Problems

Problem 1:

A 30.0 kg chair is attached to a spring and allowed to oscillate.  When the chair is empty it takes 0.8 s to make one complete vibration.  But with a person sitting in it, with her feet off of the floor, the chair now takes 1.5 s for one cycle.  What is the mass of the person?

Solution:

Concepts, principles, relations that apply to the problem:
A mass on a spring, the harmonic oscillator
Why do they apply?
For a mass m on a spring with spring constant k, the period of oscillation is proportional to the square root of m.
How do they apply?
w = (k/m)1/2 = 2p/t.  t = 2p(m/k)1/2.  t2/t1 = (m2/m1)1/2.
m2 = 30 kg *(1.5/0.8)2  = 105.47 kg.  mperson = m2 – 30kg = 75.47 kg.
Details of the calculation:
None

Problem 2:

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 3:

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 4:

A mass on a spring moves in one dimension and is subject to a velocity-dependent force.  The net force is F = -kx – bv.  Assuming b is small, what fraction of energy is dissipated in each cycle?

Solution:

Concepts, principles, relations that apply to the problem:
The underdamped harmonic oscillator
Why do they apply?
The oscillator is underdamped, since it goes through many cycles.
How do they apply?
md2x/dt2 = -kx - bdx/dt,  d2x/dt2 = -(k/m)x - (b/m)dx/dt,
Let w02 = k/m, b = b/2m. 
Then 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.
Details of the calculations:
The energy is proportional to the square of the effective amplitude, (A exp(-bt))2.  After one cycle the amplitude decreases by a factor of exp(-bt), where  t = 2p/w1.  The energy decreases by a factor of exp(-2bt). 
If b is small, then exp(-2bt) = 1 - 2bt.
The fraction of the energy dissipated in each cycle is
2bt = (b/m)(4pm)/(4km-b2)1/2 = (4pb)/(4km-b2)1/2.

Problem 5:

A triple pendulum consists of masses am, m and m attached to a single light string at distances a, 2a, and 3a respectively from its points of suspension.  Consider only motion in a plane.
(a)  Determine the value of a such that one of the normal frequencies of the system will equal the frequency of a simple pendulum of length a/2 and mass m.  You may assume the displacements of the masses from equilibrium are small.
(b)  Find the mode corresponding to this frequency and sketch it.

Solution:

Concepts, principles, relations that apply to the problem:
Small oscillations, coupled oscillations, normal modes

Solutions of the form qj = Re(Ajeiwt) can be found. 
We can find the w2 from det(kij - w2Tij) = 0.
Why do they apply?
We are asked to find the normal mode frequencies for small displacements.
How do they apply?

(a)  Let f1, f2, and f3 be the generalized coordinates.  Then
x1 = a sinf1, x2 = a(sinf1 + sinf2), x3 = a(sinf1 + sinf2 + sinf3),
y1 = a cosf1, y2 = a(cosf1 + cosf2), x3 = a(cosf1 + cosf2 + cosf3).
For small displacements we have sinf = f, cosf = 1 - f2/2.
The kinetic energy of the system is
.
Let us only keep terms up to second order in f and df/dt.  Then
.
.
The potential energy of the system is
U = -amgy1 - mgy2 - mgy3.
.
We have:
.
T11 = ma2(a + 2), T22 = 2ma2, T33 = ma2, T12 = T21 = 2ma2, T13 = T31 = T23 = T32 = ma2.
k11 = mga(a + 2), k22 = 2mga, k33 = mga, kij(i ¹ j) = 0.

.

.
Details of the calculations:
For a simple pendulum with length a/2 we have w = (2g/a)1/2.  We are looking for a mode with w2 = 2g/a, or l = 2.
.
(b)  The mode corresponding to this frequency is found from

.

A3 = -2A1, A2 = 0.

