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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 440-Hz source of sound. The microphone vibrates up and down in simple harmonic motion with a period of 2.0 s. The difference between the maximum and minimum sound frequencies detected by the microphone is 2.1 Hz. Ignoring any reflections of sound in the room and using 343 m /s for the speed of sound, determine the amplitude of the simple harmonic motion.

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Answer:

[tex]0.261\ \text{m}[/tex]

Explanation:

[tex]\Delta f[/tex] = Change in frequency = 2.1 Hz

[tex]f[/tex] = Frequency of source of sound = 440 Hz

[tex]v_m[/tex]= Maximum of the microphone

[tex]v[/tex] = Speed of sound = 343 m/s

[tex]T[/tex] = Time period = 2 s

We have the relation

[tex]\Delta f=2f\dfrac{v_m}{v}\\\Rightarrow v_m=\dfrac{\Delta fv}{2f}\\\Rightarrow v_m=\dfrac{2.1\times 343}{2\times 440}\\\Rightarrow v_m=0.8185\ \text{m/s}[/tex]

Amplitude is given by

[tex]A=\dfrac{v_m T}{2\pi}\\\Rightarrow A=\dfrac{0.8185\times 2}{2\pi}\\\Rightarrow A=0.261\ \text{m}[/tex]

The amplitude of the simple harmonic motion is [tex]0.261\ \text{m}[/tex].

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