Question

Two sound sources \[{{S}_{1}}\] and \[{{S}_{2}}\] of frequencies 324 Hz and 320 Hz are placed at certain separation. An observer is moving away from \[{{S}_{1}}\] towards \[{{S}_{2}}\] online joining them. If he hears no beats then speed of observer is \[\left( v=344\dfrac{m}{s} \right)\] :
 \[A.\,20\dfrac{m}{s}\]
 \[B.\,10\dfrac{m}{s}\]
 \[C.\,50\dfrac{m}{s}\]
 \[D.\,2.1\dfrac{m}{s}\]

Answer 1

Hint: The formula for calculating the apparent frequency of the sound, when the observer moves away and towards the source should be used. Both these equations should be equated to find the value of the speed of the observer when he hears no sound.

Formula used:
 \[f={{f}_{0}}\left[ \dfrac{v-{{v}_{0}}}{v} \right]\]
 \[f={{f}_{0}}\left[ \dfrac{v+{{v}_{0}}}{v} \right]\]

Complete step-by-step answer:
From the data, we have the data as follows.
Two sound sources \[{{S}_{1}}\] and \[{{S}_{2}}\] of frequencies 324 Hz and 320 Hz are placed at certain separation.
The formulae used are:
 \[f={{f}_{0}}\left[ \dfrac{v-{{v}_{0}}}{v} \right]\]
 \[f={{f}_{0}}\left[ \dfrac{v+{{v}_{0}}}{v} \right]\]
Where f is the apparent frequency, \[{{f}_{0}}\] is the source frequency, v is the velocity of the sound and \[{{v}_{0}}\] is the velocity of the observer.
Here, we need to consider 2 cases. In one case, we will find the value of the frequency when the observer moves away from the sound source \[{{S}_{1}}\] . In the other case, we will find the value of the frequency when the observer moves towards the sound source \[{{S}_{2}}\] .
Let us consider the cases one by one.
Case I:
The value of the frequency when the observer moves away from the sound source \[{{S}_{1}}\] .
The formula for calculating the apparent frequency when the observer moves away from the source is given as follows.
 \[{{f}_{1}}=324\left[ \dfrac{344-{{v}_{0}}}{344} \right]\] …… (1)
Case II:
The value of the frequency when the observer moves towards the sound source \[{{S}_{2}}\] .
The formula for calculating the apparent frequency when the observer moves towards the source is given as follows.
 \[{{f}_{2}}=320\left[ \dfrac{344+{{v}_{0}}}{344} \right]\] …… (2)
As the observer hears no beats, so, we can equate the equations (1) and (2),
 \[\begin{align}
  & 324\left[ \dfrac{344-{{v}_{0}}}{344} \right]=320\left[ \dfrac{344+{{v}_{0}}}{344} \right] \\
 & \Rightarrow 324(344-{{v}_{0}})=320(344+{{v}_{0}}) \\
\end{align}\]
Continue the further calculation.
 \[\begin{align}
  & 111456-324{{v}_{0}}=110080+320{{v}_{0}} \\
 & \Rightarrow 1376=644{{v}_{0}} \\
 & \Rightarrow {{v}_{0}}=2.13{m}/{s}\; \\
\end{align}\]
As the value of the speed of the observer when he hears no sound is \[2.13{m}/{s}\;\] , thus, the option (D) is correct.

So, the correct answer is “Option (D)”.

Note: The formulae are different for the different cases, that is, when the observer moves away from the source, then, the velocity of the observer should be subtracted and when the observer moves towards the source, then, the velocity of the observer should be added. The units of the parameters should be taken care of.