation depends on the properties of the pendulum
The period of oscillation is the amount of time it takes for the pendulum to take a full swing going from the original angle to the other side and
3 0
∘
The acceleration is zero when the angle is either or
The acceleration is zero when the angle is
The acceleration is never equal to zero as it swings back and forth
+ 3 0
∘
minus 3 0
∘
0
∘
0
∘
0
∘
0
∘
The tension is equal to the force of gravity
The tension is greater than the force of gravity only if it is swinging really fast
The tension is less than the force of gravity
The tension is greater than the force of gravity
0
∘
9 0
∘
The tension is zero at the angles and
The tension is zero when the angle is and
The tension is zero when the angle is
The tension is never zero
+ 9 0
∘
minus 9 0
∘
+ 4 5
∘
minus 4 5
∘
0
∘
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 713
returning to the original angle You can determine the period by selecting other tools which gives you a stopwatch With the pendulum swinging yo
can start the stopwatch when the pendulum is at its original angle and time how long it takes to complete 10 swings The period will be this time
interval divided by 10 (this method is more accurate than trying to time one swing)
Set the length of the pendulum to 10 and the mass to 10 Click Reset and then drag the pendulum to an angle (with respect to the vertical) o
and release it What is the period of oscillation
ANSWER
Correct
A - - pendulum completes one osc illation in 20
Part G
How does the period of oscillation depend on the initial angle of the pendulum when released (Be sure to measure the period for initial angles much
greater than )
ANSWER
Correct
Unlike a harmonic oscillator such as a mass on a spring the period actually depends on the initial angle For small angles (eg ) it is a
pretty good approximation that the period doesnrsquot change but for larger angles the period does in fact increase
Part H
Keeping the length of the pendulum fixed determine the period for a few different masses (Alternatively you can set up two pendulums by selecting
Show 2nd pendulum Adjust the lengths to be the same and have one pendulum with a higher mass You can release one and then release the
other with the same angle when the first one is back at that angle)
How does the period of the pendulum depend on mass
ANSWER
Correct
The period of a pendulum does not depend on mass The reason for this result is very similar to the reason that without air resistance all objects
fall to the ground at the same rate (changing the mass changes both the inertia and the force of gravity by the same amount)
Part I
Now keep the mass fixed to any value you choose and measure the period for several different pendulum lengths
How does the period of the pendulum depend on the length
ANSWER
m k g
3 0
∘
40
15
200
20
10
05
s
s
s
s
s
s
1 0 m 1 0 k g s
3 0
∘
The period is longer when the initial angle is greater
The period is shorter when the initial angle is greater
The period is independent of the initial angle
lt 3 0
∘
A heavier pendulum has a shorter period
The period is independent of the pendulumrsquos mass
A heavier pendulum has a longer period
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 813
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 913
Express your answer to two significant figures and include the appropriate units
ANSWER
All attempts used correct answer withheld by instructor
Problem 1480
The figure shows a 240 uniform rod pivoted at one end The other end is attached to a horizontal spring The spring is neither stretched nor compresse
when the rod hangs straight down
Part A
What is the rods oscillation period You can assume that the rods angle from vertical is always small
Express your answer to two significant figures and include the appropriate units
ANSWER
Correct
Problem 1458
A geologist needs to determine the local value of Unfortunately his only tools are a meter s tick a saw and a stopwatch He s tarts by hanging the me
stic k from one end and measuring its frequency as it swings He then saws off 20 mdashusing the centimeter markingsmdashand measures the frequency
again After two more cuts these are his data
Length Frequency
100 061
80 067
60 079
40 096
Part A
You want to determine the local value of by graphing the data Select the appropriate variables to graph on each axis that will produce a straight-lin
graph with either a slope or intercept that is related to
Sort all variables into the appropriate bins
Hint 1 How to approach the problem
= 4410^-21983149
D N A
k g
g
= 060T s
983143
c m
L ( c m ) 983142 ( H z )
983143
983143
I L
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1013
A physical pendulum has moment of inertia and dist ance between the pivot and the center of mass Its oscillation frequency is
ANSWER
ANSWER
Correct
Part B
Graphing either versus or versus gives a straight line In the graph shown we chose to plot on the vertical axis and on th
horizontal axis From the equation for the line of best fit given determine the local value of
Express your answer to three significant figures and include the appropriate units
ANSWER
I L
983142 =
1
2 π
M 983143 L
I
983142 =
1
2 π
I
M 983143 L
983142 =
1
2 π
M 983143 L
I
minus minus minus minus
radic
983142 =
1
2 π
I
M 983143 L
minus minus minus minus
radic
983142
2
1 L 1 L 983142
2
983142
2
1 L
983143
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1113
Correct
Conceptual Question 1410
Suppose the damping constant of an oscillator increases
Part A
Is the medium more resistive or less resistive
ANSWER
Correct
Part B
Do the oscillations damp out more quickly or less quickly
ANSWER
Correct
Part C
Is the time constant increased or decreased
ANSWER
Correct
Problem 1432
The two graphs in the figure are for two different vertical mass-spring systems
= 976983143
m
s
2
b
The medium is more resistive
The medium is less resistive
The oscillations will damp out more quickly
The oscillations will damp out less quickly
τ
is increased
is decreased
τ
τ
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 213
ANSWER
Correct
Part C
Recall that in the equilibrium position the upward force of the spring balances the force of gravity on the weight Use this concept to estimate the
spring constant of spring 1
Express your answer in to one significant figure
Hint 1 How to approach the problem
The spring constant is used in Hookersquos law for the force exerted by the spring where is the amount the spring is stretched
When the spring is stretched to the equilibrium position the force exerted by the spring pulling up on the weight is equal to the force of gravity
pulling down Use this relationship to solve for
ANSWER
Correct
Part D
Now for parts D-F yoursquoll investigate the energetics of the spring
Select 1 in the Show Energy of box which shows an energy bar diagram Select the g = 0 option (under the planet names) which simulates what
happens without any gravitational forces (and consequently removes gravitational potential energy from the energetics) Adjust the friction slider to
none (this prevents any thermal energy from being generated) Place a weight on spring 1 stretch it and release it Watch how the kinetic energy
and elastic potential energy vary with time (You can slow down or stop time using the buttons next to the list of planets)
When is the elastic potential energy of the spring a maximum
ANSWER
983147
N m
983147 F = minus 983147 983160 983160
983147
= 10983147 N m
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 313
Correct
The elastic potential energy depends on the magnitude of the change in the length of the spring Mathematically the elastic potential energy is
given by where is the spring constant and is the difference between the length of the spring and its unweighted length A
compression of therefore results in the same elastic potential energy as a stretch of
Part E
When is the kinetic energy of the mass a maximum
ANSWER
Correct
The total energy of the system is equal to the kinetic energy of the mass (since the spring has negligible mass) plus the elastic potential energy
Since this total energy is conserved the kinetic energy is a maximum when the elastic potential energy is a minimum which occurs when the
spring is at its unweighted length
Part F
Select Earth in the menu box so that there is now a force of gravity Now the total energy of the massspring system is the sum of the kinetic energy
the elastic potential energy and the gravitational potential energy
When is the kinetic energy a maximum (It may help to watch the simulation in slow motion - 116 time)
ANSWER
Correct
The total potential energy (the gravitational potential energy plus elastic potential energy) is a minimum at the equilibrium position even though
there is some elastic potential energy when the mass is at this location The kinetic energy is always a maximum when the total potential energy
is a minimum (since the total energy is conserved)
Part G
Now for parts G-I yoursquoll investigate what determines the frequency of oscillation For these parts turn off the friction using the slider bar
Select the stopwatch and time how long it takes for a weight to oscillate back and forth 10 times The period of oscillation is this time divided by 10
The frequency of oscillation is one divided by the period
How does the frequency of oscillation depend on the mass of the weight
ANSWER
When the spring isnrsquot stretched or compressed
When the spring is most stretched
When the spring is most compressed
Both when the spring is most compressed and when the spring is most stretched
= ( 1 2 ) 983147 U
e l a s
983160
2
983147 983160
983160 983160
When the spring is most compressed
Both when the spring is most compressed and when the spring is most stretched
When the spring is most stretched
When the spring is at its unweighted length (when it isnrsquot stretched or compressed)
When the spring is at its unweighted length (its equilibrium position without the weight attached)
When the mass is at the equilibrium position
When the spring is most stretched or most compressed
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 413
Correct
