Problem 1:
A rope has a mass of 2 kg and a length of 10 m. It is stretched with a tension of 50 N and fixed at both ends. What is the frequency of the first harmonic on this rope?
Solution:
 | Concepts, principles, relations
that apply to the problem: Standing waves on a string |
| Why do they apply? We are asked to find the frequency of a standing wave on a string. |
| How do they apply? An integer number of half-wavelengths have to fit into the length L of the string. nl/2 = L, l = 2L/n. fn = nv/(2L) = nf1. v = (F/m)1/2, F = tension in the string, m = mass per unit length. m = m/L = 0.2 kg/m, v2 = F/m = 50N/(0.2 kg/m) = 250(m/s)2, v = 15.8m/s. The wavelength of the first harmonic is 20m. The frequency f is f = v/l = (15.8/20)Hz = 0.79Hz. |
| Details of
the calculations: None |
Problem 2:
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”. If the amplitude of the damped oscillator decreases to 1/e of its initial value after n periods, show that the frequency of the oscillator must be approximately [ 1 - (8p2n2)-1] times the frequency of the corresponding undamped oscillator.
Solution:
| Concepts, principles, relations
that apply to the problem: The underdamped harmonic oscillator |
| Why do they apply? The oscillator is underdamped, since it crosses the equilibrium position many times. |
| How do they apply? 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. |
| Details of the calculation: After n periods, nT = 2nT1, the amplitude of the oscillator decreases to A/e. So exp(-1) = exp(-bnT), bnT = 1, b = 1/nT. With T = 2p/w1 we have b = w1/n2p. w12 = w02 – (w1/n2p.)2, w12 + w12/(4n2p2) = w02. w1 = w0 [1 + 1/(4n2p2)]–˝ . (1 + x)-˝ = 1 – ˝ x + … . w1 @ w0 [1 - 1/(8n2p2)] . |
Problem 3:
When the system shown in the diagram is in equilibrium, the right spring is stretched by x1. The coefficient of static friction between the blocks is µs. There is no friction between the bottom block and the supporting surface. The force constants of the springs are k and 3k (see the diagram). The blocks have equal mass m. Find the maximum amplitude of the oscillations of the system shown in the diagram that does not allow the top block to slide on the bottom.
Solution:
| Concepts, principles, relations
that apply to the problem: Simple harmonic motion |
| Why do they apply? As long as the blocks stick together, they will execute simple harmonic motion about an equilibrium position. For a displacement x from the equilibrium position, the force on each block then will obey Hooke's law, F = -ax. We can find the equilibrium position and a, since we know the spring constants. But the force on each block is also the sum of the force exerted on the block by the spring it is attached to and the frictional force. Equating the two expressions for the force we can find an expression for the frictional force f, and setting if equal to its maximum value we can find the maximum allowed amplitude for the blocks to not start sliding with respect to each other. |
| How do they apply? Let displacements to the right be positive and displacements to the left be negative. In equilibrium: Force on upper block = -3kx2 + f = 0 Force on lower block = -kx1 - f = 0 ŕ x2 = -x1/3 (Note: x1 is negative, x2 is positive.) The left spring is stretched by x1/3. Assume
the masses are stuck together. What is the equation of motion? But we also have for the upper block F = -3k(x + x2)
+ f. |
| Details of the calculation: None |
Problem 4:
A train has a whistle, which emits a 400 Hz sound. You are stationary and you hear the whistle, but the pitch is 440 Hz. How fast is train moving towards or away from you?
Solution:
| Concepts, principles, relations that apply to the
problem: The Doppler shift |
| Why do they apply? The frequency of the train's whistle is shifted, because the train is moving with respect to the observer. |
| How do they apply? The pitch is higher, so the train is moving towards you. The train is a moving source. Its speed relative to you is found from f = f0v/(v-vs). We have (v-vs) = f0v/f = (400/s)(340 m/s)/(440/s) = 309m/s. Therefore vs = 340m/s – 309m/s = 31m/s = 69mph. |
| Details
of the calculations: None |
Problem 5:
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 1cm3 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 6:
Use Lagrange's equations to find the normal modes and normal frequencies for linear vibrations of the CO2 molecule shown below.
Solution:
| Concepts, principles, relations that apply to the
problem: Small 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 modes of the system. | |||||||||
| How do they apply? We are only considering linear vibrations. Let qi denote the displacement of the ith particle from its equilibrium position. qi = xi - xi0. Then T = (1/2)[mo(dq1/dt)2 + mc(dq2/dt)2 + mo(dq3/dt)2]. U = (1/2)k1[(q2 - q1)2 + (q3 - q2)2] + (1/2)k2[(q3 - q1)2]. U = (1/2)(k1 + k2)q12 + (1/2)(k1 + k2)q32 + k1q22 - (1/2)k1(q2q1 + q1q2 + q3q2 + q2q3) - (1/2)k2(q3q1 + q1q3). T11 = T33 = mo, T22 = mc, Tij(i ą j) = 0. k11 = k33 = k1 + k2, k22 = 2k1, k12 = k21 = k32 = k23 = -k1, k31 = k13 = -k2. | |||||||||
| Details
of the calculations: det(kij - w2Tij) = 0.  .w2 = 0, or  .To find the relative displacements we solve the system of equations  for each w.
|
Problem 7:
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. |