Problems

Problem 1:

A particle that hangs from a spring oscillates with an angular frequency of 2 rad/s.  The spring is suspended from the ceiling of an elevator car and hangs motionless (relative to the car) as the car descends at a constant speed of 1.5 m/s.  The car then suddenly stops.  Neglect the mass of the spring.
(a) With what amplitude does the particle oscillate?
(b) What is the equation of motion for the particle?  (Choose the upward direction to be positive.)

Solution:

Concepts, principles, relations that apply to the problem:
Simple harmonic motion
Why do they apply?
After the elevator stops, we have a mass on a spring that moves through the equilibrium position with a given velocity.
How do they apply?
(a) When traveling in the elevator at constant speed, the total force on the mass is zero.  The force exerted by the spring is equal in magnitude to the gravitational force on the mass, the spring has the equilibrium length of a vertical spring.  When the elevator suddenly stops, the end of the spring attached to the ceiling stops.  The mass, however has momentum, p = mv, and therefore starts stretching the spring.  It moves through the equilibrium position of the vertical spring with its maximum velocity vmax = 1.5m/s.
Its velocity as a function of time is v(t) = -wAsin(wt+f).
Since vmax = wA and w = 2/s, the amplitude of the oscillations is A = 0.75m.
(b) The equation of motion for the particle is d2x/dt2 = -(k/m)x = -w2x.  Its solution is
x(t) = Acos(wt+f) = (0.75m)cos((2/s)t+f).
If we choose the t = 0 to be the time the elevator stops and let the upward direction be positive, then x(0) = 0, and v(0) = -1.5m/s. We therefore need f to be p/2.
Details of the calculation:
None

Problem 2:

One end of a spring of negligible mass is attached to the ceiling.  When a 250g mass is placed on the free end (without stretching the spring) and then released, the mass descends 20.0cm before it changes direction and begins to ascend.  What is the spring constant k (in N/m)?

Solution:

Concepts, principles, relations that apply to the problem:
Simple harmonic motion
Why do they apply?
The force exerted by a spring obeys Hooke's law.
How do they apply?
When the mass is hanging on the spring in equilibrium, i.e. when all oscillations have damped out, then the magnitude of the force F = kx from the spring exactly cancels the magnitude of the gravitational force F = mg.
We have mg = kx, or k = mg/x.
m = 0.25 kg is given, we need to know the equilibrium stretch x.
The mass initially oscillates between x = 0 and x = 20 cm.  If we wait long enough the oscillation will damp out and it will come to rest at x = 0.1 m.
k = (0.25kg)(9.8 m/s2)/(0.1 m) = 24.5 N/m.
Details of the calculation:
None

Problem 3:

A simple pendulum is attached to the ceiling of a boxcar which accelerates at a constant rate a.  Find the equilibrium angle of the pendulum, and also the frequency of small oscillations.

Solution:

Concepts, principles, relations that apply to the problem:
Motion in an accelerating frame, small oscillations
Why do they apply?
The boxcar is not an inertial frame.
How do they apply?

The pendulum bob is in equilibrium in the accelerating frame.  It is accelerating in an inertial frame with acceleration a = ai.  The forces on the bob are Fg = mg and T due to the tension in the string.  We have:
Tcosq = mg,  Tsinq = ma, tanq = a/g.
An observer in the boxcar experiences a fictitious force -mai.  The total force observed in the accelerating frame is therefore -m(gj + ai).  The magnitude of this force is mg', with g' = g/cosq. or g' = (g2 + a2)1/2.  Small angular displacements d from the equilibrium angle q will result in a restoring force proportional to d. d2d/dt2 = -(g'/l)d.  The frequency of small oscillations is w = (g'/l)1/2.
Details of the calculation:
None

Problem 4:

A thin rod of linear density r and length L is balanced on a cylindrical arc of radius R as shown.

If one end of the rod is tapped, find the frequency of small oscillations for the rod.

