Question

A car is moving towards a huge wall with a speed $\dfrac{c}{{10}}$, where $c = $ speed of sound in still air. A wind is also blowing parallel to the velocity of the car in the same direction and with the same speed. If the car sounds a horn of frequency $f$, then the frequency of the reflected sound of the horn heard by the driver of the car is $\dfrac{{xf}}{9}$. Find the value of $x$.

Answer 1

Hint: In this problem, we need to first find the frequency of the sound of the horn as perceived by the wall by using the Doppler Effect formula and then by considering the wall as a source, the frequency of the sound as heard by the driver. In both the cases the velocity of the wall will be $\dfrac{c}{{10}}$ as the car is in rest in the wind frame and the wall is moving in the opposite direction with the same velocity.

Formula Used In this solution we will be using the following formula,
$f' = \dfrac{{v + {v_o}}}{{v - {v_s}}}f$
where $f'$ is the modified frequency and $f$ is the real frequency
$v$ is the velocity of the sound waves,
${v_o}$ is the velocity of the observer and ${v_s}$ is the velocity of the source.

Complete Step by Step Solution:
It is said in the problem that the car and the wind are moving in the same direction, that is towards the wall which is stationary, with the same velocity $\dfrac{c}{{10}}$. So in the rest frame of the wind, the car will be at rest and the wall will be moving towards the car with the velocity of $\dfrac{c}{{10}}$.
Hence we have the wall as the observer and the car as the source, which is at rest. The velocity of the sound waves in air is given to be $c$.
Therefore we can use the formula for the Doppler Effect. It is given by,
$f' = \dfrac{{v + {v_o}}}{{v - {v_s}}}f$
Now in the first case when the sound wave travels from the car to the wall we have,
$f$ is the real frequency,
${v_o} = \dfrac{c}{{10}}$ , ${v_s} = 0$ and $v = c$
Therefore, substituting we get,
$f' = \dfrac{{c + \dfrac{c}{{10}}}}{{c - 0}}f$
On taking LCM in the numerator as 10,
$f' = \dfrac{{10c + c}}{{10c}}f$
The $c$ gets cancelled and we have,
$\Rightarrow f' = \dfrac{{11f}}{{10}}$
Now in the second case, the sound gets reflected from the wall and reaches the driver. So the wall is the source and driver is the observer.
Hence $f' = \dfrac{{11f}}{{10}}$, ${v_s} = \dfrac{c}{{10}}$ and ${v_o} = 0$
Substituting these values again in the same formula, $f'' = \dfrac{{v + {v_o}}}{{v - {v_s}}}f'$ we have
\[\Rightarrow f'' = \left( {\dfrac{{c + 0}}{{c - \dfrac{c}{{10}}}}} \right)\dfrac{{11f}}{{10}}\]
On doing the LCM in the denominator as 10,
\[\Rightarrow f'' = \left( {\dfrac{{10c}}{{9c}}} \right)\dfrac{{11f}}{{10}}\]
Cancelling the $c$ and 10 we get,
\[\Rightarrow f'' = \dfrac{{11f}}{9}\]
In the question we are given that the frequency of the reflected sound is, $\dfrac{{xf}}{9}$. So comparing $\dfrac{{xf}}{9}$ with \[\dfrac{{11f}}{9}\] we see that the value of the variable $x$ is 11.

So the answer is $x = 11$.

Note When both the source of a sound wave and the observer that hears the sound are moving relative to each other, then the frequency of the sound observer by the observer is different from the frequency of the source. This is called the Doppler Effect. However this difference is not because the frequency of the sound wave changes. This Doppler Effect can be observed in both sound and light waves.