A train blows a whistle of frequency 640 Hz in air. Find the difference in apparent frequencies of the whistle for a stationary observer
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/hour.
(Speed of sound in air = 340 m/s)
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Given: vs = 72 km/ hr = 20 m/s, n0 = 640 Hz,
v = 340 m/s
To find: Difference in apparent frequencies (nA – n’A)
Formulae:
i. When the train moves towards the stationary observer then,
nA = n0 \((\frac{v}{v-v_s})\)
ii. When the train moves away the stationary observer then,
n’A = n0 \((\frac{v}{v+v_s})\)
Calculation: From formula (i),
nA = 640 \((\frac{340}{340-20})\)
∴ nA = 680 Hz
From formula (ii),
n’A = 640 \((\frac{340}{340+20})\)
∴ n’A = 604.4 Hz
Difference between nA and n’A
= nA – n’A = 75.6 Hz
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