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Question
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.
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Solution 1
(i)
(a)Frequency of the whistle, ν = 400 Hz
Speed of the train, vT= 10 m/s
Speed of sound, v = 340 m/s
The apparent frequency (v') of the whistle as the train approaches the platform is given by the relation:
`v' = (v/(v - v'))v`
`= ((340)/(340 - 10))xx 400 = 412.12 Hz`
b) The apparent frequency (v'')of the whistle as the train recedes from the platform is given by the relation:
`v" = ((v)/(v+v_T))v`
`= (340/(340+ 10))xx 400 = 388.57` Hz
(ii) The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of sound. Therefore, the speed of sound in air in both the cases remains the same, i.e., 340 m/s
Solution 2
Frequency of whistle, v = 400 Hz; speed of sound, υ= 340 ms-1 speed of train, υs= 10 ms1
(i) (a) When the train approaches the platform (i.e., the observer at rest),
`v' = v/(v - v_s)xx v = 340/(340 - 10)xx 400 = 412 Hz`
(b) When the train recedes from the platform (i.e., from the observer at rest),
`v' = v/(v + v_s) xx v = 340/(340+10)xx400 = 389 Hz`
(ii) The speed of sound in each case does not change.It is 340 ms-1 in each case.
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