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A train, standing at the outer signal of railway station blows a whistle of frequency $400 \mathrm{~Hz}$ 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 $10 \mathrm{~m} / \mathrm{s}$
(b) recedes from the platform with a speed of $10 \mathrm{~m} / \mathrm{s}$ ?
(ii) What is the speed of sound in each case?
[Speed of sound in still air $=340 \mathrm{~m} / \mathrm{s}$ ]
(a) approaches the platform with a speed of $10 \mathrm{~m} / \mathrm{s}$
(b) recedes from the platform with a speed of $10 \mathrm{~m} / \mathrm{s}$ ?
(ii) What is the speed of sound in each case?
[Speed of sound in still air $=340 \mathrm{~m} / \mathrm{s}$ ]
Solution:
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Verified Answer
(i) Given, $v=400 \mathrm{~Hz}, v=340 \mathrm{~m} / \mathrm{s}$
(a) When the train approaches the platform $v_s=10 \mathrm{~m} / \mathrm{s}$ $v^{\prime}=\frac{v}{v-v_s} \times v=\frac{340 \times 400}{340-10}=412.12 \mathrm{~Hz}$
(b) When the train recedes $\quad v^{\prime}=\frac{v \times v}{v+v_s}$ $=\frac{340 \times 400}{340+10}=388.6 \mathrm{~Hz}$
(ii) The speed of sound remains same in both cases i.e. $340 \mathrm{~m} / \mathrm{s}$.
(a) When the train approaches the platform $v_s=10 \mathrm{~m} / \mathrm{s}$ $v^{\prime}=\frac{v}{v-v_s} \times v=\frac{340 \times 400}{340-10}=412.12 \mathrm{~Hz}$
(b) When the train recedes $\quad v^{\prime}=\frac{v \times v}{v+v_s}$ $=\frac{340 \times 400}{340+10}=388.6 \mathrm{~Hz}$
(ii) The speed of sound remains same in both cases i.e. $340 \mathrm{~m} / \mathrm{s}$.
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