A greater mass results in a lower frequency and a longer period of oscillation
Part H
The amplitude of oscillation is the maximum distance between the oscillating weight and the equilibrium position Determine the frequency of
oscillation for several different amplitudes by pulling the weight down different amounts
How does the frequency depend on the amplitude of oscillation
ANSWER
CorrectEven though the weight has to travel farther each oscillation if the amplitude is greater the spring on average exerts a stronger force causing a
greater acceleration and a greater average speed The effects of the longer distance and faster speed cancel out so that the period of oscillation
doesnrsquot change
Part I
The spring constant of spring 3 can be adjusted with the softness spring 3 slider bar (harder means a greater spring constant or stiffer spring)
How does the frequency of oscillation depend on the spring constant
Hint 1 How to approach the problem
You can place a weight on spring 3 with the softness of spring 3 set to a very low (soft) value and place another weight with the same masson spring 1 Then release the two weights and compare their frequencies of oscillation
ANSWER
Correct
It turns out that the frequency of oscillation depends on the square root of the ratio of the spring constant to mass where
is the frequency A stiffer spring constant causes the frequency to increase Sports cars use stiff springs whereas large plush Cadillacs use softsprings for their suspension
PhET Interactive Simulations
University of Colorado
httpphetcoloradoedu
PhET Tutorial Pendulum Lab
Learning Goal
To understand the relationships of the energetics forces acceleration and velocity of an oscillating pendulum and to determine how the motion of a
pendulum depends on the mass the length of the string and the acceleration due to gravity
The frequency increases as the mass increases
The frequency is independent of the mass
The frequency decreases as the mass increases
The frequency increases as the amplitude increases
The frequency decreases as the amplitude increases
The frequency is independent of the amplitude
The frequency increases as the spring constant increases
The frequency is independent of the spring constant
The frequency decreases as the spring constant increases
983142 = 2 π ( 983147 983149 )
minus minus minus minus minus
radic 983142
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 513
For this tutorial use the PhET simulation Pendulum Lab This simulation mimics a real pendulum and allows you to adjust the initial position the mass
and the length of the pendulum
Start the simulation You can drag the pendulum to an arbitrary initial angle and release it from rest You can adjust the length and the mass of the
pendulum using the slider bars at the top of the green panel Velocity and acceleration vectors can be selected to be shown as well as the forms of
energy
Feel free to play around with the simulation When you are done click the Reset button
Part A
Select to show the energy of pendulum 1 Be sure that friction is set to none
Drag the pendulum to an angle (with respect to the vertical) of and then release itWhen the pendulum is at what form(s) of energy does it have
Check all that apply
ANSWER
Correct
The pendulum starts off with no kinetic energy since it is released from rest so it initially only has potential energy When the pendulum is at
it is just as high above the ground as when it started so it must have the same amount of potential energy as it initially had Since thetotal energy is conserved it canrsquot have other forms of energy at (if it did it would have more energy there than it initially had) so it again has
only potential energy
Part B
Drag the pendulum to an angle (with respect to the vertical) of and then release it
Where is the pendulum swinging the fastest
ANSWER
Correct
The pendulum has the least potential energy at this location since it is at the lowest point in the arc (in fact for this simulation the potential
energy reference location is here so it has no potential energy) This means that the kinetic energy is greatest here so the pendulum is moving
the fastest
Part C
3 0
∘
minus 3 0
∘
Kinetic energy
Thermal energy
Potential energy
minus 3 0
∘
minus 3 0
∘
3 0
∘
at
at
at
at
1 5
∘
minus 3 0
∘
0
∘
3 0
∘
3 0
∘
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 613
Drag the pendulum to an angle (with respect to the vertical) of and then release it Select to show the acceleration vector
With the pendulum swinging back and forth at which locations is the acceleration equal to zero
ANSWER
Correct
The pendulum is moving in a circular path so its velocity is never constant In fact for most locations the acceleration has both a radial
component (the centripetal acceleration which is directed along the rope) and a tangential component (due to the speed changing directed along
the path the only place the tangential acceleration is zero is when the angle is )
Part D
With the pendulum swinging back and forth how does the tension of the rope compare to the force of gravity when the angle is
Hint 1 How to approach the problem
Look at the direction of the acceleration when the angle is (you can slow down or pause the simulation to see this more clearly) and think
about the relationship between the net force acting on the pendulum and the acceleration (Newtonrsquos 2nd law of motion) It would probably help todraw a free-body diagram for the mass
ANSWER
Correct
Since the acceleration of the pendulum is directed up when the angle is the net force must be directed up (Newtonrsquos 2nd law) This meansthat the upward force of tension must be stronger than the downward force of gravity
Part E
Drag the pendulum to an angle (with respect to the vertical) of and then release it
With the pendulum swinging back and forth where is the tension equal to zero
ANSWER
Correct
At these locations the acceleration is solely due to gravity and directed downward Thus the net force acting on t he pendulum i s also directed
downward meaning there are no horizontal forces This requires the tension to be zero
Part F
Now for parts F-I you will investigate how the period of oscillation depends on the properties of the pendulum
The period of oscillation is the amount of time it takes for the pendulum to take a full swing going from the original angle to the other side and
3 0
∘
The acceleration is zero when the angle is either or
The acceleration is zero when the angle is
The acceleration is never equal to zero as it swings back and forth
+ 3 0
∘
minus 3 0
∘
0
∘
0
∘
0
∘
0
∘
The tension is equal to the force of gravity
The tension is greater than the force of gravity only if it is swinging really fast
The tension is less than the force of gravity
The tension is greater than the force of gravity
0
∘
9 0
∘
The tension is zero at the angles and
The tension is zero when the angle is and
The tension is zero when the angle is
The tension is never zero
+ 9 0
∘
minus 9 0
∘
+ 4 5
∘
minus 4 5
∘
0
∘
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 713
returning to the original angle You can determine the period by selecting other tools which gives you a stopwatch With the pendulum swinging yo
can start the stopwatch when the pendulum is at its original angle and time how long it takes to complete 10 swings The period will be this time
interval divided by 10 (this method is more accurate than trying to time one swing)
Set the length of the pendulum to 10 and the mass to 10 Click Reset and then drag the pendulum to an angle (with respect to the vertical) o
and release it What is the period of oscillation
ANSWER
Correct
A - - pendulum completes one osc illation in 20
Part G
How does the period of oscillation depend on the initial angle of the pendulum when released (Be sure to measure the period for initial angles much
greater than )
ANSWER
Correct
Unlike a harmonic oscillator such as a mass on a spring the period actually depends on the initial angle For small angles (eg ) it is a
pretty good approximation that the period doesnrsquot change but for larger angles the period does in fact increase
Part H
Keeping the length of the pendulum fixed determine the period for a few different masses (Alternatively you can set up two pendulums by selecting
Show 2nd pendulum Adjust the lengths to be the same and have one pendulum with a higher mass You can release one and then release the
other with the same angle when the first one is back at that angle)
How does the period of the pendulum depend on mass
ANSWER
Correct
The period of a pendulum does not depend on mass The reason for this result is very similar to the reason that without air resistance all objects
fall to the ground at the same rate (changing the mass changes both the inertia and the force of gravity by the same amount)
Part I
Now keep the mass fixed to any value you choose and measure the period for several different pendulum lengths
How does the period of the pendulum depend on the length
ANSWER
m k g
3 0
∘
40
15
200
20
10
05
s
s
s
s
s
s
1 0 m 1 0 k g s
3 0
∘
The period is longer when the initial angle is greater
The period is shorter when the initial angle is greater
The period is independent of the initial angle
lt 3 0
∘
A heavier pendulum has a shorter period
The period is independent of the pendulumrsquos mass
A heavier pendulum has a longer period
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 813
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 913
Express your answer to two significant figures and include the appropriate units
ANSWER
All attempts used correct answer withheld by instructor
Problem 1480
The figure shows a 240 uniform rod pivoted at one end The other end is attached to a horizontal spring The spring is neither stretched nor compresse
when the rod hangs straight down
Part A
What is the rods oscillation period You can assume that the rods angle from vertical is always small
Express your answer to two significant figures and include the appropriate units
ANSWER
Correct
Problem 1458
A geologist needs to determine the local value of Unfortunately his only tools are a meter s tick a saw and a stopwatch He s tarts by hanging the me
