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Question: A train is moving on a straight line with speed \(20m{{s}^{-1}}\). It is blowing its whistle at the ...

A train is moving on a straight line with speed 20ms−120m{{s}^{-1}}. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320ms−1320m{{s}^{-1}}) close to :
A. 6%
B. 12%
C. 18%
D. 24%

Explanation

Solution

When the source of sound and the observer are in motion, the frequency of the sound heard by the observer is different from the frequency of the actual frequency of the sound.

Use the formulae for the apparent frequency when the source is moving towards the stationary observer and when the source is moving away from the observer.

Formula used:

f=f0(vv−vs)f={{f}_{0}}\left( \dfrac{v}{v-{{v}_{s}}} \right)
f=f0(vv+vs)f={{f}_{0}}\left( \dfrac{v}{v+{{v}_{s}}} \right)

Complete step by step answer:
Suppose the frequency of sound heard by a stationary observer, emitted by a stationary source is f0{{f}_{0}}. If any one or both of them move, the frequency heard by the observer will not be equal to f0{{f}_{0}}. This different frequency is called apparent frequency.

When the source of sound moves towards a stationary observer, the apparent frequency is equal to f=f0(vv−vs)f={{f}_{0}}\left( \dfrac{v}{v-{{v}_{s}}} \right),

where v is the speed of the sound and vs{{v}_{s}} is the speed of the source.

In this case, v=320ms−1v=320m{{s}^{-1}}, vs=20ms−1{{v}_{s}}=20m{{s}^{-1}} and f0=100Hz{{f}_{0}}=100Hz.

Therefore, when the comes closer to the person, the apparent frequency heard by the person is equal to f1=1000(320320−20)=1000(320300)Hz{{f}_{1}}=1000\left( \dfrac{320}{320-20} \right)=1000\left( \dfrac{320}{300} \right)Hz.

When the source is moving away from the stationary observer than the apparent frequency is equal to f=f0(vv+vs)f={{f}_{0}}\left( \dfrac{v}{v+{{v}_{s}}} \right)

Therefore, when the train passes the person and moves away from him, the apparent frequency heard by him is equal to f1=1000(320320+20)=1000(320340)Hz{{f}_{1}}=1000\left( \dfrac{320}{320+20} \right)=1000\left( \dfrac{320}{340} \right)Hz.

Now, the percentage change in the frequency will be equal to f1−f2f1×100\dfrac{{{f}_{1}}- {{f}_{2}}}{{{f}_{1}}}\times 100.

Substitute the known values.
⇒f1−f2f1×100=1000(320300)−1000(320340)1000(320300)×100\Rightarrow \dfrac{{{f}_{1}}-{{f}_{2}}}{{{f}_{1}}}\times 100=\dfrac{1000\left( \dfrac{320}{300} \right)-1000\left( \dfrac{320}{340} \right)}{1000\left( \dfrac{320}{300} \right)}\times 100
⇒f1−f2f1×100=[1−(320340)(320300)]×100=(1−0.88)×100=0.12×100=12\Rightarrow \dfrac{{{f}_{1}}-{{f}_{2}}}{{{f}_{1}}}\times 100=\left[ 1-\dfrac{\left( \dfrac{320}{340} \right)}{\left( \dfrac{320}{300} \right)} \right]\times 100=\left( 1-0.88 \right)\times 100=0.12\times 100=12%.

Therefore, the percentage change in the frequency heard by a person standing near the track as the train passes him is 12 %.

Hence, the correct option is B.

Note: Note that the formulae given formulae for the apparent frequency is valid only the motion of the source is along the line joining the source and the observer.

That is why it is mentioned that the train is moving on a straight track and the person is standing very close to the track.