hw4 physics 311 mechanics - 12000.org
TRANSCRIPT
HW4 Physics 311 Mechanics
Fall 2015Physics department
University of Wisconsin, Madison
Instructor: Professor Stefan Westerhoff
By
Nasser M. Abbasi
November 28, 2019
Contents
0.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.1.1 part(1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.1.2 part(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.1.3 Part(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.1.4 Part(4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
0.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
0.3.1 SOLUTION method one . . . . . . . . . . . . . . . . . . . . . . . . . . 80.3.2 another solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
0.4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120.4.1 case (1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130.4.2 case (2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140.4.3 case(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
0.5 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150.5.1 Part (1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160.5.2 Part(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160.5.3 Part(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2
3
0.1 Problem 1
Mechanics
Physics 311Fall 2015
Homework 4 (10/2/15, due 10/9/15)
1. (5 points)The damping factor Ξ» of a spring suspension system is one-tenth the critical value. LetΟ0 be the undamped frequency. Find (i) the resonant frequency, (ii) the quality factor Q,(iii) the phase angle Ξ¦ when the system is driven at frequency Ο = Ο0/2, and (iv) thesteady-state amplitude at this frequency.
2. (10 points)A string of length 2 l is suspended at points A and B located on a horizontal line. Thedistance between A and B is 2 d, with d < l. A small, heavy bead can slide on the stringwithout friction. Find the period of the small-amplitude oscillations of the bead in thevertical plane containing the suspension points.Hint: The trajectory of the bead is a section of an ellipse (why?). Move the origin to theequilibrium point and use a Taylor expansion to get an approximate expression for thetrajectory around the equilibrium point. Apply Lagrange.
...continued on next page...
SOLUTION:
Note that πππππ‘ππππ = π0. We are told that π = 0.1π0 in this problem.
0.1.1 part(1)
The resonant frequency (for this case of under-damped) occurs when the steady state am-plitude is maximum
π =ππ
οΏ½οΏ½π20 β π2οΏ½
2+ 4π2π2
This happens when the denominator is minimum. Taking derivative of the denominator w.r.t.π and setting the result to zero gives
πππ
οΏ½οΏ½π20 β π2οΏ½
2+ 4π2π2οΏ½ = 0
2 οΏ½π20 β π2οΏ½ (β2π) + 8π2π = 08π2π + 4π3 β 4ππ2
0 = 02π2 + π2 β π2
0 = 0π2 = π2
0 β 2π2
Taking the positive root (since π must be positive) gives
π = οΏ½π20 β 2π2
When π = 0.1π0 the above becomes
π =οΏ½π20 β 2 οΏ½
110π0οΏ½
2
=οΏ½98100
π20
= 0.98995π0 rad/sec
4
0.1.2 part(2)
Quality factor π is defined as
π =ππ2π
= οΏ½π20 β π2
2π
= οΏ½π20 β (0.1π0)
2
2 (0.1π0)
=π0β1 β 0.12
0.2π0
= β1 β 0.12
0.2Therefore
π = 4.975
0.1.3 Part(3)
Given
π₯β²β² (π‘) + 2ππ₯β² + π20π₯ =
ππππππ‘ (1)
Assuming the particular solution is π₯π (π‘) = π΅ππππ‘ where π΅ = ππππ is the complex amplitudeand π is the amplitude and π is the phase of π΅. We want to find the phase. Plugging π₯π (π‘)into (1) and simplifying gives
π΅ =ππ
π20 β π2 + 2πππ
Hence
π = 0 β tanβ1 οΏ½2ππ
π20 β π2 οΏ½
= tanβ1 οΏ½β2πππ20 β π2 οΏ½
Since π = 0.1π0 and π = π02 the above becomes
π = tanβ1
βββββββββ2 (0.1π0)
π02
π20 β οΏ½
π02οΏ½2
ββββββββ
= tanβ1 (β0.13333)= β0.13255 rad
5
0.1.4 Part(4)
The steady state amplitude is π from above, which is found as follows
π2 = π΅π΅β
Where π΅β is the complex conjugate of π΅ =ππ
π20βπ2+2πππ
. Therefore
π =ππ
οΏ½οΏ½π20 β π2οΏ½
2+ 4π2π2
=ππ
1
οΏ½οΏ½π2
0 β οΏ½π02οΏ½2οΏ½2+ 4 (0.1π0)
2 οΏ½π02οΏ½2
=ππ
1
οΏ½0.572 5π40
= 1.3216π
ππ20
But ππ20 = π, the stiοΏ½ness, hence the above is
π = 1.3216ππ
0.2 Problem 2
Mechanics
Physics 311Fall 2015
Homework 4 (10/2/15, due 10/9/15)
1. (5 points)The damping factor Ξ» of a spring suspension system is one-tenth the critical value. LetΟ0 be the undamped frequency. Find (i) the resonant frequency, (ii) the quality factor Q,(iii) the phase angle Ξ¦ when the system is driven at frequency Ο = Ο0/2, and (iv) thesteady-state amplitude at this frequency.