stic k from one end and measuring its frequency as it swings He then saws off 20 mdashusing the centimeter markingsmdashand measures the frequency
again After two more cuts these are his data
Length Frequency
100 061
80 067
60 079
40 096
Part A
You want to determine the local value of by graphing the data Select the appropriate variables to graph on each axis that will produce a straight-lin
graph with either a slope or intercept that is related to
Sort all variables into the appropriate bins
Hint 1 How to approach the problem
= 4410^-21983149
D N A
k g
g
= 060T s
983143
c m
L ( c m ) 983142 ( H z )
983143
983143
I L
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1013
A physical pendulum has moment of inertia and dist ance between the pivot and the center of mass Its oscillation frequency is
ANSWER
ANSWER
Correct
Part B
Graphing either versus or versus gives a straight line In the graph shown we chose to plot on the vertical axis and on th
horizontal axis From the equation for the line of best fit given determine the local value of
Express your answer to three significant figures and include the appropriate units
ANSWER
I L
983142 =
1
2 π
M 983143 L
I
983142 =
1
2 π
I
M 983143 L
983142 =
1
2 π
M 983143 L
I
minus minus minus minus
radic
983142 =
1
2 π
I
M 983143 L
minus minus minus minus
radic
983142
2
1 L 1 L 983142
2
983142
2
1 L
983143
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1113
Correct
Conceptual Question 1410
Suppose the damping constant of an oscillator increases
Part A
Is the medium more resistive or less resistive
ANSWER
Correct
Part B
Do the oscillations damp out more quickly or less quickly
ANSWER
Correct
Part C
Is the time constant increased or decreased
ANSWER
Correct
Problem 1432
The two graphs in the figure are for two different vertical mass-spring systems
= 976983143
m
s
2
b
The medium is more resistive
The medium is less resistive
The oscillations will damp out more quickly
The oscillations will damp out less quickly
τ
is increased
is decreased
τ
τ
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 313
Correct
The elastic potential energy depends on the magnitude of the change in the length of the spring Mathematically the elastic potential energy is
given by where is the spring constant and is the difference between the length of the spring and its unweighted length A
compression of therefore results in the same elastic potential energy as a stretch of
Part E
When is the kinetic energy of the mass a maximum
ANSWER
Correct
The total energy of the system is equal to the kinetic energy of the mass (since the spring has negligible mass) plus the elastic potential energy
Since this total energy is conserved the kinetic energy is a maximum when the elastic potential energy is a minimum which occurs when the
spring is at its unweighted length
Part F
Select Earth in the menu box so that there is now a force of gravity Now the total energy of the massspring system is the sum of the kinetic energy
the elastic potential energy and the gravitational potential energy
When is the kinetic energy a maximum (It may help to watch the simulation in slow motion - 116 time)
ANSWER
Correct
The total potential energy (the gravitational potential energy plus elastic potential energy) is a minimum at the equilibrium position even though
there is some elastic potential energy when the mass is at this location The kinetic energy is always a maximum when the total potential energy
is a minimum (since the total energy is conserved)
Part G
Now for parts G-I yoursquoll investigate what determines the frequency of oscillation For these parts turn off the friction using the slider bar
Select the stopwatch and time how long it takes for a weight to oscillate back and forth 10 times The period of oscillation is this time divided by 10
The frequency of oscillation is one divided by the period
How does the frequency of oscillation depend on the mass of the weight
ANSWER
When the spring isnrsquot stretched or compressed
When the spring is most stretched
When the spring is most compressed
Both when the spring is most compressed and when the spring is most stretched
= ( 1 2 ) 983147 U
e l a s
983160
2
983147 983160
983160 983160
When the spring is most compressed
Both when the spring is most compressed and when the spring is most stretched
When the spring is most stretched
When the spring is at its unweighted length (when it isnrsquot stretched or compressed)
When the spring is at its unweighted length (its equilibrium position without the weight attached)
When the mass is at the equilibrium position
When the spring is most stretched or most compressed
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 413
Correct
A greater mass results in a lower frequency and a longer period of oscillation
Part H
The amplitude of oscillation is the maximum distance between the oscillating weight and the equilibrium position Determine the frequency of
oscillation for several different amplitudes by pulling the weight down different amounts
How does the frequency depend on the amplitude of oscillation
ANSWER
CorrectEven though the weight has to travel farther each oscillation if the amplitude is greater the spring on average exerts a stronger force causing a
greater acceleration and a greater average speed The effects of the longer distance and faster speed cancel out so that the period of oscillation
doesnrsquot change
Part I
The spring constant of spring 3 can be adjusted with the softness spring 3 slider bar (harder means a greater spring constant or stiffer spring)
How does the frequency of oscillation depend on the spring constant
Hint 1 How to approach the problem
You can place a weight on spring 3 with the softness of spring 3 set to a very low (soft) value and place another weight with the same masson spring 1 Then release the two weights and compare their frequencies of oscillation
ANSWER
Correct
It turns out that the frequency of oscillation depends on the square root of the ratio of the spring constant to mass where
is the frequency A stiffer spring constant causes the frequency to increase Sports cars use stiff springs whereas large plush Cadillacs use softsprings for their suspension
PhET Interactive Simulations
University of Colorado
httpphetcoloradoedu
PhET Tutorial Pendulum Lab
Learning Goal
To understand the relationships of the energetics forces acceleration and velocity of an oscillating pendulum and to determine how the motion of a
pendulum depends on the mass the length of the string and the acceleration due to gravity
The frequency increases as the mass increases
The frequency is independent of the mass
The frequency decreases as the mass increases
The frequency increases as the amplitude increases
The frequency decreases as the amplitude increases
The frequency is independent of the amplitude
The frequency increases as the spring constant increases
The frequency is independent of the spring constant
The frequency decreases as the spring constant increases
983142 = 2 π ( 983147 983149 )
minus minus minus minus minus
radic 983142
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 513
For this tutorial use the PhET simulation Pendulum Lab This simulation mimics a real pendulum and allows you to adjust the initial position the mass
and the length of the pendulum
Start the simulation You can drag the pendulum to an arbitrary initial angle and release it from rest You can adjust the length and the mass of the
pendulum using the slider bars at the top of the green panel Velocity and acceleration vectors can be selected to be shown as well as the forms of
energy
Feel free to play around with the simulation When you are done click the Reset button
Part A
Select to show the energy of pendulum 1 Be sure that friction is set to none
Drag the pendulum to an angle (with respect to the vertical) of and then release itWhen the pendulum is at what form(s) of energy does it have
Check all that apply
ANSWER
Correct
The pendulum starts off with no kinetic energy since it is released from rest so it initially only has potential energy When the pendulum is at
it is just as high above the ground as when it started so it must have the same amount of potential energy as it initially had Since thetotal energy is conserved it canrsquot have other forms of energy at (if it did it would have more energy there than it initially had) so it again has
only potential energy
Part B
Drag the pendulum to an angle (with respect to the vertical) of and then release it
Where is the pendulum swinging the fastest
ANSWER
Correct
The pendulum has the least potential energy at this location since it is at the lowest point in the arc (in fact for this simulation the potential
energy reference location is here so it has no potential energy) This means that the kinetic energy is greatest here so the pendulum is moving
the fastest
Part C
3 0
∘
minus 3 0
∘
Kinetic energy
Thermal energy
Potential energy
minus 3 0
∘
minus 3 0
∘
3 0
∘
at
at
at
at
1 5
∘
minus 3 0
∘
0
∘
3 0
∘
3 0
∘
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 613
Drag the pendulum to an angle (with respect to the vertical) of and then release it Select to show the acceleration vector
With the pendulum swinging back and forth at which locations is the acceleration equal to zero
ANSWER
Correct
The pendulum is moving in a circular path so its velocity is never constant In fact for most locations the acceleration has both a radial
component (the centripetal acceleration which is directed along the rope) and a tangential component (due to the speed changing directed along
the path the only place the tangential acceleration is zero is when the angle is )
Part D
With the pendulum swinging back and forth how does the tension of the rope compare to the force of gravity when the angle is
Hint 1 How to approach the problem
Look at the direction of the acceleration when the angle is (you can slow down or pause the simulation to see this more clearly) and think
about the relationship between the net force acting on the pendulum and the acceleration (Newtonrsquos 2nd law of motion) It would probably help todraw a free-body diagram for the mass
