Question

A microphone is attaced to a spring that is suspended from

Answer 1

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

$f_{max} - f_{min} = 2 f_s (\frac{v_0}{v})$

Here,

$f_s$= Source Frequency

$v_0$ = Speed of microphone

v = Speed sound

$f_{max} - f_{min} = 2 f_s (\frac{v_0}{v})$

Maximum speed of the microphone is

$v_0 = f_{max} - f_{min} * \frac{v}{2f_s}$

$v_0= \frac{2.1Hz*343m/s}{2*440Hz}$

$v_0= 0.8185 m/s$

Now the amplitude is

$A = \frac{v_0}{2\pi/T}$

Here T means the Period, then

$A= \frac{0.8185 * 2.0}{2\pi}$

A= 0.2605m

Therefore the amplitude of the simple harmonic motion is 0.2605m