2. (10 points)A string of length 2 l is suspended at points A and B located on a horizontal line. Thedistance between A and B is 2 d, with d < l. A small, heavy bead can slide on the stringwithout friction. Find the period of the small-amplitude oscillations of the bead in thevertical plane containing the suspension points.Hint: The trajectory of the bead is a section of an ellipse (why?). Move the origin to theequilibrium point and use a Taylor expansion to get an approximate expression for thetrajectory around the equilibrium point. Apply Lagrange.
...continued on next page...
SOLUTION:
The locus the bead describes is an ellipse, since in an ellipse the total distance from anypoint on it to the points π΄,π΅ is always the same
6
In an ellipse, these twosegments always add tosame length. In this ex-ample, this is 2l
ellipse
To obtain the potential energy, we move the bead a little from the origin and find how muchthe bead moved above the origin, as shown in the following diagram
s 2l β s
d+ x dβ x
originx
AB
H =βl2 β d2
h
y = H β h
s2 = h2 + (d+ x)2
(2l β s)2 = h2 + (dβ x)2
From the above, we see that, by applying pythagoras triangle theorem to the left and to theright triangles, we obtain two equations which we solve for β in order to obtain the potentialenergy
π 2 = β2 + (π + π₯)2
(2π β π )2 = β2 + (π β π₯)2
Solving for β gives
β =οΏ½1 β
π2
π2βπ2 β π₯2
Therefore
π¦ = π» β β
= π» βοΏ½1 β
π2
π2βπ2 β π₯2
7
Hence
π = πππ¦
= ππββββββπ» β
οΏ½1 β
π2
π2βπ2 β π₯2
ββββββ
The kinetic energy is
π =12π οΏ½οΏ½οΏ½2 + οΏ½οΏ½2οΏ½
Therefore the Lagrangian is
πΏ = π β π
=12π οΏ½οΏ½οΏ½2 + οΏ½οΏ½2οΏ½ β ππ
ββββββπ» β
οΏ½1 β
π2
π2βπ2 β π₯2
ββββββ
The equation of motion in the π₯ coordinate is now found. From
ππΏππ₯
=12πποΏ½1 β
π2
π2(β2π₯)
βπ2 β π₯2
= βπποΏ½1 β
π2
π2π₯
βπ2 β π₯2And
πππ‘ππΏππ₯
= ποΏ½οΏ½
Applying Euler-Lagrangian equation gives
πππ‘ππΏππ₯
βππΏππ₯
= 0
οΏ½οΏ½ + ποΏ½1 β
π2
π2π₯
βπ2 β π₯2= 0
For very small π₯, we drop the π₯2 term and the above reduces to
οΏ½οΏ½ + ποΏ½1 β
π2
π2π₯π= 0
Hence the undamped natural frequency is
π20 =
ππ οΏ½
1 βπ2
π2or
π0 =οΏ½πποΏ½
1 βπ2
π2
8
The period of small oscillation is therefore
π =2ππ0
= 2π1
οΏ½πποΏ½1 β
π2
π2
0.3 Problem 3
3. (10 points)A rod of length L rotates in a plane with a constant angular velocity Ο about an axis fixedat one end of the rod and perpendicular to the plane of rotation. A bead of mass m isinitially at the stationary end of the rod. It is given a slight push so that its initial speedalong the rod is ΟL. Find the time it takes the bead to reach the other end of the rod.
4. (10 points)Consider a harmonic oscillator with Ο0 = 0.5 sβ1. Let x0 = 1.0m be the initial amplitudeat t = 0 and assume that the oscillator is released with zero initial velocity. Use a computerto plot the phase-space plot (x versus x) for the following damping coefficients Ξ».(1) Ξ» = 0.05 sβ1 (weak damping)(2) Ξ» = 0.25 sβ1 (strong damping)(3) Ξ» = Ο0 (critical damping).
5. (15 points)A damped harmonic oscillator has a period of free oscillation (with no damping) of T0 =1.0 s. The oscillator is initially displaced by an amount x0 = 0.1m and released with zeroinitial velocity.(1) Consider the case that the oscillator is critically damped. Determine the displacementx as a function of time and use a computer program to plot x(t) for 0 β€ t β€ 2 s.(2) Now consider the case that the system is overdamped. Determine the displacementas a function of time and use a computer program to plot x(t) for damping coefficients(i) Ξ» = 2.2 Οsβ1, (ii) Ξ» = 4 Οsβ1, and (iii) Ξ» = 10 Οsβ1 for 0 β€ t β€ 2 s. Compare to thecritically damped case.(3) Now consider the case that the system is underdamped. Determine the displacementas a function of time and use a computer program to plot x(t) for damping coefficients (i)Ξ» = 5.0 sβ1, (ii) Ξ» = 1.0 sβ1, and (iii) Ξ» = 0.1 sβ1 for 0 β€ t β€ 2 s. Compare to the criticallydamped case.
