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 523-Hz source of sound. The microphone vibrates up and down in simple harmonic motion with a period of 2.72 s. The difference between the maximum and minimum sound frequencies detected by the microphone is 2.43 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

given

f = 523 m/s

speed of sound, v = 343 m/s

let vo is the maximum speed of microphone at mean position,

so,

fmax = f*(v + vo)/v

fmin = f*(v - vo)/v

given, fmax - fmin = 2.43 hz

f*(v + vo)/v - f*(v - vo)/v = 2.43

2*f*vo/v = 2.43

vo = 2.43*v/(2*f)

= 2.43*343/(2*523)

= 0.797 m/s

angular frequency of motion of microphone, w = 2*pi/T

= 2*pi/2.72

= 2.31 rad/s

now Apply, vo = A*w

==> A = vo/w

= 0.797/2.31

= 0.345 m <<<<<<<-----------------Answer

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