Question

A person with a vibrating tuning fork of frequency $338Hz$ is moving towards a vertical wall with a speed of $2m{s^{ - 1}}$. Velocity of sound in air is $340m{s^{ - 1}}$. The number of beats heard by that person per second is
A)$2$
B)$4$
C)$6$
D)$8$

Answer 1

Hint This question involves Doppler effect because of the relative motion that is present between the source of sound wave and the observer. The relation between initial frequency and changed frequency is
$f' = \left( {\dfrac{{v + {v_0}}}{{v - {v_s}}}} \right)f$
Where $f'$ is the changed frequency
$f$ is the initial frequency
$v$ is the velocity of sound wave
${v_0}$ is the velocity of observer
${v_s}$ is the velocity of source.
We have to solve this question with the same formula in two parts. First case will be when a sound wave travels from the tuning fork to the wall. Second case will be when a sound wave will reach the man after striking the wall.

Complete step-by-step solution:
First of all, we will consider the case when sound from the tuning fork will strike the wall. We need to first determine the source and the receiver or the observer and know their velocities. The sound wave is generated by the tuning fork, so it is the source. The man carrying the tuning fork is moving with the velocity $2m{s^{ - 1}}$. So velocity of the source,
${v_s} = 2m{s^{ - 1}}$
In this case, the sound waves are received by the wall, so it's the observer. We can then say that the velocity of the observer, ${v_0} = 0m{s^{ - 1}}$.
Since, the tuning fork is the source, so, frequency of the sound wave will be the frequency of tuning fork. So,
$f = 338Hz$
We know that velocity of sound in air is $340m{s^{ - 1}}$.
$v = 340m{s^{ - 1}}$
The phenomena that we will observe here is the Doppler effect. Doppler effect is the change in the frequency of a wave in relation to an observer who is moving relative to the wave source.
The mathematical expression for the same is
$f' = \left( {\dfrac{{v + {v_0}}}{{v - {v_s}}}} \right)f$
Where $f'$ is the changed frequency
$f$ is the initial frequency
Substituting all the values in this relation, we get the frequency received by the wall, say, ${f_1}$ as
$
  {f_1} = \left( {\dfrac{{340 + 0}}{{340 - 2}}} \right) \times 338 \\
   = \dfrac{{340}}{{338}} \times 338 \\
   = 340Hz \\
 $
Now we will consider the second case where the sound wave will reflect off the wall and reach the man. Now the source is the wall with velocity ${v_s} = 0m{s^{ - 1}}$ and receiver or observer is the man with velocity ${v_0} = 2m{s^{ - 1}}$. The frequency of the sound wave will be ${f_1} = 340Hz$. Keeping these values in the relation to get the final frequency, ${f_2}$:
$
  {f_2} = \left( {\dfrac{{v + {v_0}}}{{v - {v_s}}}} \right) \times {f_1} \\
   = \left( {\dfrac{{340 + 2}}{{340 - 0}}} \right) \times 340 \\
   = 342Hz \\
 $
So, the final frequency is $342Hz$.
The initial frequency, that is the frequency of the tuning fork is $338Hz$.
Number of beats heard by the man will be the difference of these two frequencies.
Number of beats$ = 342 - 338 = 4beats/\sec $
So, option B is correct.

Note:-
Beats can only be calculated when the given frequencies have similar values like in the given case. When the difference in the frequencies is large, beats are not perceived by the human ear. Not much conceptual knowledge but a good practice is required so as to visualise such questions.