0.3.1 SOLUTION method one
The velocity of the particle is as shown in the following diagram
ΞΈ
velocity diagram
x
yΟ = ΞΈ
r
rΟ Vx = r cos ΞΈ β rΟ sin ΞΈ
Vy = r sin ΞΈ + rΟ cos ΞΈ
There is no potential energy, and the Lagrangian only comes from kinetic energy.
π£2 = π2π₯ + π2
π¦
= (οΏ½οΏ½ cosπ β ππ sinπ)2 + (οΏ½οΏ½ sinπ + ππ cosπ)2
Exapnding and simplifying gives
π£2 = οΏ½οΏ½2 + π2π2
Hence
πΏ =12π οΏ½οΏ½οΏ½2 + π2π2οΏ½
9
And the equation of motion in the radial π direction is
πππ‘ππΏποΏ½οΏ½
βππΏππ
= 0
πππ‘ποΏ½οΏ½ β πππ2 = 0
Hence the equation of motion is
οΏ½οΏ½ β ππ2 = 0 (1)
The roots of the characteristic equation are Β±π, hence the solution is
π (π‘) = π1πππ‘ + π2πβππ‘
At π‘ = 0, π (0) = 0 and οΏ½οΏ½ (π‘) = πΏπ. Using these we can find π1, π2.
0 = π1 + π2 (2)
But οΏ½οΏ½ (π‘) = ππ1πππ‘ β ππ2πβππ‘ and at π‘ = 0 this becomes
πΏπ = ππ1 β ππ2 (3)
From (2,3) we solve for π1, π2. From (2), π1 = βπ2 and (3) becomes
πΏπ = βππ2 β ππ2
π2 =πΏπβ2π
=β12πΏ
Hence π1 =12πΏ and the solution is
π (π‘) = π1πππ‘ + π2πβππ‘
=12πΏπππ‘ β
12πΏπβππ‘
= πΏ οΏ½πππ‘ β πβππ‘
2 οΏ½
Or
π (π‘) = πΏ (sinhππ‘)
To find the time it takes to reach end of rod, we solve for π‘π from
πΏ = πΏ οΏ½sinhππ‘ποΏ½1 = sinhππ‘π
Hence
ππ‘π = sinhβ1 (1)= 0.88137
Therefore
π‘π =0.88137
π sec
10
0.3.2 another solution
Let the local coordinate frame rotate with the bar, where the bar is oriented along the π₯axis of the local body coordinate frame as shown below.
X
Y
x
y
~rm
x, y is Rotating (body) frame of reference
X,Y is inertial (fixed) frame of reference
~i~j
~v = ~rrel + ~Ο Γ ~r
Ο
~r = r~i
~Ο = Ο~k
The position vector of the particle is π = ππ where π is unit vector along the π₯ axis. Takingtime derivative, and using the rotating vector time derivative rule which says that ππ¨
ππ‘ =
οΏ½ππ¨ππ‘οΏ½πππππ‘ππ£π
+ π Γ π¨ where π is the angular velocity of the rotating frame then
οΏ½οΏ½ = οΏ½οΏ½πππ + π Γ π (1)
To find the acceleration of the particle, we take time derivative one more timeπππ‘οΏ½οΏ½ =
πππ‘(οΏ½οΏ½πππ) + οΏ½οΏ½ Γπ + π Γ οΏ½οΏ½
But πππ‘(οΏ½οΏ½πππ) = οΏ½οΏ½πππ + π Γ οΏ½οΏ½πππ by applying the rule of time derivative of rotating vector again.
