Question

A train blows a whistle of frequency $$640\,Hz$$ in air. Find the difference in apparent frequencies of the whistle for a stationary observer, when the train moves towards and away from the observer with the speed of $$72\,km\,h{r^{ - 1}}$$
[Speed of sound in air $ = \,340\,m\,{s^{ - 1}}$ ]

Answer 1

Hint: When an observer moves towards or away from the source of sound, different frequencies are observed due to Doppler Effect. When the source moves towards the observer, the frequency increases and when the source moves away from the observer, the frequency decreases.

Complete step by step answer:
Doppler effect is observed in everyday life, when standing on a station; the approaching train’s siren feels to be having greater frequency. When the same trains pass us, the frequency seems to be decreasing.
The formula to calculate the apparent frequency is:
$$f = \left( {\dfrac{{v \pm {v_o}}}{{v \pm {v_s}}}} \right){f_0}$$
Here, $${f_0}$$ is the actual frequency that the source produces.
$$f$$ is the apparent frequency observed by the observer;
$$v$$ is the velocity of sound;
$${v_o}$$ is the velocity of the observer;
$${v_s}$$ is the velocity of the source.
The given values are:
$${v_o} = 0$$ as the observer is at rest;
$$v = 340\,m\,{s^{ - 1}}$$ given speed of sound in air;
$${f_0} = 60\,Hz$$ frequency at which the whistle is blown;
$${v_s} = 72\,km\,h{r^{ - 1}}\, = \,20\,m\,{s^1}$$ speed at which the train is moving towards the observer.
As the source is moving towards the observed, hence greater frequency is observed. Therefore, the above formula is modified as:
$$f = \left( {\dfrac{{v \pm {v_o}}}{{v - {v_s}}}} \right){f_0}$$
A decrease in denominator will increase the resultant frequency.
Substituting these values in above equation, we have:
 $$f = \left( {\dfrac{{340 \pm 0}}{{340 - 20}}} \right)640$$
$$ \Rightarrow f = \dfrac{{340}}{{320}} \times 640$$
$$ \Rightarrow f = 680\,Hz$$
This is the frequency when the train is moving towards the observer.
When the train is moving away from the observer, the apparent frequency must decrease. Therefore, the formula is modified as:
$$f = \left( {\dfrac{{v \pm {v_o}}}{{v + {v_s}}}} \right){f_0}$$
As the denominator will increase, the overall value will decrease.
Substituting the values, we have:
$$f = \left( {\dfrac{{340 \pm 0}}{{340 + 20}}} \right) \times 640$$
$$ \Rightarrow f = \left( {\dfrac{{340}}{{360}}} \right) \times 640$$
$$ \Rightarrow f = 604.4\,Hz$$
This is the frequency when the train is moving away from the observer.

The frequency when the train moves towards and away from the observer is $$680\,Hz$$ and $$604.4\,Hz$$ respectively.

Note:
The Doppler effect is observed as the source starts moving towards the observer, the distance between them decreases and hence the sound wave reaches faster. While solving, convert the speed of the train in SI units. Also remember that if the observer would have been moving then we would have modified the equation accordingly.