Question

A stationary observer receives a sound of frequency $f_0=2000Hz$ when the source is at rest. When the source starts moving with a constant velocity along a line from starting from a large distance, the apparent frequency f varies with time as shown in figure. Speed of sound =300 m/s. Choose the CORRECT alternative(s):-

• A. Speed of source is 66.7 m/s
• B. $f_m$ shown in figure can be 2500 Hz
• C. Speed of source is 33.33 m/s
• D. $f_m$ shown in figure cannot be greater than 2250 Hz

Answer 1

Answer: (C), (D)

$f_m$ shown in figure cannot be greater than 2250 Hz

The graph shows the situation shown in figure below. The observed frequency will initially be more than the natural frequency. When the source is at P, observed frequency is equal to its natural frequency i.e., 2000 Hz.

For region AP: $f=f_0\left(\frac{V}{V-V_{s}cos\theta }\right)$

For PB: $f=f_0\left(\frac{V}{V+V_{s}cos\theta }\right)$
Minimum value of f will be:
$f_{min}=f_0\left(\frac{V}{V+V_s}\right)\text{when}\cos \theta =1$
or $1800=2000\left(\frac{300}{300+V_s}\right)$
Solving this we get, $V_s$=33.33 m/s and maximum value of f can be $f_{max}=2000\left(\frac{300}{300-33.33}\right)$
$f_{max}=2250Hz$