Question

One train is approaching an observer at rest and another is receding him with the same velocity 4 m/s. Both the trains blow whistles of the same frequency of 243 Hz. The beat frequency in Hz as heard by the observer is: (Speed of sound in air = 320 m/s)
A. 10
B. 6
C. 4
D. 1

Answer 1

Hint:The beat frequency is the difference between the apparent frequency when the source is approaching and the apparent frequency when the source is leaving the stationary observer. Determine the apparent frequency heard by the observer in the both cases using Doppler’s equations. Take the difference of the apparent frequencies to determine the beat frequency.

Formula used:
Doppler’s formula when the source is moving and observer is stationary,
\[f' = f\left( {\dfrac{v}{{v \mp {v_s}}}} \right)\]
Here, v is the speed of source and \[{v_s}\] is the speed of sound.
The negative sign is for the source is approaching the stationary observer and the positive sign is for the source is leaving the stationary observer.

Complete step by step answer:
We have the frequency of sound emitted by both the trains is \[f = 243\,{\text{Hz}}\] and the velocity of both the trains is \[v = 4\,{\text{m/s}}\]. We can use Doppler’s equation to determine the apparent frequency heard by the observer when the first train is approaching the observer. We have the expression for the apparent frequency when the source is approaching a stationary observer,
\[{f_1} = f\left( {\dfrac{v}{{v - {v_s}}}} \right)\] …… (1)
Here, f is the original frequency, v is the velocity of the source (train) and \[{v_s}\] is the velocity of sound.

We also have the expression for the apparent frequency when the source is leaving a stationary observer,
\[{f_2} = f\left( {\dfrac{v}{{v + {v_s}}}} \right)\] …… (2)
We know that the beat frequency is the difference between the apparent frequency when the source is approaching and the apparent frequency when the source is leaving the stationary observer.
\[{f_b} = {f_1} - {f_2}\]
Using equation (1) and (2) in the above equation, we get,
\[{f_b} = f\left( {\dfrac{v}{{v - {v_s}}}} \right) - f\left( {\dfrac{v}{{v + {v_s}}}} \right)\]
\[ \Rightarrow {f_b} = fv\left( {\dfrac{1}{{v - {v_s}}} - \dfrac{1}{{v + {v_s}}}} \right)\]
\[ \Rightarrow {f_b} = \left( {\dfrac{{2v\,{v_s}}}{{v_s^2 - {v^2}}}} \right)f\]

Substituting \[f = 243\,{\text{Hz}}\], \[v = 4\,{\text{m/s}}\] and\[{v_s} = 320\,{\text{m/s}}\] in the above equation, we get,
\[{f_b} = \left( {\dfrac{{2\left( 4 \right)\,\left( {320} \right)}}{{{{\left( {320} \right)}^2} - {{\left( 4 \right)}^2}}}} \right)\left( {243} \right)\]
\[ \Rightarrow {f_b} = \left( {\dfrac{{2560}}{{102384}}} \right)\left( {243} \right)\]
\[ \therefore {f_b} = 6\]
Therefore, the beat frequency is 6 Hz.

So, the correct answer is option B.

Note: Always remember, when the source approaches a stationary observer, the apparent frequency heard by the observer increases and therefore, in the expression, \[{f_1} = f\left( {\dfrac{v}{{v - {v_s}}}} \right)\], the denominator is less than the numerator. When the source leaves the stationary observer, the apparent frequency decreases and hence, \[{f_2} = f\left( {\dfrac{v}{{v + {v_s}}}} \right)\], the denominator is greater than the numerator. The Doppler equations take different forms when both source and observer moves relative to each other and students must remember all the formulae regarding these cases.