ANSWER
Correct
Since the acceleration of the pendulum is directed up when the angle is the net force must be directed up (Newtonrsquos 2nd law) This meansthat the upward force of tension must be stronger than the downward force of gravity
Part E
Drag the pendulum to an angle (with respect to the vertical) of and then release it
With the pendulum swinging back and forth where is the tension equal to zero
ANSWER
Correct
At these locations the acceleration is solely due to gravity and directed downward Thus the net force acting on t he pendulum i s also directed
downward meaning there are no horizontal forces This requires the tension to be zero
Part F
Now for parts F-I you will investigate how the period of oscillation depends on the properties of the pendulum
The period of oscillation is the amount of time it takes for the pendulum to take a full swing going from the original angle to the other side and
3 0
∘
The acceleration is zero when the angle is either or
The acceleration is zero when the angle is
The acceleration is never equal to zero as it swings back and forth
+ 3 0
∘
minus 3 0
∘
0
∘
0
∘
0
∘
0
∘
The tension is equal to the force of gravity
The tension is greater than the force of gravity only if it is swinging really fast
The tension is less than the force of gravity
The tension is greater than the force of gravity
0
∘
9 0
∘
The tension is zero at the angles and
The tension is zero when the angle is and
The tension is zero when the angle is
The tension is never zero
+ 9 0
∘
minus 9 0
∘
+ 4 5
∘
minus 4 5
∘
0
∘
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 713
returning to the original angle You can determine the period by selecting other tools which gives you a stopwatch With the pendulum swinging yo
can start the stopwatch when the pendulum is at its original angle and time how long it takes to complete 10 swings The period will be this time
interval divided by 10 (this method is more accurate than trying to time one swing)
Set the length of the pendulum to 10 and the mass to 10 Click Reset and then drag the pendulum to an angle (with respect to the vertical) o
and release it What is the period of oscillation
ANSWER
Correct
A - - pendulum completes one osc illation in 20
Part G
How does the period of oscillation depend on the initial angle of the pendulum when released (Be sure to measure the period for initial angles much
greater than )
ANSWER
Correct
Unlike a harmonic oscillator such as a mass on a spring the period actually depends on the initial angle For small angles (eg ) it is a
pretty good approximation that the period doesnrsquot change but for larger angles the period does in fact increase
Part H
Keeping the length of the pendulum fixed determine the period for a few different masses (Alternatively you can set up two pendulums by selecting
Show 2nd pendulum Adjust the lengths to be the same and have one pendulum with a higher mass You can release one and then release the
other with the same angle when the first one is back at that angle)
How does the period of the pendulum depend on mass
ANSWER
Correct
The period of a pendulum does not depend on mass The reason for this result is very similar to the reason that without air resistance all objects
fall to the ground at the same rate (changing the mass changes both the inertia and the force of gravity by the same amount)
Part I
Now keep the mass fixed to any value you choose and measure the period for several different pendulum lengths
How does the period of the pendulum depend on the length
ANSWER
m k g
3 0
∘
40
15
200
20
10
05
s
s
s
s
s
s
1 0 m 1 0 k g s
3 0
∘
The period is longer when the initial angle is greater
The period is shorter when the initial angle is greater
The period is independent of the initial angle
lt 3 0
∘
A heavier pendulum has a shorter period
The period is independent of the pendulumrsquos mass
A heavier pendulum has a longer period
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 813
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 913
Express your answer to two significant figures and include the appropriate units
ANSWER
All attempts used correct answer withheld by instructor
Problem 1480
The figure shows a 240 uniform rod pivoted at one end The other end is attached to a horizontal spring The spring is neither stretched nor compresse
when the rod hangs straight down
Part A
What is the rods oscillation period You can assume that the rods angle from vertical is always small
Express your answer to two significant figures and include the appropriate units
ANSWER
Correct
Problem 1458
A geologist needs to determine the local value of Unfortunately his only tools are a meter s tick a saw and a stopwatch He s tarts by hanging the me
stic k from one end and measuring its frequency as it swings He then saws off 20 mdashusing the centimeter markingsmdashand measures the frequency
again After two more cuts these are his data
Length Frequency
100 061
80 067
60 079
40 096
Part A
You want to determine the local value of by graphing the data Select the appropriate variables to graph on each axis that will produce a straight-lin
graph with either a slope or intercept that is related to
Sort all variables into the appropriate bins
Hint 1 How to approach the problem
= 4410^-21983149
D N A
k g
g
= 060T s
983143
c m
L ( c m ) 983142 ( H z )
983143
983143
I L
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1013
A physical pendulum has moment of inertia and dist ance between the pivot and the center of mass Its oscillation frequency is
ANSWER
ANSWER
Correct
Part B
Graphing either versus or versus gives a straight line In the graph shown we chose to plot on the vertical axis and on th
horizontal axis From the equation for the line of best fit given determine the local value of
Express your answer to three significant figures and include the appropriate units
ANSWER
I L
983142 =
1
2 π
M 983143 L
I
983142 =
1
2 π
I
M 983143 L
983142 =
1
2 π
M 983143 L
I
minus minus minus minus
radic
983142 =
1
2 π
I
M 983143 L
minus minus minus minus
radic
983142
2
1 L 1 L 983142
2
983142
2
1 L
983143
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1113
Correct
Conceptual Question 1410
Suppose the damping constant of an oscillator increases
Part A
Is the medium more resistive or less resistive
ANSWER
Correct
Part B
Do the oscillations damp out more quickly or less quickly
ANSWER
Correct
Part C
Is the time constant increased or decreased
ANSWER
Correct
Problem 1432
The two graphs in the figure are for two different vertical mass-spring systems
= 976983143
m
s
2
b
The medium is more resistive
The medium is less resistive
The oscillations will damp out more quickly
The oscillations will damp out less quickly
τ
is increased
is decreased
τ
τ
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 413
Correct
A greater mass results in a lower frequency and a longer period of oscillation
Part H
The amplitude of oscillation is the maximum distance between the oscillating weight and the equilibrium position Determine the frequency of
oscillation for several different amplitudes by pulling the weight down different amounts
How does the frequency depend on the amplitude of oscillation
ANSWER
CorrectEven though the weight has to travel farther each oscillation if the amplitude is greater the spring on average exerts a stronger force causing a
greater acceleration and a greater average speed The effects of the longer distance and faster speed cancel out so that the period of oscillation
doesnrsquot change
Part I
The spring constant of spring 3 can be adjusted with the softness spring 3 slider bar (harder means a greater spring constant or stiffer spring)
How does the frequency of oscillation depend on the spring constant
Hint 1 How to approach the problem
You can place a weight on spring 3 with the softness of spring 3 set to a very low (soft) value and place another weight with the same masson spring 1 Then release the two weights and compare their frequencies of oscillation
ANSWER
Correct
It turns out that the frequency of oscillation depends on the square root of the ratio of the spring constant to mass where
is the frequency A stiffer spring constant causes the frequency to increase Sports cars use stiff springs whereas large plush Cadillacs use softsprings for their suspension
PhET Interactive Simulations
University of Colorado
httpphetcoloradoedu
PhET Tutorial Pendulum Lab
Learning Goal
To understand the relationships of the energetics forces acceleration and velocity of an oscillating pendulum and to determine how the motion of a
pendulum depends on the mass the length of the string and the acceleration due to gravity
The frequency increases as the mass increases
The frequency is independent of the mass
The frequency decreases as the mass increases
The frequency increases as the amplitude increases
The frequency decreases as the amplitude increases
The frequency is independent of the amplitude
The frequency increases as the spring constant increases
The frequency is independent of the spring constant
The frequency decreases as the spring constant increases
983142 = 2 π ( 983147 983149 )
minus minus minus minus minus
radic 983142
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 513
For this tutorial use the PhET simulation Pendulum Lab This simulation mimics a real pendulum and allows you to adjust the initial position the mass
and the length of the pendulum
Start the simulation You can drag the pendulum to an arbitrary initial angle and release it from rest You can adjust the length and the mass of the
pendulum using the slider bars at the top of the green panel Velocity and acceleration vectors can be selected to be shown as well as the forms of
energy
Feel free to play around with the simulation When you are done click the Reset button
Part A
Select to show the energy of pendulum 1 Be sure that friction is set to none
Drag the pendulum to an angle (with respect to the vertical) of and then release itWhen the pendulum is at what form(s) of energy does it have
Check all that apply
ANSWER
Correct
The pendulum starts off with no kinetic energy since it is released from rest so it initially only has potential energy When the pendulum is at
