Question

A wall is moving with a velocity $u$ and a source of sound moves with a velocity $\dfrac{u}{2}$ in the same direction as shown in the figure. Assuming that the sound travels with a velocity $10u$ , the ratio of incident sound wavelength on the wall to the reflected sound wavelength by the wall is equal to
(A) $9:11$
(B) $11:9$
(C) $4:5$
(D) $5:4$

Answer 1

Hint: - Doppler Effect can be considered as a property of the sound wave. It is a variation in sound wave frequency produced by the relative movement of either source, listener, or both. Use the equation of the Doppler Effect to find the ratio of incident sound wavelength on the wall to the reflected sound wavelength by the wall.
Formula used: The equation for the Doppler Effect is:
$f' = f\left( {\dfrac{{v \pm {v_o}}}{{v \pm {v_s}}}} \right)$
where $f'$ represents the frequency received by the observer and $f$ represents the original frequency of the source.

Complete step-by-step answer:
When a source of sound reaches us the pitch becomes higher and as it moves away from the listener, the pitch becomes lower. It is due to the change in the number of waves passing the listener, the pitch gets lower. It is due to the change in the number of waves passing over per unit area per unit time, i.e. due to the frequency. As someone reaches a stationary listener the sound waves compress which increases the frequency resulting in a higher pitch. Likewise, as the source moves away, the number of waves decreases as it spreads apart therefore frequency will decrease giving a lower pitch. Similarly, when the source and listener approach each other the frequency increases, and as they move apart the frequency decreases.
Let $\lambda '$ be the wavelength of the sound incident on the wall.
and $f$ be the source frequency of the sound.
$\lambda ' = \left( {\dfrac{{{v_{sound}} - {v_{source}}}}{f}} \right)$
On putting the given values in the above equation we get,
$\lambda ' = \dfrac{{10u - \dfrac{u}{2}}}{f} = \dfrac{{19u}}{{2f}}$ ............... $\left( 1 \right)$

Now let $\lambda ''$be the wavelength of the sound reflected by the wall.
And $f'$ be the frequency of the sound received by the wall.
$\lambda '' = \left( {\dfrac{{{v_{sound}} + {v_{wall}}}}{{f'}}} \right)$
On putting the given values in the above equation we get,
$\lambda '' = \left( {\dfrac{{10u + u}}{{f'}}} \right) = \left( {\dfrac{{11u}}{{f'}}} \right)$ ............. $\left( 2 \right)$
Considering the influencing factors such as the velocity of the source or the listener either it is approaching or going further away the equation of Doppler Effect can be written as
$f' = f\left( {\dfrac{{{v_{sound}} - {v_{wall}}}}{{{v_{sound}} - {v_{source}}}}} \right)$ ……….. $\left( 3 \right)$
As given in the question, the velocity of the wall is $u$ , the velocity of the source of sound is $\dfrac{u}{2}$ and the velocity is $10u$ .
Put the values in the equation $\left( 3 \right)$ we get,
$f' = f\left( {\dfrac{{10u - u}}{{10u - \dfrac{u}{2}}}} \right)$
$ \Rightarrow f' = f\left( {\dfrac{{9u}}{{\dfrac{{19u}}{2}}}} \right) = f\left( {\dfrac{{18}}{{19}}} \right)$
Put the value of $f'$ in equation $\left( 2 \right)$ we get,
$\lambda '' = \dfrac{{11u}}{{f'}}$
$ \Rightarrow \lambda '' = \dfrac{{11u}}{{\left( {\dfrac{{18f}}{{19}}} \right)}} = \dfrac{{209u}}{{18f}}$ ………….. $\left( 4 \right)$
As we know that, the wavelength is inversely proportional to the frequency
Using equation $\left( 1 \right)$ and $\left( 4 \right)$ the ratio of wavelengths will become,
$\dfrac{{\lambda '}}{{\lambda ''}} = \dfrac{{\left( {\dfrac{{19u}}{{2f}}} \right)}}{{\left( {\dfrac{{209u}}{{18f}}} \right)}}$
On further solving the equation we get,
$\dfrac{{\lambda '}}{{\lambda ''}} = \left( {\dfrac{{19u}}{{2f}}} \right) \times \left( {\dfrac{{18f}}{{209u}}} \right) = \dfrac{9}{{11}}$
Therefore, the ratio of incident sound’s wavelength on the wall to the reflected sound’s wavelength by the wall is $9:11$ .

So, the correct answer is option (A) $9:11$ .

Note: Since the speed of all the Electromagnetic waves is relatively very large as compared to the speed of source or listener in our day to day life, the apparent change in frequency is not easily noticed. Thus, the apparent frequency is almost equal to its original frequency.