Question

A sound wave of frequency f travels horizontally to the right. It is reflected from a large vertical plane surface moving to left with a speed v. The speed of sound in medium is c. Then.
A. The number of waves striking the surface per second is $f\dfrac{\left( c+v \right)}{c}$.
B. The wavelength of reflected wave is $\dfrac{c\left( c+v \right)}{f\left( c+v \right)}$
C. The frequency of reflected wave is $f\dfrac{\left( c-v \right)}{\left( c-v \right)}$
D. The number of beats heard by a stationary listener to the left to the reflecting surface is $\dfrac{vf}{c-v}$

Answer 1

We are given with a source emitting frequency wave f and vertical plan moving towards the observer. After that we will check and calculate all the options for solving the above given question. We have to keep in mind that the source is at rest and the plane is moving towards the observer.

Complete step by step answer:
Firstly we will check the first option. Number of waves striking means frequency of the wave striking.
Frequency of the waves striking $={{n}^{1}}=\dfrac{c+v}{\lambda }=\dfrac{f\left( c+v \right)}{c}$
This is the correct option for the given question.
Then we will check the second option
Frequency of the waves reflecting$\begin{aligned} & ={{n}^{2}}=\dfrac{c}{c-v}{{n}^{1}}=f\dfrac{\left( c+v \right)}{\left( c-v \right)} \\\ & \\\ \end{aligned}$
It does not match with the option given hence, it is the wrong option.
Number of waves encountered by the moving plane per unit time
$\begin{aligned} & =\dfrac{\text{distance travelled}}{\text{wavelength}} \\\ & =\dfrac{c+v}{\lambda }=\dfrac{c}{\lambda }\left( 1+\dfrac{v}{c} \right)=f\left( 1+\dfrac{v}{c} \right) \\\ \end{aligned}$
f′ is the frequency of the incident wave and f'' is the frequency of the reflected wave that a stationary observer meets.
$\begin{aligned} & f''=\dfrac{f'}{\left( 1-\dfrac{v}{c} \right)}=\dfrac{f\left( 1+\dfrac{v}{c} \right)}{\left( 1-\dfrac{v}{c} \right)}=\dfrac{f\left( c+v \right)}{\left( c-v \right)} \\\ & \lambda ''=\dfrac{c}{f''}=\dfrac{c}{f}\dfrac{\left( c+v \right)}{\left( c-v \right)} \\\ \end{aligned}$
Beat frequency = f”= f
$\dfrac{f\left( 1+\dfrac{v}{c} \right)}{\left( 1-\dfrac{v}{c} \right)}-f=\dfrac{2vf}{c-v}$
Therefore the given option D is wrong.

So, the correct answer is “Option A”.

Additional Information: All waves can be described by their two basic related properties: their wavelength and frequency. Wavelength is the distance between two adjacent (next to each other) and identical parts of the wave, such as between two wave crests (peak). Frequency can be defined as the number of wave crests that pass a given point per second.

Note: We can say that if the source of waves is moving away from the observer, every wave is emitted from a position farther from the observer than the previous wave, so the striking time between successive waves is increased which results in reducing the frequency. The distance between successive wave fronts is increased.
It is the application of Doppler’s effect. Let us see how the Doppler effect is defined The Doppler effect is about how we perceive waves depending on how we're moving in relation to them.