it is just as high above the ground as when it started so it must have the same amount of potential energy as it initially had Since thetotal energy is conserved it canrsquot have other forms of energy at (if it did it would have more energy there than it initially had) so it again has
only potential energy
Part B
Drag the pendulum to an angle (with respect to the vertical) of and then release it
Where is the pendulum swinging the fastest
ANSWER
Correct
The pendulum has the least potential energy at this location since it is at the lowest point in the arc (in fact for this simulation the potential
energy reference location is here so it has no potential energy) This means that the kinetic energy is greatest here so the pendulum is moving
the fastest
Part C
3 0
∘
minus 3 0
∘
Kinetic energy
Thermal energy
Potential energy
minus 3 0
∘
minus 3 0
∘
3 0
∘
at
at
at
at
1 5
∘
minus 3 0
∘
0
∘
3 0
∘
3 0
∘
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 613
Drag the pendulum to an angle (with respect to the vertical) of and then release it Select to show the acceleration vector
With the pendulum swinging back and forth at which locations is the acceleration equal to zero
ANSWER
Correct
The pendulum is moving in a circular path so its velocity is never constant In fact for most locations the acceleration has both a radial
component (the centripetal acceleration which is directed along the rope) and a tangential component (due to the speed changing directed along
the path the only place the tangential acceleration is zero is when the angle is )
Part D
With the pendulum swinging back and forth how does the tension of the rope compare to the force of gravity when the angle is
Hint 1 How to approach the problem
Look at the direction of the acceleration when the angle is (you can slow down or pause the simulation to see this more clearly) and think
about the relationship between the net force acting on the pendulum and the acceleration (Newtonrsquos 2nd law of motion) It would probably help todraw a free-body diagram for the mass
ANSWER
Correct
Since the acceleration of the pendulum is directed up when the angle is the net force must be directed up (Newtonrsquos 2nd law) This meansthat the upward force of tension must be stronger than the downward force of gravity
Part E
Drag the pendulum to an angle (with respect to the vertical) of and then release it
With the pendulum swinging back and forth where is the tension equal to zero
ANSWER
Correct
At these locations the acceleration is solely due to gravity and directed downward Thus the net force acting on t he pendulum i s also directed
downward meaning there are no horizontal forces This requires the tension to be zero
Part F
Now for parts F-I you will investigate how the period of oscillation depends on the properties of the pendulum
The period of oscillation is the amount of time it takes for the pendulum to take a full swing going from the original angle to the other side and
3 0
∘
The acceleration is zero when the angle is either or
The acceleration is zero when the angle is
The acceleration is never equal to zero as it swings back and forth
+ 3 0
∘
minus 3 0
∘
0
∘
0
∘
0
∘
0
∘
The tension is equal to the force of gravity
The tension is greater than the force of gravity only if it is swinging really fast
The tension is less than the force of gravity
The tension is greater than the force of gravity
0
∘
9 0
∘
The tension is zero at the angles and
The tension is zero when the angle is and
The tension is zero when the angle is
The tension is never zero
+ 9 0
∘
minus 9 0
∘
+ 4 5
∘
minus 4 5
∘
0
∘
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 713
returning to the original angle You can determine the period by selecting other tools which gives you a stopwatch With the pendulum swinging yo
can start the stopwatch when the pendulum is at its original angle and time how long it takes to complete 10 swings The period will be this time
interval divided by 10 (this method is more accurate than trying to time one swing)
Set the length of the pendulum to 10 and the mass to 10 Click Reset and then drag the pendulum to an angle (with respect to the vertical) o
and release it What is the period of oscillation
ANSWER
Correct
A - - pendulum completes one osc illation in 20
Part G
How does the period of oscillation depend on the initial angle of the pendulum when released (Be sure to measure the period for initial angles much
greater than )
ANSWER
Correct
Unlike a harmonic oscillator such as a mass on a spring the period actually depends on the initial angle For small angles (eg ) it is a
pretty good approximation that the period doesnrsquot change but for larger angles the period does in fact increase
Part H
Keeping the length of the pendulum fixed determine the period for a few different masses (Alternatively you can set up two pendulums by selecting
Show 2nd pendulum Adjust the lengths to be the same and have one pendulum with a higher mass You can release one and then release the
other with the same angle when the first one is back at that angle)
How does the period of the pendulum depend on mass
ANSWER
Correct
The period of a pendulum does not depend on mass The reason for this result is very similar to the reason that without air resistance all objects
fall to the ground at the same rate (changing the mass changes both the inertia and the force of gravity by the same amount)
Part I
Now keep the mass fixed to any value you choose and measure the period for several different pendulum lengths
How does the period of the pendulum depend on the length
ANSWER
m k g
3 0
∘
40
15
200
20
10
05
s
s
s
s
s
s
1 0 m 1 0 k g s
3 0
∘
The period is longer when the initial angle is greater
The period is shorter when the initial angle is greater
The period is independent of the initial angle
lt 3 0
∘
A heavier pendulum has a shorter period
The period is independent of the pendulumrsquos mass
A heavier pendulum has a longer period
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 813
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 913
Express your answer to two significant figures and include the appropriate units
ANSWER
All attempts used correct answer withheld by instructor
Problem 1480
The figure shows a 240 uniform rod pivoted at one end The other end is attached to a horizontal spring The spring is neither stretched nor compresse
when the rod hangs straight down
Part A
What is the rods oscillation period You can assume that the rods angle from vertical is always small
Express your answer to two significant figures and include the appropriate units
ANSWER
Correct
Problem 1458
A geologist needs to determine the local value of Unfortunately his only tools are a meter s tick a saw and a stopwatch He s tarts by hanging the me
stic k from one end and measuring its frequency as it swings He then saws off 20 mdashusing the centimeter markingsmdashand measures the frequency
again After two more cuts these are his data
Length Frequency
100 061
80 067
60 079
40 096
Part A
You want to determine the local value of by graphing the data Select the appropriate variables to graph on each axis that will produce a straight-lin
graph with either a slope or intercept that is related to
Sort all variables into the appropriate bins
Hint 1 How to approach the problem
= 4410^-21983149
D N A
k g
g
= 060T s
983143
c m
L ( c m ) 983142 ( H z )
983143
983143
I L
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1013
A physical pendulum has moment of inertia and dist ance between the pivot and the center of mass Its oscillation frequency is
ANSWER
ANSWER
Correct
Part B
Graphing either versus or versus gives a straight line In the graph shown we chose to plot on the vertical axis and on th
horizontal axis From the equation for the line of best fit given determine the local value of
Express your answer to three significant figures and include the appropriate units
ANSWER
I L
983142 =
1
2 π
M 983143 L
I
983142 =
1
2 π
I
M 983143 L
983142 =
1
2 π
M 983143 L
I
minus minus minus minus
radic
983142 =
1
2 π
I
M 983143 L
minus minus minus minus
radic
983142
2
1 L 1 L 983142
2
983142
2
1 L
983143
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1113
Correct
Conceptual Question 1410
Suppose the damping constant of an oscillator increases
Part A
Is the medium more resistive or less resistive
ANSWER
Correct
Part B
Do the oscillations damp out more quickly or less quickly
ANSWER
Correct
Part C
Is the time constant increased or decreased
ANSWER
Correct
Problem 1432
The two graphs in the figure are for two different vertical mass-spring systems
= 976983143
m
s
2
b
The medium is more resistive
The medium is less resistive
The oscillations will damp out more quickly
The oscillations will damp out less quickly
τ
is increased
is decreased
τ
τ
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 513
For this tutorial use the PhET simulation Pendulum Lab This simulation mimics a real pendulum and allows you to adjust the initial position the mass
and the length of the pendulum
Start the simulation You can drag the pendulum to an arbitrary initial angle and release it from rest You can adjust the length and the mass of the
pendulum using the slider bars at the top of the green panel Velocity and acceleration vectors can be selected to be shown as well as the forms of
energy
Feel free to play around with the simulation When you are done click the Reset button
Part A
Select to show the energy of pendulum 1 Be sure that friction is set to none
Drag the pendulum to an angle (with respect to the vertical) of and then release itWhen the pendulum is at what form(s) of energy does it have
Check all that apply
ANSWER
Correct
The pendulum starts off with no kinetic energy since it is released from rest so it initially only has potential energy When the pendulum is at
it is just as high above the ground as when it started so it must have the same amount of potential energy as it initially had Since thetotal energy is conserved it canrsquot have other forms of energy at (if it did it would have more energy there than it initially had) so it again has