Therefore the above equation becomesπππ‘οΏ½οΏ½ = οΏ½οΏ½πππ + π Γ οΏ½οΏ½πππ+ οΏ½οΏ½ Γπ + π Γ οΏ½οΏ½
Replacing οΏ½οΏ½ in the above from its value in (1) gives
οΏ½οΏ½ = οΏ½οΏ½πππ + π Γ οΏ½οΏ½πππ+ οΏ½οΏ½ Γπ + π Γ (οΏ½οΏ½πππ + π Γ π)= οΏ½οΏ½πππ + π Γ οΏ½οΏ½πππ+ οΏ½οΏ½ Γπ + π Γ οΏ½οΏ½πππ +π Γ (π Γ π)= οΏ½οΏ½πππ + 2 (π Γ οΏ½οΏ½πππ) + οΏ½οΏ½ Γπ + π Γ (π Γ π)
But π is constant (bar rotate with constant angular speed), hence the term οΏ½οΏ½ above is zero,and the above reduces to
οΏ½οΏ½ = οΏ½οΏ½πππ + 2 (π Γ οΏ½οΏ½πππ) + π Γ (π Γ π) (2)
The above is the acceleration of the particle as seen in the inertial frame. Now we calculatethis acceleration by preforming the vector operations above, noting that π = ππ, π = ππ,
11
hence (2) becomes
οΏ½οΏ½ = ποΏ½οΏ½πππ + 2 (ππ Γ ποΏ½οΏ½πππ) + ππ Γ (ππ Γ ππ)
= ποΏ½οΏ½πππ + 2 οΏ½πποΏ½οΏ½ππποΏ½ + ππ Γ οΏ½ππποΏ½
= ποΏ½οΏ½πππ + 2 οΏ½πποΏ½οΏ½ππποΏ½ β ππ2π
= π οΏ½οΏ½οΏ½πππ β π2ποΏ½ + π (2ποΏ½οΏ½πππ)
The particle has an acceleration along π₯ axis and an acceleration along π¦ axis. We areinterested in the acceleration along π₯ since this is where the rod is oriented along. Thescalar version of the acceleration in the π₯ direction is
ππ₯ = οΏ½οΏ½πππ β π2π
Using πΉπ₯ = πππ₯ and since πΉπ₯ = 0 (there is no force on the particle) then the equation ofmotion along the bar (π₯ axis) is
οΏ½οΏ½πππ β π2π = 0
The roots of the characteristic equation is Β±π, hence the solution is
π (π‘) = π1πππ‘ + π2πβππ‘
At π‘ = 0, π (0) = 0 and οΏ½οΏ½ (π‘) = πΏπ. Using these we can find π1, π2.
0 = π1 + π2 (3)
But οΏ½οΏ½ (π‘) = ππ1πππ‘ β ππ2πβππ‘ and at π‘ = 0 this becomes
πΏπ = ππ1 β ππ2 (4)
From (3,4) we solve for π1, π2. From (3), π1 = βπ2 and (4) becomes
πΏπ = βππ2 β ππ2
π2 =πΏπβ2π
=β12πΏ
Hence π1 =12πΏ and the solution is
π (π‘) = π1πππ‘ + π2πβππ‘
=12πΏπππ‘ β
12πΏπβππ‘
= πΏ οΏ½πππ‘ β πβππ‘
2 οΏ½
= πΏ (sinhππ‘)To find the time it takes to reach end of rod, we solve for π‘π from
πΏ = πΏ οΏ½sinhππ‘ποΏ½1 = sinhππ‘π
Hence
ππ‘π = sinhβ1 (1)= 0.88137
12
Therefore
π‘π =0.88137
π sec
0.4 Problem 4
3. (10 points)A rod of length L rotates in a plane with a constant angular velocity Ο about an axis fixedat one end of the rod and perpendicular to the plane of rotation. A bead of mass m isinitially at the stationary end of the rod. It is given a slight push so that its initial speedalong the rod is ΟL. Find the time it takes the bead to reach the other end of the rod.
4. (10 points)Consider a harmonic oscillator with Ο0 = 0.5 sβ1. Let x0 = 1.0m be the initial amplitudeat t = 0 and assume that the oscillator is released with zero initial velocity. Use a computerto plot the phase-space plot (x versus x) for the following damping coefficients Ξ».(1) Ξ» = 0.05 sβ1 (weak damping)(2) Ξ» = 0.25 sβ1 (strong damping)(3) Ξ» = Ο0 (critical damping).
5. (15 points)A damped harmonic oscillator has a period of free oscillation (with no damping) of T0 =1.0 s. The oscillator is initially displaced by an amount x0 = 0.1m and released with zeroinitial velocity.(1) Consider the case that the oscillator is critically damped. Determine the displacementx as a function of time and use a computer program to plot x(t) for 0 β€ t β€ 2 s.(2) Now consider the case that the system is overdamped. Determine the displacementas a function of time and use a computer program to plot x(t) for damping coefficients(i) Ξ» = 2.2 Οsβ1, (ii) Ξ» = 4 Οsβ1, and (iii) Ξ» = 10 Οsβ1 for 0 β€ t β€ 2 s. Compare to thecritically damped case.(3) Now consider the case that the system is underdamped. Determine the displacementas a function of time and use a computer program to plot x(t) for damping coefficients (i)Ξ» = 5.0 sβ1, (ii) Ξ» = 1.0 sβ1, and (iii) Ξ» = 0.1 sβ1 for 0 β€ t β€ 2 s. Compare to the criticallydamped case.
