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A string of length 2 m fixed at both ends vibrates in its 7th overtone. The standing wave is given by y = A sin(kx) cos(omega*t + pi/3), with A = 1 cm and omega = 100*pi rad/s. If t0 is the time (after t = 0) at which the total energy of vibration is completely potential energy for the second time, find 120*t0. (Use pi² = 10.)

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

For a standing wave y = A sin(kx) cos(omega*t + phi), all energy is potential (KE = 0) when the velocity dy/dt = 0 everywhere, i.e., sin(omega*t + phi) = 0, which means cos(omega*t + phi) = +/-1. With phi = pi/3 and omega = 100*pi: omega*t + pi/3 = n*pi → t = (n*pi - pi/3)/(100*pi) = (n - 1/3)/100. For n=1: t = (2/3)/100 = 1/150 s (first time). For n=2: t0 = (5/3)/100 = 1/60 s (second time). 120*t0 = 120*(1/60) = 2.

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