only potential energy
Part B
Drag the pendulum to an angle (with respect to the vertical) of and then release it
Where is the pendulum swinging the fastest
ANSWER
Correct
The pendulum has the least potential energy at this location since it is at the lowest point in the arc (in fact for this simulation the potential
energy reference location is here so it has no potential energy) This means that the kinetic energy is greatest here so the pendulum is moving
the fastest
Part C
3 0
∘
minus 3 0
∘
Kinetic energy
Thermal energy
Potential energy
minus 3 0
∘
minus 3 0
∘
3 0
∘
at
at
at
at
1 5
∘
minus 3 0
∘
0
∘
3 0
∘
3 0
∘
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 613
Drag the pendulum to an angle (with respect to the vertical) of and then release it Select to show the acceleration vector
With the pendulum swinging back and forth at which locations is the acceleration equal to zero
ANSWER
Correct
The pendulum is moving in a circular path so its velocity is never constant In fact for most locations the acceleration has both a radial
component (the centripetal acceleration which is directed along the rope) and a tangential component (due to the speed changing directed along
the path the only place the tangential acceleration is zero is when the angle is )
Part D
With the pendulum swinging back and forth how does the tension of the rope compare to the force of gravity when the angle is
Hint 1 How to approach the problem
Look at the direction of the acceleration when the angle is (you can slow down or pause the simulation to see this more clearly) and think
about the relationship between the net force acting on the pendulum and the acceleration (Newtonrsquos 2nd law of motion) It would probably help todraw a free-body diagram for the mass
ANSWER
Correct
Since the acceleration of the pendulum is directed up when the angle is the net force must be directed up (Newtonrsquos 2nd law) This meansthat the upward force of tension must be stronger than the downward force of gravity
Part E
Drag the pendulum to an angle (with respect to the vertical) of and then release it
With the pendulum swinging back and forth where is the tension equal to zero
ANSWER
Correct
At these locations the acceleration is solely due to gravity and directed downward Thus the net force acting on t he pendulum i s also directed
downward meaning there are no horizontal forces This requires the tension to be zero
Part F
Now for parts F-I you will investigate how the period of oscillation depends on the properties of the pendulum
The period of oscillation is the amount of time it takes for the pendulum to take a full swing going from the original angle to the other side and
3 0
∘
The acceleration is zero when the angle is either or
The acceleration is zero when the angle is
The acceleration is never equal to zero as it swings back and forth
+ 3 0
∘
minus 3 0
∘
0
∘
0
∘
0
∘
0
∘
The tension is equal to the force of gravity
The tension is greater than the force of gravity only if it is swinging really fast
The tension is less than the force of gravity
The tension is greater than the force of gravity
0
∘
9 0
∘
The tension is zero at the angles and
The tension is zero when the angle is and
The tension is zero when the angle is
The tension is never zero
+ 9 0
∘
minus 9 0
∘
+ 4 5
∘
minus 4 5
∘
0
∘
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 713
returning to the original angle You can determine the period by selecting other tools which gives you a stopwatch With the pendulum swinging yo
can start the stopwatch when the pendulum is at its original angle and time how long it takes to complete 10 swings The period will be this time
interval divided by 10 (this method is more accurate than trying to time one swing)
Set the length of the pendulum to 10 and the mass to 10 Click Reset and then drag the pendulum to an angle (with respect to the vertical) o
and release it What is the period of oscillation
ANSWER
Correct
A - - pendulum completes one osc illation in 20
Part G
How does the period of oscillation depend on the initial angle of the pendulum when released (Be sure to measure the period for initial angles much
greater than )
ANSWER
Correct
Unlike a harmonic oscillator such as a mass on a spring the period actually depends on the initial angle For small angles (eg ) it is a
pretty good approximation that the period doesnrsquot change but for larger angles the period does in fact increase
Part H
Keeping the length of the pendulum fixed determine the period for a few different masses (Alternatively you can set up two pendulums by selecting
Show 2nd pendulum Adjust the lengths to be the same and have one pendulum with a higher mass You can release one and then release the
other with the same angle when the first one is back at that angle)
How does the period of the pendulum depend on mass
ANSWER
Correct
The period of a pendulum does not depend on mass The reason for this result is very similar to the reason that without air resistance all objects
fall to the ground at the same rate (changing the mass changes both the inertia and the force of gravity by the same amount)
Part I
Now keep the mass fixed to any value you choose and measure the period for several different pendulum lengths
How does the period of the pendulum depend on the length
ANSWER
m k g
3 0
∘
40
15
200
20
10
05
s
s
s
s
s
s
1 0 m 1 0 k g s
3 0
∘
The period is longer when the initial angle is greater
The period is shorter when the initial angle is greater
The period is independent of the initial angle
lt 3 0
∘
A heavier pendulum has a shorter period
The period is independent of the pendulumrsquos mass
A heavier pendulum has a longer period
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 813
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 913
Express your answer to two significant figures and include the appropriate units
ANSWER
All attempts used correct answer withheld by instructor
Problem 1480
The figure shows a 240 uniform rod pivoted at one end The other end is attached to a horizontal spring The spring is neither stretched nor compresse
when the rod hangs straight down
Part A
What is the rods oscillation period You can assume that the rods angle from vertical is always small
Express your answer to two significant figures and include the appropriate units
ANSWER
Correct
Problem 1458
A geologist needs to determine the local value of Unfortunately his only tools are a meter s tick a saw and a stopwatch He s tarts by hanging the me
stic k from one end and measuring its frequency as it swings He then saws off 20 mdashusing the centimeter markingsmdashand measures the frequency
again After two more cuts these are his data
Length Frequency
100 061
80 067
60 079
40 096
Part A
You want to determine the local value of by graphing the data Select the appropriate variables to graph on each axis that will produce a straight-lin
graph with either a slope or intercept that is related to
Sort all variables into the appropriate bins
Hint 1 How to approach the problem
= 4410^-21983149
D N A
k g
g
= 060T s
983143
c m
L ( c m ) 983142 ( H z )
983143
983143
I L
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1013
A physical pendulum has moment of inertia and dist ance between the pivot and the center of mass Its oscillation frequency is
ANSWER
ANSWER
Correct
Part B
Graphing either versus or versus gives a straight line In the graph shown we chose to plot on the vertical axis and on th
horizontal axis From the equation for the line of best fit given determine the local value of
Express your answer to three significant figures and include the appropriate units
ANSWER
I L
983142 =
1
2 π
M 983143 L
I
983142 =
1
2 π
I
M 983143 L
983142 =
1
2 π
M 983143 L
I
minus minus minus minus
radic
983142 =
1
2 π
I
M 983143 L
minus minus minus minus
radic
983142
2
1 L 1 L 983142
2
983142
2
1 L
983143
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1113
Correct
Conceptual Question 1410
Suppose the damping constant of an oscillator increases
Part A
Is the medium more resistive or less resistive
ANSWER
Correct
Part B
Do the oscillations damp out more quickly or less quickly
ANSWER
Correct
Part C
Is the time constant increased or decreased
ANSWER
Correct
Problem 1432
The two graphs in the figure are for two different vertical mass-spring systems
= 976983143
m
s
2
b
The medium is more resistive
The medium is less resistive
The oscillations will damp out more quickly
The oscillations will damp out less quickly
τ
is increased
is decreased
τ
τ
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 613
Drag the pendulum to an angle (with respect to the vertical) of and then release it Select to show the acceleration vector
With the pendulum swinging back and forth at which locations is the acceleration equal to zero
ANSWER
Correct
The pendulum is moving in a circular path so its velocity is never constant In fact for most locations the acceleration has both a radial
component (the centripetal acceleration which is directed along the rope) and a tangential component (due to the speed changing directed along
the path the only place the tangential acceleration is zero is when the angle is )
Part D
With the pendulum swinging back and forth how does the tension of the rope compare to the force of gravity when the angle is
Hint 1 How to approach the problem
Look at the direction of the acceleration when the angle is (you can slow down or pause the simulation to see this more clearly) and think
about the relationship between the net force acting on the pendulum and the acceleration (Newtonrsquos 2nd law of motion) It would probably help todraw a free-body diagram for the mass
ANSWER
Correct
Since the acceleration of the pendulum is directed up when the angle is the net force must be directed up (Newtonrsquos 2nd law) This meansthat the upward force of tension must be stronger than the downward force of gravity
Part E
Drag the pendulum to an angle (with respect to the vertical) of and then release it
With the pendulum swinging back and forth where is the tension equal to zero
ANSWER
Correct
