Q70P - University Physics with Modern Physics

Question

CP A police siren of frequency ${\mathrm{f}}_{\mathrm{siren}}$ is attached to a vibrating platform. The platform and siren oscillate up and down in simple harmonic motion with amplitude ${\mathbf{A}}_{\mathbf{P}}$and frequency ${\mathbf{f}}_{\mathbf{p}}$.(a) Find the maximum and minimum sound frequencies that you would hear at a position directly above the siren. (b) At what point in the motion of the platform is the maximum frequency heard? The minimum frequency? Explain.

Step-by-Step Solution

Answer

A) The maximum and minifmum frequency are $f_{L\max }=f_{siren}\left(\frac{v}{v-2\pi fA}\right)$  and $f_{L\min }=f_{siren}\left(\frac{v}{v+2\pi fA}\right)$.

B) frequency is maximum at mean position and minimum at the edges.

STEP 1: Simple Harmonic Motion

For an oscillatory motion, if the restoring force is directly proportional to the displacement of particle from its mean position. Then the motion is called simple harmonic motion.

STEP 2: The Maximum And Minimum Frequency

The maximum velocity of a system undergoing simple harmonic motion is  $v_{\max }=2\pi fA$

 

where $v_{\max }$ is the maximum speed, f is the frequency and A is amplitude.

Frequency heard by the listener, directly over the siren, is

$f_L=f_s\left(\frac{v}{v\mp v_{\max }}\right)$

Now, $f_L$ will be maximum when the sign in the denominator is minus, which leads to 

$f_{L\max }=f_{siren}\left(\frac{v}{v-2\pi fA}\right)$

On the other hand,$f_L$  will be minimum when the sign in the denominator is plus, meaning a larger denominator, which leads to

$f_{L\min }=f_{siren}\left(\frac{v}{v+2\pi fA}\right)$

Therefore, the maximum frequency is detected when the platform reaches its maximum velocity and moving towards the listener, which happens when the platform is passing through the equilibrium position and moving up. In the same way, the minimum frequency is detected when the platform is at its edges, moving down.