A microphone is attached to a spring that is suspended from the ceiling, as the

Question

A microphone is attached to a spring that is suspended from the ceiling, as the drawing indicates. Directly below on the floor is a stationary 577-Hz source of sound. The microphone vibrates up and down in simple harmonic motion with a period of 1.73 s. The difference between the maximum and minimum sound frequencies detected by the microphone is 1.78 Hz. Ignoring any reflections of sound in the room and using 343 m/s for the speed of sound, determine the amplitude (in m) of the simple harmonic motion.?

Explanation / Answer

   The microphone performs an harmonic oscillatory motion like
x = A·cos[(2/T)·t]  
A amplitude T period
So the velocity of the microphone is
v = dx/dt = -A·(2/T)·sin[(2/T)·t]
Hence the maximum/minimum of velocity of the microphone is:
v = ± A·(2/T) =  

Due to Doppler effect frequency observed by the moving microphone shifts to:
f' = f·(1 + v/c)
f frequency at sound source c speed of sound

Using max/min velocity of the microphone you find max/min frequency observed:

f'_max = f·(1 + A·(2/T)/c)
f'_min = f·(1 - A·(2/T)/c)

Therefore
f' = f'_max - f'_min
<=>
f' = 2·f·A·(2/T)/c
<=>
f' = 4··f·A/(T·c)

Solve for A
A = f'·T·c / (4··f)
= 1.78Hz · 1.73s · 343m/s / (4 · · 577.Hz)
= 0.146m

= 14.6cm

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