SOLUTION:
Starting with the equation of motion for damped oscillator
π₯β²β² + 2ππ₯β² + π20π₯ = 0
The solution for cases 1,2 (both are underdamped) is
π₯ = πβππ‘ (π΄ cosπππ‘ + π΅ sinπππ‘) (1)
Where ππ = οΏ½π20 β π2. While the solution for case (3), the critical damped case is
π₯ = (π΄ + π‘π΅) πβππ‘ (2)
For (1) above, at π‘ = 0 we obtain
1 = π΄
Hence (1) becomes π₯ = πβππ‘ (cosπππ‘ + π΅ sinπππ‘), and taking derivative gives
οΏ½οΏ½ = βππβππ‘ (cosπππ‘ + π΅ sinπππ‘) + πβππ‘ (βππ sinπππ‘ + π΅ππ cosπππ‘)
At π‘ = 0 we have
0 = βπ + π΅ππ
π΅ =πππ
Hence the complete solution for (1) is
π₯ = πβππ‘ οΏ½cosπππ‘ +πππ
sinπππ‘οΏ½ (3)
οΏ½οΏ½ = βππ₯ + πβππ‘ (βππ sinπππ‘ + π cosπππ‘) (4)
Now we find the solution for (2), the critical damped case. At π‘ = 0
1 = π΄
13
Hence (2) becomes π₯ = (1 + π‘π΅) πβππ‘, and taking derivative gives
οΏ½οΏ½ = π΅πβππ‘ β π (1 + π‘π΅) πβππ‘
At π‘ = 0
0 = π΅ β ππ΅ = π
Hence the solution to (2) becomes
π₯ = (1 + ππ‘) πβππ‘ (5)
οΏ½οΏ½ = ππβππ‘ β π (1 + ππ‘) πβππ‘ (6)
Now that the solutions are found, we plot the phase space using the computer, using para-metric plot command
0.4.1 case (1)
For π = 0.05, and ππ = οΏ½π20 β π2 = β0.52 β 0.052 = 0.4975, then equations (3,4) become
π₯ = πβ0.05π‘ (cos 0.4975π‘ + 0.1005 sin 0.4975π‘) (3A)
οΏ½οΏ½ = β0.05π₯ + πβ0.05π‘ (β0.4975 sin 0.4975π‘ + 0.05 cos 0.4975π‘) (4A)
Here is the plot generated, showing starting point (1, 0) with the code used
Out[59]=
-0.5 0.0 0.5 1.0
-0.4
-0.2
0.0
0.2
x(t)
v(t)
Phase plot, 50 seconds, case(1)
am = 0.05;wn = 0.5;wd = Sqrt[wn^2 - lam^2];x = Exp[-lam t] (Cos[wd t] + lam/wd Sin[wd t]);y = -lam x + Exp[-lam t] (-wd Sin[wd t] + lam Cos[lam t]);ParametricPlot[{x, y}, {t, 0, 50}, Frame -> True,GridLines -> Automatic, GridLinesStyle -> LightGray,FrameLabel -> {{"v(t)", None}, {"x(t)",
"Phase plot, 50 seconds, case(1)"}}, Epilog -> Disk[{1, 0}, .02],ImageSize -> 400]
14
0.4.2 case (2)
For π = 0.25, and ππ = οΏ½π20 β π2 = β0.52 β 0.252 = 0.433, equations (3,4) become
π₯ = πβ0.25π‘ (cos 0.433π‘ + 0.5774 sin 0.433π‘) (3A)
οΏ½οΏ½ = β0.05π₯ + πβ0.25π‘ (β0.433 sin 0.433π‘ + 0.05 cos 0.433π‘) (4A)
Here is the plot generated where the starting point was (1, 0)
Out[150]=
-0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2-0.3
-0.2
-0.1
0.0
0.1
0.2
x(t)
v(t)
Phase plot, 50 seconds, case(2)
This below is a zoomed in version of the above close to the origin
Out[151]=
-0.20 -0.15 -0.10 -0.05 0.00 0.05 0.10-0.15
-0.10
-0.05
0.00
0.05
0.10
x(t)
v(t)
Phase plot, 50 seconds, case(2), zoom in
0.4.3 case(3)
For this case, equations (5,6) are used. For π = 0.5, equations (5,6) become
π₯ = (1 + 0.5π‘) πβ0.5π‘ (5A)
οΏ½οΏ½ = 0.5πβ0.5π‘ β 0.5 (1 + 0.5π‘) πβ0.5π‘ (6A)
Here is the plot generated, showing starting point (1, 0) with the code used
15
Out[155]=
-0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2-0.3
-0.2
-0.1
0.0
0.1
0.2
x(t)v(t)
Phase plot, 50 seconds, case(3)
lam = 0.5;x = (1 + lam*t) Exp[-lam t];y = lam*Exp[-lam t] - lam*(1 + lam t) Exp[- lam t]ParametricPlot[{x, y}, {t, 0, 30}, Frame -> True,GridLines -> Automatic, GridLinesStyle -> LightGray,FrameLabel -> {{"v(t)", None}, {"x(t)",
"Phase plot, 50 seconds, case(3)"}}, Epilog -> Disk[{1, 0}, .02],ImageSize -> 500, PlotRange -> {{-.3, 1.2}, {-.3, .2}},PlotTheme -> "Classic"]
0.5 Problem 5
3. (10 points)A rod of length L rotates in a plane with a constant angular velocity Ο about an axis fixedat one end of the rod and perpendicular to the plane of rotation. A bead of mass m isinitially at the stationary end of the rod. It is given a slight push so that its initial speedalong the rod is ΟL. Find the time it takes the bead to reach the other end of the rod.
