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A microphone is attaced to a spring that is suspended from
theceiling. Directly below on the floor is a stationary 440
Hzsource of sound. The microphone vibrates up and down insimple
harmonic motiono with a period of 2.0 s. The
differnce between the maximum and minimum sound frequenciesdetected
by the microphone is 2.1 Hz. Ignorign anyreflections of sound in
the room and using 343 m/s for the speed ofsound, determine the
amplitude of the simple harmonic motion.

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cjmejiab

To solve this problem we will apply the concept related to the amplitude and the Doppler effect. The difference between the maximum and minimum frequency detected by the microphone would be given by the mathematical function

[tex]f_{max} - f_{min} = 2 f_s (\frac{v_0}{v})[/tex]

Here,

[tex]f_s[/tex]= Source Frequency

[tex]v_0[/tex] = Speed of microphone

v = Speed sound

[tex]f_{max} - f_{min} = 2 f_s (\frac{v_0}{v})[/tex]

Maximum speed of the microphone is

[tex]v_0 = f_{max} - f_{min} * \frac{v}{2f_s}[/tex]

[tex]v_0= \frac{2.1Hz*343m/s}{2*440Hz}[/tex]

[tex]v_0= 0.8185 m/s[/tex]

Now the amplitude is

[tex]A = \frac{v_0}{2\pi/T}[/tex]

Here T means the Period, then

[tex]A= \frac{0.8185 * 2.0}{2\pi}[/tex]

A= 0.2605m

Therefore the amplitude of the simple harmonic motion is 0.2605m

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