Solution:

Concepts, principles, relations that apply to the problem:
The Lagrangian formalism, small oscillations
Why do they apply?
We are asked to find the frequency of small oscillations.
How do they apply?
Assume we have a very thin rod.
L = T - U.
.
U = MgY.
X and Y are the coordinates of the CM.
Let the origin of the coordinate system be the center of the circle in the figure above.
X = Rsinq - Rqcosq.
Y = Rcosq + Rqsinq.
Small angle approximation:  In L, keep only terms up to second order in q.
X = Rq - Rq = 0.
Y = R(1-q2/2) + Rq2 = R(1+q2/2).
dX/dt = 0, dY/dt = Rqdq/dt.
Therefore
L = (1/2)MR2q2(dq/dt)2 + (1/2)M(L2/12)(dq/dt)2 - MgR(1+q2/2)
@ (1/2)M(L2/12)(dq/dt)2 - Mg R(1+q2/2).
Again:  In the small angle approximation, where we only keep terms to second order in the small quantities.
The equation of motion is 
d2q/dt2 = -(12gR/L2)q,

which yields , independent of the mass of the rod.
Details of the calculation:
None

Problem 5:

A string of length 2l is suspended at points A and B located on a horizontal line.  The distance between A and B is 2d (d < l). A small, heavy bead can slide on the string without friction.  Find the period of the small-amplitude oscillations of the bead in the vertical plane containing the suspension points.  The acceleration due to gravity is g.

Solution:

Concepts, principles, relations that apply to the problem:
 The Lagrangian formalism, small oscillations
Why do they apply?
We are asked to find the period of small oscillations.
How do they apply?
For any point on the trajectory of the bead the sum of the distances from point A and point B is 2l.  The trajectory therefore is a section of an ellipse.  The semi-major axis of this ellipse is a = l and the semi-minor axis is b = (l2 – d2)1/2.  If the origin of the coordinate system is placed at the center of the ellipse, as shown, the equation of the ellipse is x2/a2 + y2/b2 = 1.

The equilibrium position of the bead is at x = 0, y = -b.  For small displacements from equilibrium we have y = -b(1 - x2/a2)1/2 ≈ -b(1 - x2/(2a2)).
Moving the origin of the coordinate system to the equilibrium point we have y = b(x2/(2a2).  To first order the trajectory is parabolic near the equilibrium point.

 

 

Since this problem involves holonomic constraints the Lagrangian method is well suited for solving this problem.  The equation of constraint is y ≈ b(x2/(2a2)).
Using the Lagrangian method we have:

.  L = T - U.

[If we want only first order terms in the equation of motions we must keep terms up to second order in the small quantities in the Lagrangian.]

. 

d2x/dt2 = -gbx/a2. 

Solution:  x = Acos(wt + f) with w = (gb/a2)1/2.  T = 2p/w= 2pa/(gb)1/2 .

Details of the calculation:
 None

Problem 6:

Find the normal, longitudinal modes of vibration for three masses connected by identical springs of spring constant k.  The masses are collinear.  The end masses have mass m, while the inner mass has mass 2m.
(a) Calculate the normal modes of the system.
(b) Describe the relative motion of the particles for each normal mode.

Solution:

Concepts, principles, relations that apply to the problem:
Coupled oscillations, normal modes
Why do they apply?
We are asked to find the normal modes of coupled harmonic oscillators.
How do they apply?
(a)  The kinetic energy is T = (1/2)[ m(dx1/dt)2 +  2m(dx2/dt)2 +  m(dx3/dt)2],
and the potential energy is U = (k/2)[(x2 - x1)2 + (x3 - x2)2].
U = (k/2)[x12 + 2x22 + x32 - x1x2 - x2x1 - x2x3 - x3x2].
The xi are the displacements from the equilibrium positions.  We use the xi as our generalized coordinates qi.
The Lagrangian is L = T - U.  This can be put into the form
.
Here T11 = T33 = m, T22 = 2m, Tij(i¹j) = 0,
k11 = k33 = k, k22 = 2k, k12 = k21 = k23 = k32 = -k, k13 = k31 = 0.
Solutions of the form xj = Re(Ajeiw t) can be found.  For solutions of this form the equations of motion reduce to:


