JEE Main 2023 — Waves and Sound Question with Solution

The formula to calculate the frequency $\left(f\text{'}\right)$ as heard by an observer with respect to the frequency $\left(f\right)$ produced from a source is given by

$f\text{'}= \frac{v\pm v_o}{v\mp v_s}f ....\left(1\right)$

where, $v$ is the speed of sound in air, $v_o$ is the speed of the observer and $v_s$ is the speed of the source.

In the first situation, the source is the moving car and the observer is the wall. Hence, the frequency $\left(f_w\right)$ as heard on the wall can be written as 

$\begin{array}{rcl}{f}_w& =& \frac{v-0}{v-v_c}f\\ & =& \frac{v}{v-v_c}f ....\left(2\right)\end{array}$

where, $v_c$ is the speed of the car.

In the second situation, when the sound reflects back from the wall, the source of sound is the wall and the observer is the car.

Hence, the frequency $\left(f_c\right)$ of sound as heard by the car after the reflection is given by

$\begin{array}{rcl}{f}_c& =& \frac{v+v_c}{v-0}{f}_w\\ & =& \frac{v+v_c}{v}{f}_w ....\left(3\right)\end{array}$

From equation (2) and (3), it follows that

$\begin{array}{rcl}{f}_c& =& \frac{v+v_c}{v}\times \frac{v}{v-v_c}f\\ & =& \frac{v+v_c}{v-v_c}f ....\left(4\right)\end{array}$

It is given in the problem that

$f_c-f= 40 .....\left(5\right)$

Substitute the expression from equation (4) into equation (5) and solve to calculate the frequency of the horn.

$\begin{array}{rcl}\frac{v+v_c}{v-v_c}f-f& =& 40\\ & \Rightarrow & \frac{2v_c}{v-v_c}f= 40\end{array}$

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