Question

A train is approaching the platform with a speed of 4m/s. Another train is leaving the platform at the same speed. The velocity of sound is 320m/s. If both trains sound their whistles at frequency $230\, Hz$, the number of beats heard per second will be:
A. 10
B. 8
C. 7
D. 6

Answer 1

We can use the doppler effect to find the frequency of the sound of whistles heard for the approaching train and sound of the whistle of the train which is leaving. The number of beats can be found by finding the difference between both these frequencies.

Complete step by step answer:
It is given that the velocity of a train approaching platform is $4m/s$.
Another train leaving the platform has the same velocity $4m/s$.
The velocity of sound is given as $v = 320\,m/s$ .
The frequency of both the train’s whistle is given as
$f = 230\,Hz$
We need to find the number of beats.
The number of beats can be found out by taking the difference in frequency of the sounds of whistles heard.
To find the frequency that is heard we need to use the Doppler effect.
Doppler effect should be used to find the frequency whenever there is relative motion involved.
According to Doppler effect,
$f' = f\left( {\dfrac{{v + {v_0}}}{{v - {v_s}}}} \right)$
Where $f$ is the source frequency, $v$ is the velocity of sound, ${v_0}$ is the velocity of the observer and ${v_s}$ is the velocity of source.
Let the observer be standing still at the station.
Then ${v_0} = 0$
In the case of a train approaching the station velocity of the source, that is the velocity of the tarin can be taken as positive. Thus, we get
${v_s} = 4\,m/s$
So, the frequency of sound produced by the whistle of first train heard by him will be
$\Rightarrow {f_1}^\prime = f\left( {\dfrac{{v + 0}}{{v - 4}}} \right)$
$\Rightarrow {f_1}^\prime = 230\left( {\dfrac{{v + 0}}{{v - 4}}} \right)$
In the case of a train leaving the station velocity of the source can be taken as negative. Thus, we get
${v_s} = - 4\,m/s$
Thus, we get the frequency of second train heard by him as
$\Rightarrow {f_2}^\prime = f\left( {\dfrac{{v + 0}}{{v - \left( { - 4} \right)}}} \right)$
$\Rightarrow {f'_2} = 230\left( {\dfrac{{v + 0}}{{v + 4}}} \right)$
Now let us calculate the number of beats.
$n = \left| {{f_1}^\prime - {f_2}^\prime } \right|$
On substituting the corresponding values,
$\Rightarrow n = \left| {230\left( {\dfrac{v}{{v - 4}}} \right) - 230\left( {\dfrac{v}{{v + 4}}} \right)} \right|$
$\Rightarrow n = \left| {230v\left[ {\dfrac{1}{{v - 4}} - \dfrac{1}{{v + 4}}} \right]} \right|$
On simplification,
$\Rightarrow n = \left| {230v\left[ {\dfrac{{v + 4 - v + 4}}{{{v^2} - {4^2}}}} \right]} \right|$
On simplifications,
$\Rightarrow n = \left| {230v\left[ {\dfrac{8}{{{v^2} - {4^2}}}} \right]} \right|$
Substituting the velocity of sound, we get
$\Rightarrow n = \left| {\dfrac{{230 \times 320 \times 8}}{{{{320}^2} - {4^2}}}} \right|$
$\Rightarrow n = 5 \cdot 75$
Approximating the number of beats,
$\Rightarrow n \approx 6$

$\therefore$ The number of beats is 6. The Correct answer is option D.

Note:
Always remember to use the doppler effect to find the frequency in cases where there is relative motion between the source and the observer. Also, take care of the sign of velocity to be used. In the case when the source is approaching towards the observer, we take a positive sign for velocity. So, in the case when the source is moving away from the observer take a negative sign.