adapted from holt book on physics

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Simple harmonic motionChapter 10

Adapted from Holt book on physics

Section 1 Simple Harmonic Motion

Section 2 Springs

Section 3 Pendulum

Section 4 Review

Physics 101: Lecture 22, Pg 2

Simple Harmonic MotionSimple Harmonic Motion

Period = T (seconds per cycle)

Frequency = f = 1/T (cycles per second)

Angular frequency = = 2f = 2/T

Defn: any periodic motion that is the result of a restoring force

that is proportional to displacement.



Chapter 10

Simple Harmonic Motion




Chapter 10

Amplitude in SHM

• In SHM, the maximum displacement from equilibrium is defined as the amplitude of the vibration.

– A pendulum’s amplitude = swing angle from the vertical (= radians)

– For a mass-spring system, the amplitude is the maximum amount the spring is stretched or compressed from its equilibrium position. (x = meters)

Section 2 Measuring Simple Harmonic Motion



Chapter 10

Measures of Simple Harmonic Motion




Chapter 11

Amplitude of a pendulum




Chapter 11

Simple Harmonic Motion




Chapter 11

Hooke’s Law

• One type of periodic motion is the motion of a mass attached to a spring.

• The direction of the force acting on the mass (Felastic) is always opposite the direction of the mass’s displacement from equilibrium (x = 0).




Chapter 11

Hooke’s Law, continued

At equilibrium:• The spring force and the mass’s acceleration

become zero.• The speed reaches a maximum.

At maximum displacement:• The spring force and the mass’s acceleration reach

a maximum.• The speed becomes zero.




Chapter 11

Hooke’s Law, continued

• Measurements show that the spring force, or restoring force, is directly proportional to the displacement of the mass.

• This relationship is known as Hooke’s Law:

Felastic = –kx

spring force = –(spring constant displacement)

• The quantity k is a positive constant called the spring constant.




Chapter 11

Spring Constant




Chapter 11

Sample Problem

Hooke’s Law

If a mass of 0.55 kg attached to a vertical spring stretches the spring 2.0 cm from its original equilibrium position, what is the spring constant?




Chapter 11

Sample Problem, continued


Unknown:

k = ?

1. DefineGiven: m = 0.55 kg x = –2.0 cm = –0.20 m g = 9.81 m/s2

Diagram:



Chapter 11



2. PlanChoose an equation or situation: When the mass is attached to the spring,the equilibrium position changes. At the new equilibrium position, the net force acting on the mass is zero. So the spring force (given by Hooke’s law) must be equal and opposite to the weight of the mass.

Fnet = 0 = Felastic + Fg

Felastic = –kx

Fg = –mg

–kx – mg = 0



Chapter 11



2. Plan, continued Rearrange the equation to isolate the unknown:

kx mg 0

kx mg

k mg

x



Chapter 11



3. Calculate Substitute the values into the equation and

solve:

k mg

x

(0.55 kg)(9.81 m/s2 )

–0.020 m

k 270 N/m

4. Evaluate The value of k implies that 270 N of force is

required to displace the spring 1 m.



Chapter 11

The Simple Pendulum

• A simple pendulum consists of a mass called a bob, which is attached to a fixed string.


The forces acting on the bob at any point are the force exerted by the string and the gravitational force.

• At any displacement from equilibrium, the weight of the bob (Fg) can be resolved into two components.

• The x component (Fg,x = Fg sin ) is the only force acting on the bob in the direction of its motion and thus is the restoring force.



Chapter 11

The Simple Pendulum, continued

• The magnitude of the restoring force (Fg,x = Fg sin ) is proportional to sin .

• When the maximum angle of displacement is relatively small (<15°), sin is approximately equal to in radians.


• As a result, the restoring force is very nearly proportional to the displacement.

• Thus, the pendulum’s motion is an excellent approximation of simple harmonic motion.



