Question

A source of sound $A$ emitting waves of frequency $1800Hz$ is falling towards ground with a terminal speed $v$. The observer $B$ on the ground directly beneath the source receives waves of frequency $2150Hz$. The source A receives waves, reflected from ground, of frequency nearly : (Speed of sound = $343m/s$)

• A. 2150 Hz
• B. 2500 Hz
• C. 1800 Hz
• D. 2400 Hz

Answer 1

Answer: (B)

Given $f_A=1800Hz$
$v_t=v$
$f_B=2150Hz$
Reflected wave frequency received by $A,f_A^{\prime }=?$
Applying doppler’s effect of sound, $f^{\prime }=\frac{v_{s}f}{v_s-v_t}$
Here, $v_t=v_s\left(1-\frac{f_A}{f_B}\right)=343\left(1-\frac{1800}{2150}\right)$
$v_t=55.8372m/s$
Now, for the reflected wave,
$\therefore f_A^{\prime }=\left(\frac{v_s+v_t}{v_s-v_t}\right)f_A$
$=\left(\frac{343+55.83}{343-55.83}\right)\times 1800$
$=2499.44\approx 2500Hz$