Problem 6:

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 7:

Phonons are quantized lattice vibrations, and many aspects of these excitations can be understood in terms of simple mode counting.
(a)  Estimate the number of phonon modes in 1 cm3 of a crystalline material with an inter-atomic spacing of 2 Angstrom.
(b)  Assuming that in thermal equilibrium each phonon mode has kBT of energy, give a numerical estimate of the heat capacity DE/DT of this 1 cm3 of material, in [J/K].

Solution:

Concepts, principles, relations that apply to the problem:
Normal modes of coupled oscillators
Why do they apply?
We are asked to estimate the number of normal modes of a large number of coupled harmonic oscillators.
How do they apply?
(a) We have N atoms, N coupled oscillators.  We therefore have 3N normal modes.  (We may want to neglect translation and rotation of the crystal as a whole, but since N is so large, it does not make any difference.)
N = n3, n = 10-2m/(2*10-10m) = 5*107.  N = 1.25*1023.
(b)  kB = 1.380658 ´ 10-23 J/K.
E = 3NkBT,  DE/DT = 3NkB = 5.19 J/K  for the 1 cm3 piece of material.
This is a reasonable number.  For water the specific heat is 4.186 J/(g K).
Details of the calculations:
None

Problem 8:

A pendulum is composed of two masses, 3m and m, and two strings of equal length L as shown.  

At t = 0 the system is released from rest with the upper (heavier) mass not displaced from its equilibrium position and the lower mass displaced to the right a distance a.

x2(0) = 0, x1(0) = a << L.

Find an expression for the subsequent motion of the lower mass, x1(t).

Solution

Concepts, principles, relations that apply to the problem:
Small, coupled oscillations; normal modes

Solutions of the form qj = Re(Ajeiwt) can be found.  We can find the w2 from det(kij - w2Tij) = 0.
Why do they apply?
We are asked to find the motion of the masses for small displacements.
How do they apply?
Let f1 and f2 be the generalized coordinates.  Then
x2 = Lsinf2, x1 = L(sinf1 + sinf2),
y2 = Lcosf1, y1 = L(cosf1 + cosf2).
For small displacements we have sinf = f, cosf = 1 - f2/2.
The kinetic energy of the system is
T = (3/2)m[(dx2/dt)2 + (dy2/dt)2] + ½m[(dx1/dt)2 + (dy1/dt)2]
Let us only keep terms up to second order in f and df/dt.  Then
(dx2/dt)2 + (dy2/dt)2 = L2(df2/dt)2,
(dx1/dt)2 + (dy1/dt)2 = L2[(df2/dt)2 + (df1/dt)2 + 2(df2/dt)(df2/dt)].
T = (3/2)mL2(df2/dt)2 + ½mL2[(df2/dt)2 +(df1/dt)2 + 2(df2/dt)(df2/dt)].
The potential energy of the system is
U = -mgy1 - 3mgy2  = (mgL/2)(f12 + 4f22) + constant.
We have:
.

T11 = mL2,  T22 = 4mL2, T12 = T21 = mL2,
k11 = mgL, k22 = 4mgL, kij(i ¹ j) = 0.
,
w2 = (g/L)(8 ± 4)/6, wa2 = (g/L)2, wb2 = (g/L)2/3.

For mode a:  f2 = -f1/2
For mode b:  f2 = f1/2
Most general motion for a system starting from rest:
f1 = Aexp(iwat) + Bexp(iwbt),  f2 = -(A/2)exp(iwat) + (B/2)exp(iwbt).
x1 = L(A/2)exp(iwat) + L(3B/2)exp(iwbt),  x2 = -L(A/2)exp(iwat) + L(B/2)exp(iwbt),
x1 = A’exp(iwat) + 3B’exp(iwbt),  x2 = -A’exp(iwat) + B’exp(iwbt).
Initial conditions:  A’ + 3 B’ = a,  -A’ + B’ = 0  à  A’ = B’ = a/4.
Subsequent motion of the lower mass: x1(t) = (a/4)cos wat + (3a/4)cos wbt.
Details of the calculation:
None

Problem 9:

(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 10:

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 = 0.5 m 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