4. (10 points)Consider a harmonic oscillator with Ο0 = 0.5 sβ1. Let x0 = 1.0m be the initial amplitudeat t = 0 and assume that the oscillator is released with zero initial velocity. Use a computerto plot the phase-space plot (x versus x) for the following damping coefficients Ξ».(1) Ξ» = 0.05 sβ1 (weak damping)(2) Ξ» = 0.25 sβ1 (strong damping)(3) Ξ» = Ο0 (critical damping).
5. (15 points)A damped harmonic oscillator has a period of free oscillation (with no damping) of T0 =1.0 s. The oscillator is initially displaced by an amount x0 = 0.1m and released with zeroinitial velocity.(1) Consider the case that the oscillator is critically damped. Determine the displacementx as a function of time and use a computer program to plot x(t) for 0 β€ t β€ 2 s.(2) Now consider the case that the system is overdamped. Determine the displacementas a function of time and use a computer program to plot x(t) for damping coefficients(i) Ξ» = 2.2 Οsβ1, (ii) Ξ» = 4 Οsβ1, and (iii) Ξ» = 10 Οsβ1 for 0 β€ t β€ 2 s. Compare to thecritically damped case.(3) Now consider the case that the system is underdamped. Determine the displacementas a function of time and use a computer program to plot x(t) for damping coefficients (i)Ξ» = 5.0 sβ1, (ii) Ξ» = 1.0 sβ1, and (iii) Ξ» = 0.1 sβ1 for 0 β€ t β€ 2 s. Compare to the criticallydamped case.
SOLUTION:
Since π0 =2ππ0, then π0 =
2π1 = 2π.
16
0.5.1 Part (1)
For critical damping π = π0 and the solution is
π₯ (π‘) = (π΄ + π΅π‘) πβππ‘ (1)
οΏ½οΏ½ (π‘) = π΅πβππ‘ β π (π΄ + π΅π‘) πβππ‘ (2)
Initial conditions are now used to find π΄,π΅. At π‘ = 0, π₯ (0) = π₯0 = 0.1. From (1) we obtain
π₯0 = π΄
And since οΏ½οΏ½ (0) = 0, then from (2)
0 = π΅ β ππ΄π΅ = ππ΄= ππ₯0
Putting values found for π΄,π΅, back into (1) gives
π₯ (π‘) = (π₯0 + ππ₯0π‘) πβππ‘
Since this is critical damping, then π = π0 = 2π, hence
π₯ (π‘) = (π₯0 + 2ππ₯0π‘) πβ2ππ‘
Finally, since π₯0 = 0.1 meter, then
π₯ (π‘) = οΏ½110+2π10π‘οΏ½ πβ2ππ‘
A plot of the above for 0 β€ π‘ β€ 2π is given below
Out[179]=
0.0 0.5 1.0 1.5 2.0
0.00
0.02
0.04
0.06
0.08
0.10
time (sec)
x(t)
Part(1) critical damped
0.5.2 Part(2)
For overdamped, π > π0 the two roots of the characteristic polynomial are real, hence nooscillation occur. The solution is given by
π₯ (π‘) = π΄ποΏ½βπ+οΏ½π2βπ2
0οΏ½π‘ + π΅ποΏ½βπβοΏ½π2βπ2
0οΏ½π‘ (1)
17
π΄,π΅ are found from initial conditions. When π‘ = 0 the above becomes
π₯0 = π΄ + π΅ (2)
Taking derivative of (1) gives
οΏ½οΏ½ (π‘) = π΄ οΏ½βπ +οΏ½π2 β π2
0οΏ½ ποΏ½βπ+οΏ½π2βπ2
0οΏ½π‘ + π΅ οΏ½βπ β οΏ½π2 β π2
0οΏ½ ποΏ½βπβοΏ½π2βπ2
0οΏ½π‘
At π‘ = 0 the above becomes
0 = οΏ½βπ +οΏ½π2 β π2
0οΏ½π΄ + οΏ½βπ β οΏ½π2 β π2
0οΏ½ π΅ (3)
We have two equations (2,3) which we solve for π΄,π΅. From (2), π΄ = π₯0 β π΅, and (3) becomes