We can find the w2 from det(kij-w 2Tij) = 0.  For a system with n degrees of freedom, n characteristic frequencies wa can be found.  Our system has 3 degrees of freedom.
Details of the calculation:

2(k-w2m)3 - 2k2(k-w2m) = 0.
Solution 1: k-w2m = 0,  w = (k/m)1/2.
Solution 2: k-w2m ¹ 0, then (k-w2m)2 = k2, (k-w2m) = ±k.
For (k-w2m) = +k, w = 0.
Solution 3: (k-w2m) = -k, w = (2k/m)1/2.

(b)  The displacements for each mode are determined from the equations of motion.
Solution 1: w = (k/m)1/2.
Equation 1 yields kA2 = 0, equation 2 then yields kA1 + kA3 = 0.
A1 = - A3, A2 = 0, the central mass is stationary, m1 and m3 move in opposite directions with equal amplitudes.
Solution 2: w = 0.
A1 = A2 = A3, translation of the CM, no relative motion.
Solution 3: w = (2k/m)1/2.
Equation 1 yields -kA1 - kA2 = 0, A1 = -A2.
Equation 3 yields -kA2 - kA3 = 0, A3 = -A2.
A1 = A3 = -A2, the central mass move in a direction opposite to the direction of the outer masses.  All masses oscillate with equal amplitudes.

Problem 7:

A uniform horizontal rectangular plate (mass M, length L, width W) rests with its corners on four similar vertical springs with spring constant k.  Find the normal modes of vibration and prove that their frequencies are in the ratio

Solution:

Concepts, principles, relations that apply to the problem:
Small oscillations, normal modes
Why do they apply?
We are asked to find the 3 normal modes of the plate. 
How do they apply?

Assume that the CM of the plate is restricted to move along a vertical line.
The symmetry of the setup implies that the three normal modes are:
(a)  The plate moves up and down and does not rotate.
(b)  The plate rotates about the line x2, the CM is fixed.
(c)  The plate rotates about the line x1, The CM is fixed.
Details of the calculation:
(a)  T = (1/2)M(dz/dt)2.  Here z is the displacement from its equilibrium position in the gravitational field.
U = (1/2)4kz2,  d2z/dt2 = -4(k/m)z,  wa2 = 4k/M.
(b)  T = (1/2)Ib(dq/dt)2,  q = 2z/L for small oscillations.  U = (1/2)4k(L2/4)q2.
.
Ib = ML2/12.
d2q/dt2 = -12 (k/m)q,  wb2 = 12(k/m),  wb = (3)1/2wa.
(c)  T = (1/2)Ic(df/dt)2,  f = 2z/W for small oscillations.  U = (1/2)4k(W2/4)f2.
Ic = MW2/12.
d2f/dt2 = -12 (k/m)f,  wc2 = 12(k/m),  wc = (3)1/2wa.

Problem 8:

A 0.500kg mass is attached to a spring of constant 150N/m.  A driving force F(t) = (12.0N) cos(wt) is applied to the mass, and the damping coefficient b is 6.00Ns/m.  What is the amplitude (in cm) of the steady-state motion if w is equal to half of the natural frequency w0 of the system?

