A train standing at the outer signal of a railway station blows a whistle of frequency 400Hz in still air.
(i) what is the frequency of whistle for a platform observer when the train
(a) approaches the platform with speed of 10 1 ms−
(b) recedes from the platform with the speed of 10 1 ms−
(ii) what is the speed of sound in each case.
[The speed of sound in still air = 340 1 ms− ]
Answers
Explanation:
(i) (a)Frequency of the whistle,
v=400Hz
v=400Hz
Speed of the train,
v_T=10\,m/s
v
T
=10m/s
Speed of sound,
v=340\,m/s
v=340m/s
The apparent frequency
(v')
(v
′
)
of the whistle as the train approaches the platform is given by the
relation:
\displaystyle v'=\left(\frac{v}{v-v_r}\right)v
v
′
=(
v−v
r
v
)v
=\displaystyle \left(\frac{340}{340-10}\right)\times 400=412.12Hz
=(
340−10
340
)×400=412.12Hz
(b) The apparent frequency
(v')
(v
′
)
of the whistle as the train recedes from the platform is given by the relation:
\displaystyle v''=\left(\frac{v}{v+v_r}\right)v
v
′′
=(
v+v
r
v
)v
\displaystyle =\left(\frac{340}{340+10}\right)\times 400=388.57\,Hz
=(
340+10
340
)×400=388.57Hz
(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.