0 = οΏ½βπ +οΏ½π2 β π2
0οΏ½ (π₯0 β π΅) + οΏ½βπ β οΏ½π2 β π2
0οΏ½ π΅
0 = οΏ½βπ +οΏ½π2 β π2
0οΏ½ π₯0 β π΅ οΏ½βπ +οΏ½π2 β π2
0οΏ½ + οΏ½βπ β οΏ½π2 β π2
0οΏ½ π΅
0 = οΏ½βπ +οΏ½π2 β π2
0οΏ½ π₯0 β 2π΅οΏ½π2 β π2
0
π΅ =οΏ½βπ +οΏ½π
2 β π20οΏ½ π₯0
2οΏ½π2 β π2
0
(4)
Using π΅ found in (4) then (3) now gives π΄ as
π΄ = π₯0 β π΅
= π₯0 βοΏ½βπ +οΏ½π
2 β π20οΏ½ π₯0
2οΏ½π2 β π2
0
= π₯0
ββββββββββ1 β
οΏ½βπ +οΏ½π2 β π2
0οΏ½
2οΏ½π2 β π2
0
ββββββββββ
= π₯0
βββββββββ
π + οΏ½π2 β π2
0
2οΏ½π2 β π2
0
βββββββββ
Hence the complete solution from (1) becomes
π₯ (π‘) = π₯0
βββββββββ
π + οΏ½π2 β π2
0
2οΏ½π2 β π2
0
βββββββββ ποΏ½βπ+οΏ½π2βπ2
0οΏ½π‘ + π₯0
βββββββββ
βπ +οΏ½π2 β π2
0
2οΏ½π2 β π2
0
βββββββββ ποΏ½βπβοΏ½π2βπ2
0οΏ½π‘ (5)
The above is now used for each case below to plot the solution..
18
case (i)
π = 2.2π,π0 = 2π, π₯0 = 0.1, hence (5) becomes
π₯ (π‘) = 0.1
βββββββββ
2.2π + οΏ½(2.2π)2 β (2π)2
2οΏ½(2.2π)2 β (2π)2
βββββββββ ποΏ½β2.2π+οΏ½(2.2π)2β(2π)2οΏ½π‘
+ 0.1
βββββββββ
β2.2π + οΏ½(2.2π)2 β (2π)2
2οΏ½(2.2π)2 β (2π)2
βββββββββ ποΏ½β2.2πβοΏ½(2.2π)2β(2π)2οΏ½π‘
= 0.17πβ4.032 2π‘ β 0.07 πβ9.791π‘
A plot of the above for 0 β€ π‘ β€ 2π is given below
Out[297]=
0.0 0.5 1.0 1.5 2.0
0.00
0.02
0.04
0.06
0.08
0.10
time (sec)
x(t)
Part(2.1) overdamped, Ξ»=2.2 Ο
case (ii)
π = 4π,π0 = 2π, π₯0 = 0.1, hence (5) becomes
π₯ (π‘) = 0.1
βββββββββ
4π + οΏ½(4π)2 β (2π)2
2οΏ½(4π)2 β (2π)2
βββββββββ ποΏ½β4π+οΏ½(4π)2β(2π)2οΏ½π‘
+ 0.1
βββββββββ
β4π +οΏ½(4π)2 β (2π)2
2οΏ½(4π)2 β (2π)2
βββββββββ ποΏ½β4πβοΏ½(4π)2β(2π)2οΏ½π‘
= 0.1077πβ1.6836π‘ β 0.00774πβ23.449π‘
A plot of the above for 0 β€ π‘ β€ 2π is given below
19
Out[310]=
0.0 0.5 1.0 1.5 2.0
0.00
0.02
0.04
0.06
0.08
0.10
time (sec)
x(t)
Part(2.2) overdamped, Ξ»=4 Ο
case (iii)
π = 10π,π0 = 2π, π₯0 = 0.1, hence (5) becomes
π₯ (π‘) = 0.1
βββββββββ
10π + οΏ½(10π)2 β (2π)2
2οΏ½(10π)2 β (2π)2
βββββββββ ποΏ½β10π+οΏ½(10π)2β(2π)2οΏ½π‘
+ 0.1
βββββββββ
β10π +οΏ½(10π)2 β (2π)2
2οΏ½(10π)2 β (2π)2
βββββββββ ποΏ½β10πβοΏ½(10π)2β(2π)2οΏ½π‘
= 0.101 πβ0.634 73π‘ β 0.001034πβ62.197π‘
A plot of the above for 0 β€ π‘ β€ 2π is given below
Out[314]=
0.0 0.5 1.0 1.5 2.0
0.04
0.06
0.08
0.10
time (sec)
x(t)
Part(2.3) overdamped, Ξ»=10 Ο
To compare to the critical damped case, the above three plots are plotted on the same figureagainst the critical damped case in order to get a better picture and be able to compare theresults
20
critical
damping
2.2 pi
4 pi
10 pi
increased damping
0.0 0.5 1.0 1.5 2.0
0.00
0.02
0.04
0.06
0.08
0.10
time (sec)
x(t)
Comparing critical damped response to overdamped
From the above we see that critical damping has the fastest decay of the response π₯ (π‘). Asthe damping increases, it takes longer for the response to decay.