Solution:

Concepts, principles, relations that apply to the problem:
Forced oscillations
Why do they apply?
The spring constant, driving force, and damping coefficient for a forced harmonic oscillator are given and we are asked to find the steady-state amplitude.
How do they apply?
 The equation of motion for the harmonically driven harmonic oscillator [F(t) = F0cos(wt)] with damping coefficient b = 2mb is d2x/dt2 + 2bdx/dt + w02x = Acoswt, with A = F0/m.
The stationary solution  x(t) = Dcos(wt - d) with tand  = 2wb/(w02-w2) and D =  A/[(w02-w2)2 + 4w2b2]1/2.
[The expressions for tand and D are found by substituting  x(t) = Dcos(wt - d) into the differential equation d2x/dt2 + 2bdx/dt + w02x = Acoswt.}
In this problem
D = (F0/m)/[(w02 - w2)2 + (bw/m)2]1/2,
w02 = k/m = 300,  w = w0/2,  w2 = 300/4 = 75.
F0/m = 12/0.5 = 24.
D = 24/[(300 - 75)2 + 36*75/0.25]1/2 = 0.0968 m = 9.7 cm is the amplitude of the steady state motion.
Details of the calculation:
None

Problem 9: 

Consider a damped harmonic oscillator.  Let us define T1 as the time between adjacent zero crossings, 2T1 as its “period”, and w1 = 2p/(2T1) as its “angular frequency”.  The damped harmonic oscillator is characterized by the quality factor Q = w1/(2b), where 1/b is the relaxation time, i.e. the time in which the amplitude of the oscillation is reduced by a factor of 1/e.

(a)  After 8.6 seconds and 5 periods of oscillations, the amplitude of a damped oscillator decreased to 17% of its originally set value.  Determine the quality factor Q of the system.

(b)  A motor now supplies an external driving term M coswt.  Determine the stationary solution for the now driven and damped oscillator.

(c)  For a system with an equation of motion

 

the quality factor Q is defined as Q = wR/(2b), where wR is the angular frequency for amplitude resonance.  Compare the shape of a plot of the amplitude of the oscillations versus the driving frequency W for the above driven and damped oscillator with a Lorentzian atomic line shape for Q = 4 and Q = 107.

Solution:

Concepts, principles, relations that apply to the problem:
The underdamped harmonic oscillator, the driven oscillator
Why do they apply?
The oscillator in part (a) is underdamped, since it crosses the equilibrium position many times.  For part (b) a harmonic driving force is given.
How do they apply?
(a)  The equation of motion for the damped harmonic oscillator is d2x/dt2 + 2bdx/dt + w02x = 0.  The solution for underdamped motion is x(t) = A exp(-bt) cos(w1t - d), with w12 = w02 – b2.  Here w0 is the angular frequency of an undamped oscillator with the same spring constant.
(b) The equation of motion for the harmonically driven harmonic oscillator (with damping) is d2x/dt2 + 2bdx/dt + w02x = Acoswt, with A = M/m.  The stationary solution  x(t) = Dcos(wt - d) with tand  = 2wb/(w02-w2) and D =  A/[(w02-w2)2 + 4w2b2]1/2.
[The expressions for tand and D are found by substituting  x(t) = Dcos(wt - d) into the differential equation d2x/dt2 + 2bdx/dt + w02x = Acoswt.]
Details of the calculation:
(a)  Aexp(-b 8.6s) = 0.17 A, -b 8.6s = ln(0.17) , b = 0.206/s.
5(2T1) = 8.6s, w1 = 2p/(2T1) = 3.651, Q = w1/(2b) = 8.87.
(b)  The stationary solution of the equation of motion is x(t) = Dcos(wt - d) with  D =  A/[(w02-w2)2 + 4w2b2]1/2.  When  w = wR = [w02 - 2b2]1/2 we have amplitude resonance.  [dD/dw = 0 when evaluated at wR.]
For an undamped oscillator wR = w0 and Q = infinity.  The amplitude increases without limit when the driving frequency w approaches w0.
(c)  When Q = 4, we have wR = 8b, so at amplitude resonance is D @ A/(2wRb) = 4A/wR2.  
When Q = 107, we have wR =  2 107 b, so at amplitude resonance D @  A/(2wRb) = 107A/wR2.  The shape of the resonance curve approaches that of an undamped oscillator.
The figure below shows plots of amplitude versus driving frequency for different values of b.  (A = 1, w0 = 10, b = 0.01 (blue), 0.1 (pink), 0.5 (orange).)
Additional information:
For a lightly damped oscillator we have Q @ w0/Dw, where Dw represents points on the amplitude resonance curve that are 2-1/2 = 0.707 of the maximum amplitude.
Proof:
For a lightly damped oscillator wR @ w0 and Q @ w0/(2b).  The amplitude at resonance is D @ A/(2w0b).  The driving frequency for which the amplitude is 2-1/2 of the maximum amplitude is found from A/(2-1/22wb) = A/[8w2b2]1/2 =  A/[(w02-w2)2 + 4w2b2]1/2.
8w2b2 = (w02-w2)2 + 4w2b2, 4w2b2 = (w02-w2)2, 2wb = (w02-w2) = (w0+w)(w0-w).
2wb  @ 2wDw/2, since (w0+w) @ 2w0 near resonance.  This yields 2b  @ Dw and Q @ w0/Dw.
Q becomes large when Dw becomes small compared to w0.

