Problem 1:
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.
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(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. |
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(b) A motor now supplies an external driving term M coswt. Determine the stationary solution for the now driven and damped oscillator. |
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(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 2:
Two particles of mass m and one particle of mass M are constrained to move on a line as shown. They are connected by massless springs with spring constant k.
Find the normal modes and eigenfrequencies of the system, keeping M/m finite.
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? The kinetic energy is T = (1/2)[ m(dx1/dt)2 + M(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 = M, 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:  |
| Details of the
calculation: (k-w2m)2(2k-w2M) - 2k2(k-w 2m) = 0. Solution 1: k-w 2m = 0, w = (k/m)1/2. If k-w 2m ¹ 0, then (k-w2m)(2k-w2M) = 2k2. w2(w2mM-kM-2km) = 0. Solution 2: w = 0. Solution 3: w = (k/m + 2k/M)1/2. 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 = (k/m + 2k/M)1/2. The displacements Equation 1 yields –(2m/M)A1 = A2. Equation 3 yields –(2m/M)A3 = A2. A1 = A3 = -(M/(2m))A2, the central mass move in a direction opposite to the direction of the outer masses. |