Chapter 11

Restoring Force and Simple Pendulums



PendulumPendulum

Lg

gL

T 2

2

For “small oscillation”, period does not depend on

•mass

•amplitude Demos:

M,A,L dependence


Concept QuestionConcept Question

Suppose a grandfather clock (a simple pendulum) runs slow. In order to make it run on time you should: 1. Make the pendulum shorter 2. Make the pendulum longer

Lg

gL

T 2

2

CORRECT



A pendulum is hanging vertically from the ceiling of an elevator. Initiallythe elevator is at rest and the period of the pendulum is T. Now the pendulum accelerates upward. The period of the pendulum will now be1. greater than T 2. equal to T3. less than T

Lg

gL

T 2

2

CORRECT

“Effective g” is larger when accelerating upward

(you feel heavier)



If the amplitude of the oscillation (same block and same spring) was doubled, how would the period of the oscillation change? (The period is the time it takes to make one complete oscillation) 1. The period of the oscillation would double.2. The period of the oscillation would be halved3. The period of the oscillation would stay the same

+2A

t

-2A

x

CORRECT


Potential Energy of a SpringPotential Energy of a Spring

0x

2S kx

21

PE

Where x is measured fromthe equilibrium position PES

m

xx=0


Same thing for a vertical spring:Same thing for a vertical spring:

0y

PES

m

y

2S ky

21

PE

Where y is measured fromthe equilibrium position

y=0


Chapter 11

Force and Energy in Simple Harmonic MotionForce and Energy in Simple Harmonic Motion




In Case 1 a mass on a spring oscillates back and forth. In Case 2, the mass is doubled but the spring and the amplitude of the oscillation is the same as in Case 1. In which case is the maximum kinetic energy of the mass the biggest?

1. Case 12. Case 23. Same

CORRECT



PE = 1/2kx2 KE = 0

x=0 x=+Ax=-A x=0 x=+Ax=-A

samefor both

PE = 0 KE = KEMAX

same for both


Simple Harmonic Motion:Simple Harmonic Motion:Quick ReviewQuick Review

x(t) = [A]cos(t)

v(t) = -[A]sin(t)

a(t) = -[A2]cos(t)

x(t) = [A]sin(t)

v(t) = [A]cos(t)

a(t) = -[A2]sin(t)

xmax = A

vmax = A

amax = A2

Period = T (seconds per cycle)

Frequency = f = 1/T (cycles per second)

Angular frequency = = 2f = 2/T

OR


Review: Period of a SpringReview: Period of a Spring

For simple harmonic oscillator

= 2f = 2/T

For mass M on spring with spring constant k

k

m2 T

m

k

Demos:

A,m,k dependence



Multiple Choice

Base your answers to questions 1–6 on the information below.

Standardized Test PrepChapter 11

A mass is attached to a spring and moves with simple harmonic motion on a frictionless horizontal surface.

1. In what direction does the restoring force act?

A. to the leftB. to the rightC. to the left or to the right depending on whether the

spring is stretched or compressedD. perpendicular to the motion of the mass



Multiple Choice, continued




2. If the mass is displaced –0.35 m from its equilibrium position, the restoring force is 7.0 N. What is the spring constant?

F. –5.0 10–2 N/m H. 5.0 10–2 N/mG. –2.0 101 N/m J. 2.0 101 N/m







3. In what form is the energy in the system when the mass passes through the equilibrium point?

A. elastic potential energyB. gravitational potential energyC. kinetic energyD. a combination of two or more of the above







4. In what form is the energy in the system when the mass is at maximum displacement?

F. elastic potential energyG. gravitational potential energyH. kinetic energyJ. a combination of two or more of the above







5. Which of the following does not affect the period of the mass-spring system?

A. mass B. spring constant C. amplitude of vibrationD. All of the above affect the period.







6. If the mass is 48 kg and the spring constant is 12 N/m, what is the period of the oscillation?

F. 8 s H. s G. 4 s J. s






A pendulum bob hangs from a string and moves with simple harmonic motion.

7. What is the restoring force in the pendulum?

A. the total weight of the bob B. the component of the bob’s weight tangent to the

motion of the bobC. the component of the bob’s weight perpendicular to the

motion of the bobD. the elastic force of the stretched string







8. Which of the following does not affect the period of the pendulum?

F. the length of the string G. the mass of the pendulum bobH. the free-fall acceleration at the pendulum’s locationJ. All of the above affect the period.







9. If the pendulum completes exactly 12cycles in 2.0 min, what is the frequency ofthe pendulum?

A. 0.10 Hz B. 0.17 Hz C. 6.0 Hz D. 10 Hz







10. If the pendulum’s length is 2.00 m and ag = 9.80 m/s2, how many complete oscillations does the pendulum make in 5.00 min?

F. 1.76 H. 106G. 21.6 J. 239

adapted from holt book on physics

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