Problem 10:

(a)  Derive an expression for the frequency as a function of time that you would hear if you dropped a source of frequency f0 from a tall tower.
(b)  Derive an expression for the frequency heard by an observer on the ground.

Solution:

Concepts, principles, relations that apply to the problem:
The Doppler effect
Why do they apply?
We have a moving source producing a sound wave.  The source is moving away from the observer on the tower and towards the observer on the ground.  The source is a freely falling object.  Its speed increases linearly with time, v = gt. 
How do they apply?
 Assume a source is moving towards an observer with speed vs.  It emits a beep every T seconds.  The sound travels with speed v towards the observer.  Assume beep i is emitted at t = 0, travels a distance d to the observer, and arrives at the observer at time d/v.  Beep i+1 is emitted at time T a distance d-<vs>T from the observer and arrives at the observer at time (d-<vs>T)/v + T.  The difference in the arrival times of the two beeps is 
T' = (d/v) - (<vs>T)/v + T - d/v =  (1- <vs>/v)T.
Let f = observed frequency = 1/T', f0 = frequency of source = 1/T.
f = f0v/(v-<vs>).
|<vs>| = gt, where t is the time interval from the instant the source is dropped to the instant beep i is emitted plus T/2.
(a)    <vs> is negative  f = f0*340/(340 + 9.8 t), where t is measured in seconds.
(b)   <vs> is positive  f = f0*340/(340 - 9.8 t), where t is measured in seconds.
Details of the calculation:
None

Problem 11:

You are given a microphone hooked up to a frequency analyzer and asked to measure the sound velocity in a steel bar of length L = 0.5m by measuring the frequency of sound waves generated when the bar is struck by a steel mallet. 
(a)  Assuming that the bar is supported by a rigid clamp at the center of the bar, derive the relationship between the measured frequencies and the wave velocity.  Be sure to carefully state the boundary conditions used to derive the standing wave patterns from the reflected traveling waves.
(b)  What is the sound velocity if the fundamental frequency is 3500 s-1?
(c)  When the end of the bar is struck parallel and perpendicular to the bar’s axis, different fundamental frequencies are measured.  Explain why and indicate which you expect to have the higher fundamental frequency.

Solution:

Concepts, principles, relations that apply to the problem:
Standing waves
Why do they apply?
Striking the bar sets up a standing wave with a node at the center and antinodes at the end.
How do they apply?
(a)  L = (n + ½)l, n = 0, 1, 2, … .
lf = v, Lf/(n + ½) = v.
(b) fundamental:  2Lf = v.  v = 3500 m/s.
(c)  We can set up transverse or longitudinal waves.  I expect the longitudinal (compression) waves to have the higher fundamental frequency, because they have the higher speed.
Details of the calculation:
 None