At these locations the acceleration is solely due to gravity and directed downward Thus the net force acting on t he pendulum i s also directed
downward meaning there are no horizontal forces This requires the tension to be zero
Part F
Now for parts F-I you will investigate how the period of oscillation depends on the properties of the pendulum
The period of oscillation is the amount of time it takes for the pendulum to take a full swing going from the original angle to the other side and
3 0
∘
The acceleration is zero when the angle is either or
The acceleration is zero when the angle is
The acceleration is never equal to zero as it swings back and forth
+ 3 0
∘
minus 3 0
∘
0
∘
0
∘
0
∘
0
∘
The tension is equal to the force of gravity
The tension is greater than the force of gravity only if it is swinging really fast
The tension is less than the force of gravity
The tension is greater than the force of gravity
0
∘
9 0
∘
The tension is zero at the angles and
The tension is zero when the angle is and
The tension is zero when the angle is
The tension is never zero
+ 9 0
∘
minus 9 0
∘
+ 4 5
∘
minus 4 5
∘
0
∘
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 713
returning to the original angle You can determine the period by selecting other tools which gives you a stopwatch With the pendulum swinging yo
can start the stopwatch when the pendulum is at its original angle and time how long it takes to complete 10 swings The period will be this time
interval divided by 10 (this method is more accurate than trying to time one swing)
Set the length of the pendulum to 10 and the mass to 10 Click Reset and then drag the pendulum to an angle (with respect to the vertical) o
and release it What is the period of oscillation
ANSWER
Correct
A - - pendulum completes one osc illation in 20
Part G
How does the period of oscillation depend on the initial angle of the pendulum when released (Be sure to measure the period for initial angles much
greater than )
ANSWER
Correct
Unlike a harmonic oscillator such as a mass on a spring the period actually depends on the initial angle For small angles (eg ) it is a
pretty good approximation that the period doesnrsquot change but for larger angles the period does in fact increase
Part H
Keeping the length of the pendulum fixed determine the period for a few different masses (Alternatively you can set up two pendulums by selecting
Show 2nd pendulum Adjust the lengths to be the same and have one pendulum with a higher mass You can release one and then release the
other with the same angle when the first one is back at that angle)
How does the period of the pendulum depend on mass
ANSWER
Correct
The period of a pendulum does not depend on mass The reason for this result is very similar to the reason that without air resistance all objects
fall to the ground at the same rate (changing the mass changes both the inertia and the force of gravity by the same amount)
Part I
Now keep the mass fixed to any value you choose and measure the period for several different pendulum lengths
How does the period of the pendulum depend on the length
ANSWER
m k g
3 0
∘
40
15
200
20
10
05
s
s
s
s
s
s
1 0 m 1 0 k g s
3 0
∘
The period is longer when the initial angle is greater
The period is shorter when the initial angle is greater
The period is independent of the initial angle
lt 3 0
∘
A heavier pendulum has a shorter period
The period is independent of the pendulumrsquos mass
A heavier pendulum has a longer period
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 813
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 913
Express your answer to two significant figures and include the appropriate units
ANSWER
All attempts used correct answer withheld by instructor
Problem 1480
The figure shows a 240 uniform rod pivoted at one end The other end is attached to a horizontal spring The spring is neither stretched nor compresse
when the rod hangs straight down
Part A
What is the rods oscillation period You can assume that the rods angle from vertical is always small
Express your answer to two significant figures and include the appropriate units
ANSWER
Correct
Problem 1458
A geologist needs to determine the local value of Unfortunately his only tools are a meter s tick a saw and a stopwatch He s tarts by hanging the me
stic k from one end and measuring its frequency as it swings He then saws off 20 mdashusing the centimeter markingsmdashand measures the frequency
again After two more cuts these are his data
Length Frequency
100 061
80 067
60 079
40 096
Part A
You want to determine the local value of by graphing the data Select the appropriate variables to graph on each axis that will produce a straight-lin
graph with either a slope or intercept that is related to
Sort all variables into the appropriate bins
Hint 1 How to approach the problem
= 4410^-21983149
D N A
k g
g
= 060T s
983143
c m
L ( c m ) 983142 ( H z )
983143
983143
I L
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1013
A physical pendulum has moment of inertia and dist ance between the pivot and the center of mass Its oscillation frequency is
ANSWER
ANSWER
Correct
Part B
Graphing either versus or versus gives a straight line In the graph shown we chose to plot on the vertical axis and on th
horizontal axis From the equation for the line of best fit given determine the local value of
Express your answer to three significant figures and include the appropriate units
ANSWER
I L
983142 =
1
2 π
M 983143 L
I
983142 =
1
2 π
I
M 983143 L
983142 =
1
2 π
M 983143 L
I
minus minus minus minus
radic
983142 =
1
2 π
I
M 983143 L
minus minus minus minus
radic
983142
2
1 L 1 L 983142
2
983142
2
1 L
983143
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1113
Correct
Conceptual Question 1410
Suppose the damping constant of an oscillator increases
Part A
Is the medium more resistive or less resistive
ANSWER
Correct
Part B
Do the oscillations damp out more quickly or less quickly
ANSWER
Correct
Part C
Is the time constant increased or decreased
ANSWER
Correct
Problem 1432
The two graphs in the figure are for two different vertical mass-spring systems
= 976983143
m
s
2
b
The medium is more resistive
The medium is less resistive
The oscillations will damp out more quickly
The oscillations will damp out less quickly
τ
is increased
is decreased
τ
τ
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 713
returning to the original angle You can determine the period by selecting other tools which gives you a stopwatch With the pendulum swinging yo
can start the stopwatch when the pendulum is at its original angle and time how long it takes to complete 10 swings The period will be this time
interval divided by 10 (this method is more accurate than trying to time one swing)
Set the length of the pendulum to 10 and the mass to 10 Click Reset and then drag the pendulum to an angle (with respect to the vertical) o
and release it What is the period of oscillation
ANSWER
Correct
A - - pendulum completes one osc illation in 20
Part G
How does the period of oscillation depend on the initial angle of the pendulum when released (Be sure to measure the period for initial angles much
greater than )
ANSWER
Correct
Unlike a harmonic oscillator such as a mass on a spring the period actually depends on the initial angle For small angles (eg ) it is a
pretty good approximation that the period doesnrsquot change but for larger angles the period does in fact increase
Part H
Keeping the length of the pendulum fixed determine the period for a few different masses (Alternatively you can set up two pendulums by selecting
Show 2nd pendulum Adjust the lengths to be the same and have one pendulum with a higher mass You can release one and then release the
other with the same angle when the first one is back at that angle)
How does the period of the pendulum depend on mass
ANSWER
Correct
The period of a pendulum does not depend on mass The reason for this result is very similar to the reason that without air resistance all objects
fall to the ground at the same rate (changing the mass changes both the inertia and the force of gravity by the same amount)
Part I
Now keep the mass fixed to any value you choose and measure the period for several different pendulum lengths
How does the period of the pendulum depend on the length
ANSWER
m k g
3 0
∘
40
15
200
20
10
05
s
s
s
s
s
s
1 0 m 1 0 k g s
3 0
∘
The period is longer when the initial angle is greater
The period is shorter when the initial angle is greater
The period is independent of the initial angle
lt 3 0
∘
A heavier pendulum has a shorter period
The period is independent of the pendulumrsquos mass
A heavier pendulum has a longer period
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 813
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 913
Express your answer to two significant figures and include the appropriate units
ANSWER
All attempts used correct answer withheld by instructor
Problem 1480
The figure shows a 240 uniform rod pivoted at one end The other end is attached to a horizontal spring The spring is neither stretched nor compresse
when the rod hangs straight down
Part A
What is the rods oscillation period You can assume that the rods angle from vertical is always small
Express your answer to two significant figures and include the appropriate units
ANSWER
Correct
Problem 1458
A geologist needs to determine the local value of Unfortunately his only tools are a meter s tick a saw and a stopwatch He s tarts by hanging the me
stic k from one end and measuring its frequency as it swings He then saws off 20 mdashusing the centimeter markingsmdashand measures the frequency
again After two more cuts these are his data
Length Frequency
100 061
80 067
60 079
40 096
Part A
You want to determine the local value of by graphing the data Select the appropriate variables to graph on each axis that will produce a straight-lin
graph with either a slope or intercept that is related to
Sort all variables into the appropriate bins
Hint 1 How to approach the problem
= 4410^-21983149
D N A
k g
g
= 060T s
983143
c m
L ( c m ) 983142 ( H z )
983143
983143
I L
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1013
