
A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.
(i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10$m{{s}^{-1}}$, (b) recedes from the platform with a speed of 10 $m{{s}^{-1}}$?
(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 $m{{s}^{-1}}$
Answer
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Hint: To answer this question, we are required to find the relation between the apparent frequency of the approaching train. Once we are done with this, we have to find the apparent frequency of the receding train. After answering both the mentioned situations, we have to consider the speed of sound in air and then find the speed of sound in both the previously mentioned questions.
Complete step by step answer:
1. (a) The frequency of the whistle or u = 400 Hz.
The speed of the train or ${{v}_{r}}$= 10 $m{{s}^{-1}}$
The speed of sound is considered as v = 340 $m{{s}^{-1}}$
So now the apparent frequency or v` of the whistle of the train approaches the platform us given by the relation:
v`=$\left( \dfrac{v}{v-{{v}_{r}}} \right)u$
Now putting the values, we get:
$\left( \dfrac{340}{340-10} \right)\times 400$=412.12Hz
(b) The apparent frequency (v`) of the whistle as the train recedes from the platform is given by the relation:
v``= $\left( \dfrac{v}{v+{{v}_{r}}} \right)u=\left( \dfrac{340}{340+10} \right)\times 400=388.57Hz$
2. The apparent change in the frequency of the sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of the sound. Therefore, the speed of the sound in air in both the cases remains the same, that is 340 $m{{s}^{-1}}$
Note: As we have dealt with the concept of speed of sound in the above mentioned answer, we have to define it as well. The distance which is travelled per unit time by a sound wave as it propagates through an elastic medium is known as the speed of sound.
Since sound requires a medium to travel, unlike light, the medium affects the speed of light. As of air, the speed of sound is 343 $m{{s}^{-1}}$(also considered as 340 $m{{s}^{-1}}$, like in this case). In water the speed of sound is 1480 $m{{s}^{-1}}$, and in solids the speed of sound is about 3000$m{{s}^{-1}}$.
Complete step by step answer:
1. (a) The frequency of the whistle or u = 400 Hz.
The speed of the train or ${{v}_{r}}$= 10 $m{{s}^{-1}}$
The speed of sound is considered as v = 340 $m{{s}^{-1}}$
So now the apparent frequency or v` of the whistle of the train approaches the platform us given by the relation:
v`=$\left( \dfrac{v}{v-{{v}_{r}}} \right)u$
Now putting the values, we get:
$\left( \dfrac{340}{340-10} \right)\times 400$=412.12Hz
(b) The apparent frequency (v`) of the whistle as the train recedes from the platform is given by the relation:
v``= $\left( \dfrac{v}{v+{{v}_{r}}} \right)u=\left( \dfrac{340}{340+10} \right)\times 400=388.57Hz$
2. The apparent change in the frequency of the sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of the sound. Therefore, the speed of the sound in air in both the cases remains the same, that is 340 $m{{s}^{-1}}$
Note: As we have dealt with the concept of speed of sound in the above mentioned answer, we have to define it as well. The distance which is travelled per unit time by a sound wave as it propagates through an elastic medium is known as the speed of sound.
Since sound requires a medium to travel, unlike light, the medium affects the speed of light. As of air, the speed of sound is 343 $m{{s}^{-1}}$(also considered as 340 $m{{s}^{-1}}$, like in this case). In water the speed of sound is 1480 $m{{s}^{-1}}$, and in solids the speed of sound is about 3000$m{{s}^{-1}}$.
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