0.5.3 Part(3)
For the underdamped case, the solution is given by
π₯ (π‘) = πβππ‘ (π΄ cosπππ‘ + π΅ sinπππ‘) (1)
Where ππ = οΏ½π20 β π2 and π΄,π΅ are constant of integration that can be found from initial
conditions. And
οΏ½οΏ½ (π‘) = βππβππ‘ (π΄ cosπππ‘ + π΅ sinπππ‘) + πβππ‘ (βπ΄ππ sinπππ‘ + π΅ππ cosπππ‘) (2)
Applying initial conditions π₯ (0) = π₯0 then (1) becomes
π₯0 = π΄
Applying initial conditions οΏ½οΏ½ (0) = 0 then (2) becomes
0 = βππ₯0 + π΅ππ
π΅ =ππ₯0ππ
Replacing π΄,π΅ back into the solution (1) gives the solution
π₯ (π‘) = πβππ‘ οΏ½π₯0 cosπππ‘ +ππ₯0ππ
sinπππ‘οΏ½ (3)
We now use the above solution for the rest of the problem
21
case(i)
π = 5π β1, π0 = 2π, π₯0 = 0.1, hence ππ = οΏ½π20 β π2 = οΏ½(2π)
2 β 52 = 3.8051 and (3) becomes
π₯ (π‘) = πβ5π‘ οΏ½0.1 cos (3.8051π‘) +(5) (0.1)3.8051
sin (3.8051π‘)οΏ½
= πβ5π‘ (0.1 cos (3.8051π‘) + 0.1314 sin (3.8051π‘))A plot of the above solution π₯ (π‘) for 0 β€ π‘ β€ 2π is given below
Out[600]=
0.0 0.5 1.0 1.5 2.0
0.00
0.02
0.04
0.06
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time (sec)
x(t)
Part(3.1) underdamped, Ξ»=5 per sec
case(ii)
π = 1π β1, π0 = 2π, π₯0 = 0.1, hence ππ = οΏ½π20 β π2 = οΏ½(2π)
2 β 12 = 6.2031 and (3) becomes
π₯ (π‘) = πβπ‘ οΏ½0.1 cos (6.2031π‘) +(1) (0.1)6.2031
sin (6.2031π‘)οΏ½
= πβπ‘ (0.1 cos (6.2031π‘) + 0.016 sin (6.2031π‘))A plot of the above solution π₯ (π‘) for 0 β€ π‘ β€ 2π is given below
22
Out[596]=
0.0 0.5 1.0 1.5 2.0
-0.05
0.00
0.05
0.10
time (sec)
x(t)
Part(3.2) underdamped, Ξ»=1 per sec
case(iii)
π = 0.1π β1, π0 = 2π, π₯0 = 0.1, hence ππ = οΏ½π20 β π2 = οΏ½(2π)
2 β 0.12 = 6.2824 and (3) becomes
π₯ (π‘) = πβ0.1π‘ οΏ½0.1 cos (6.2824π‘) +(0.1) (0.1)6.2824
sin (6.2824π‘)οΏ½
= πβ0.1π‘ (0.1 cos (6.2824π‘) + 0.001592 sin (6.2824π‘))A plot of the above solution π₯ (π‘) for 0 β€ π‘ β€ 2π is given below
Out[591]=
0.0 0.5 1.0 1.5 2.0-0.10
-0.05
0.00
0.05
0.10
time (sec)
x(t)
Part(3.3) underdamped, Ξ»=0.1 per sec
To compare to the critical damped case, the above 3 plots are now plotted on the same figureagainst the critical damped case in order to get a better picture and be able to compare theresults
23
decreased damping
critical
5
1.0
0.1
0.0 0.5 1.0 1.5 2.0
-0.10
-0.05
0.00
0.05
0.10
time (sec)
x(t)
Comparing critical damped response to underdamped
As the damping becomes smaller, more oscillation occur. The case for π = 5π β1 had thesmallest oscillation.