A physical pendulum has moment of inertia and dist ance between the pivot and the center of mass Its oscillation frequency is
ANSWER
ANSWER
Correct
Part B
Graphing either versus or versus gives a straight line In the graph shown we chose to plot on the vertical axis and on th
horizontal axis From the equation for the line of best fit given determine the local value of
Express your answer to three significant figures and include the appropriate units
ANSWER
I L
983142 =
1
2 π
M 983143 L
I
983142 =
1
2 π
I
M 983143 L
983142 =
1
2 π
M 983143 L
I
minus minus minus minus
radic
983142 =
1
2 π
I
M 983143 L
minus minus minus minus
radic
983142
2
1 L 1 L 983142
2
983142
2
1 L
983143
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1113
Correct
Conceptual Question 1410
Suppose the damping constant of an oscillator increases
Part A
Is the medium more resistive or less resistive
ANSWER
Correct
Part B
Do the oscillations damp out more quickly or less quickly
ANSWER
Correct
Part C
Is the time constant increased or decreased
ANSWER
Correct
Problem 1432
The two graphs in the figure are for two different vertical mass-spring systems
= 976983143
m
s
2
b
The medium is more resistive
The medium is less resistive
The oscillations will damp out more quickly
The oscillations will damp out less quickly
τ
is increased
is decreased
τ
τ
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 813
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 913
Express your answer to two significant figures and include the appropriate units
ANSWER
All attempts used correct answer withheld by instructor
Problem 1480
The figure shows a 240 uniform rod pivoted at one end The other end is attached to a horizontal spring The spring is neither stretched nor compresse
when the rod hangs straight down
Part A
What is the rods oscillation period You can assume that the rods angle from vertical is always small
Express your answer to two significant figures and include the appropriate units
ANSWER
Correct
Problem 1458
A geologist needs to determine the local value of Unfortunately his only tools are a meter s tick a saw and a stopwatch He s tarts by hanging the me
stic k from one end and measuring its frequency as it swings He then saws off 20 mdashusing the centimeter markingsmdashand measures the frequency
again After two more cuts these are his data
Length Frequency
100 061
80 067
60 079
40 096
Part A
You want to determine the local value of by graphing the data Select the appropriate variables to graph on each axis that will produce a straight-lin
graph with either a slope or intercept that is related to
Sort all variables into the appropriate bins
Hint 1 How to approach the problem
= 4410^-21983149
D N A
k g
g
= 060T s
983143
c m
L ( c m ) 983142 ( H z )
983143
983143
I L
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1013
A physical pendulum has moment of inertia and dist ance between the pivot and the center of mass Its oscillation frequency is
ANSWER
ANSWER
Correct
Part B
Graphing either versus or versus gives a straight line In the graph shown we chose to plot on the vertical axis and on th
horizontal axis From the equation for the line of best fit given determine the local value of
Express your answer to three significant figures and include the appropriate units
ANSWER
I L
983142 =
1
2 π
M 983143 L
I
983142 =
1
2 π
I
M 983143 L
983142 =
1
2 π
M 983143 L
I
minus minus minus minus
radic
983142 =
1
2 π
I
M 983143 L
minus minus minus minus
radic
983142
2
1 L 1 L 983142
2
983142
2
1 L
983143
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1113
Correct
Conceptual Question 1410
Suppose the damping constant of an oscillator increases
Part A
Is the medium more resistive or less resistive
ANSWER
Correct
Part B
Do the oscillations damp out more quickly or less quickly
ANSWER
Correct
Part C
Is the time constant increased or decreased
ANSWER
Correct
Problem 1432
The two graphs in the figure are for two different vertical mass-spring systems
= 976983143
m
s
2
b
The medium is more resistive
The medium is less resistive
The oscillations will damp out more quickly
The oscillations will damp out less quickly
τ
is increased
is decreased
τ
τ
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 913
Express your answer to two significant figures and include the appropriate units
ANSWER
All attempts used correct answer withheld by instructor
Problem 1480
The figure shows a 240 uniform rod pivoted at one end The other end is attached to a horizontal spring The spring is neither stretched nor compresse
when the rod hangs straight down
Part A
What is the rods oscillation period You can assume that the rods angle from vertical is always small
Express your answer to two significant figures and include the appropriate units
ANSWER
Correct
Problem 1458
A geologist needs to determine the local value of Unfortunately his only tools are a meter s tick a saw and a stopwatch He s tarts by hanging the me
stic k from one end and measuring its frequency as it swings He then saws off 20 mdashusing the centimeter markingsmdashand measures the frequency
again After two more cuts these are his data
Length Frequency
100 061
80 067
60 079
40 096
Part A
You want to determine the local value of by graphing the data Select the appropriate variables to graph on each axis that will produce a straight-lin
graph with either a slope or intercept that is related to
Sort all variables into the appropriate bins
Hint 1 How to approach the problem
= 4410^-21983149
D N A
k g
g
= 060T s
983143
c m
L ( c m ) 983142 ( H z )
983143
983143
I L
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1013
A physical pendulum has moment of inertia and dist ance between the pivot and the center of mass Its oscillation frequency is
ANSWER
ANSWER
Correct
Part B
Graphing either versus or versus gives a straight line In the graph shown we chose to plot on the vertical axis and on th
horizontal axis From the equation for the line of best fit given determine the local value of
Express your answer to three significant figures and include the appropriate units
ANSWER
I L
983142 =
1
2 π
M 983143 L
I
983142 =
1
2 π
I
M 983143 L
983142 =
1
2 π
M 983143 L
I
minus minus minus minus
radic
983142 =
1
2 π
I
M 983143 L
minus minus minus minus
radic
983142
2
1 L 1 L 983142
2
983142
2
1 L
983143
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1113
Correct
Conceptual Question 1410
Suppose the damping constant of an oscillator increases
Part A
Is the medium more resistive or less resistive
ANSWER
Correct
Part B
Do the oscillations damp out more quickly or less quickly
ANSWER
Correct
Part C
Is the time constant increased or decreased
ANSWER
Correct
Problem 1432
The two graphs in the figure are for two different vertical mass-spring systems
= 976983143
m
s
2
b
The medium is more resistive
The medium is less resistive
The oscillations will damp out more quickly
The oscillations will damp out less quickly
τ
is increased
is decreased
τ
τ
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1013
A physical pendulum has moment of inertia and dist ance between the pivot and the center of mass Its oscillation frequency is
ANSWER
ANSWER
Correct
Part B
Graphing either versus or versus gives a straight line In the graph shown we chose to plot on the vertical axis and on th
horizontal axis From the equation for the line of best fit given determine the local value of
Express your answer to three significant figures and include the appropriate units
ANSWER
I L
983142 =
1
2 π
M 983143 L
I
983142 =
1
2 π
I
M 983143 L
983142 =
1
2 π
M 983143 L
I
minus minus minus minus
radic
983142 =
1
2 π
I
M 983143 L
minus minus minus minus
radic
983142
2
1 L 1 L 983142
2
983142
2
1 L
983143
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1113
Correct
Conceptual Question 1410
Suppose the damping constant of an oscillator increases
Part A
Is the medium more resistive or less resistive
ANSWER
Correct
Part B
Do the oscillations damp out more quickly or less quickly
ANSWER
Correct
Part C
Is the time constant increased or decreased
ANSWER
Correct
Problem 1432
The two graphs in the figure are for two different vertical mass-spring systems
= 976983143
m
s
2
b
The medium is more resistive
The medium is less resistive
The oscillations will damp out more quickly
The oscillations will damp out less quickly
τ
is increased
is decreased
τ
τ
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1113
Correct
Conceptual Question 1410
Suppose the damping constant of an oscillator increases
Part A
Is the medium more resistive or less resistive
ANSWER
Correct
Part B
Do the oscillations damp out more quickly or less quickly
ANSWER
Correct
Part C
Is the time constant increased or decreased
ANSWER
Correct
Problem 1432
The two graphs in the figure are for two different vertical mass-spring systems
= 976983143
m
s
2
b
The medium is more resistive
The medium is less resistive
The oscillations will damp out more quickly
The oscillations will damp out less quickly
τ
is increased
is decreased
τ
τ
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1213
Part A
If both syst ems have the same mass what is the ratio of their spring constants
ANSWER
Correct
Problem 1431
The figure is the position-versus-time graph of a particle in simple harmonic motion
Part A
What is the phase constant
ANSWER
Correct
983147
A
983147
B
= 225 983147
A
983147
B
= r a d ϕ
0
π
3
= ϕ
0
r a d
2 π
3
= minus ϕ
0
r a d
2 π
3
= 0 r a d ϕ
0
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707
7242019 Mastering Physics HW 5 Ch 14 - Oscillations
httpslidepdfcomreaderfullmastering-physics-hw-5-ch-14-oscillations 1313
Part B
What is the velocity at = 0
Express your answer with the appropriate units
ANSWER
Correct
Part C
What is
Express your answer with the appropriate units
ANSWER
Correct
Problem 1437
Part A
When the displacement of a mass on a spring is what percentage of the energy is kinetic energy
ANSWER
Correct
Part B
At what displacement as a fraction of is the energy half kinetic and half potential
ANSWER
Correct
Score SummaryYour score on this assignment is 867
You received 781 out of a possible total of 9 points
983156 s
= -136983158 ( 0 )
c m
s
983158
m a x
= 157983158
m a x
c m
s
1
2
A
750
A
0707