Question

A stationary observer receives a sound from a sound of frequency $v_0$ moving with a constant velocity $v_s=30m/s.$ The apparent frequency varies with time as shown in figure. Velocity of sound v = 300 m/s. Then which of the following is incorrect?

• A. The minimum value of apparent frequency is 889 Hz.
• B. The natural frequency of source is 1000 Hz.
• C. The frequency-time curve corresponds to a source moving at an angle to the stationary observer.
• D. The maximum value of apparent frequency is 1111 Hz.

Answer 1

The correct option is A The minimum value of apparent frequency is 889 Hz.

This frequency–time curve corresponds to a source moving at an angle to a stationary observer.

In the region SN, the source is moving towards the observer, i.e., the apparent frequency
$n^{\prime }=n_0\left(\frac{v}{v-v_{s}cos\theta }\right)$
$n^{\prime }=n_0\left(\frac{300}{300-30cos\theta }\right)$
$When\theta =\frac{\pi }{2}.$ i.e., at N,
$n^{\prime }=n_0=1000Hz,$ i.e., natural frequency of source. In the region NS’ the source is moving away from the observer, i.e., apparent frequency
$n^{\prime }=n_0\left(\frac{300}{300-30cos\theta }\right)$
$When\theta =0,i.e.,cos\theta =1,$
$n_{max}=n_0\frac{v}{v-v_s}=\frac{\left(1000Hz\right)\left(300m/s\right)}{(300m/s-30m/s)}$
$=\frac{10}{9}\times 1000Hz=1111Hz$
$n_{min}=n_0\frac{v}{v+v_s}=\frac{1000\times 300}{330}=909Hz$