Question

A stationary observer receivers a sound of frequency 2000 Hz. The variation of apparent frequency and time is shown. Find the speed of source ; if velocity of sound is 300 m/s.

• A. 66.6 m/s
• B. 33.3 m/s
• C. 27.3 m/s
• D. 59.3 m/s

Answer 1

The correct option is B 33.3 m/s

Given:

Let the velocity of the observer be $V_o$ .

Given that the observer is at rest.

So, $V_o=0$

Let the velocity of source be $V_s$

Observer receives a sound of frequency, $f=2000Hz$

Velocity of the sound is, $v=300m/s$

To find:

The speed of source, $v_s$

Solution:

From the formula,

$f_m=\frac{(v+v_0)f}{v-v_s}$

Where,

$f_m$ is the maximum frequency

$f$ is the original frequency

From the figure we can write,

$f_m$ $=2250Hz$ (maximum frequency)

So, on substituting in the formula we get,

$2250=\frac{(300+0)2000}{300-v_s}$

$\frac{2250}{2000}=\frac{300}{300-v_s}$

$300-v_s=\frac{300}{1.125}$

$v_s=33.3m/s$

so, the correct answer is option (B)