Question

A source S of sonic oscillations of frequency 2200 Hz and a receiver R are located on the x-axis. At the moment $t=0$ the source starts moving away from the receiver with an acceleration $1.7m/{\sec }^2$. The velocity of sound in air is 340 $m/\sec$. The sound is received by this receiver after 21 seconds. What is the frequency recorded by the receiver at this time?

• A. 2000 Hz
• B. I500 Hz
• C. 1750 Hz
• D. 2200 Hz

Answer 1

Answer: (A)

2000 Hz

The source starts from R and reaches P after 21 seconds.

Let T be the total time taken by source to go to P( $t_1$) from R and the sound to go from P to R($t_2$)

Hence, $T=t_1+t_2=21$ as given

$RP=\frac{1}{2}a{t}^2$ $u=0$since the source starts from rest at R.

PR is also equal to the distance covered by sound wave in time $t_2$.

$\therefore \frac{1}{2}a{t}_1^2=v_{sound}\times t_2=v_{sound}\times (21-t_1)$

$\therefore \frac{1}{2}\times 1.7t_1^2=340\times (21-t_1)$

$\Rightarrow t_1^2+400t_1-8400=0$

$\Rightarrow (t_1-20)(t_1+420)=0$

$\Rightarrow t_1=20$

So the source travels for 20 seconds from receiver and attained a velocity

$v=u+at=0+1.7\times 20=34m/s$

The source is moving away from the point of detection R.

$\therefore f^{\prime }=f\times \frac{v}{v+v_{source}}=2200\times \frac{340}{340+34}=2200\times \frac{10}{11}=2000Hz$