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JEE Advance - Physics (2007 - Paper 2 Offline - No. 16)

Two trains $$A$$ and $$B$$ are moving with speeds $$20 \mathrm{~m} / \mathrm{s}$$ and $$30 \mathrm{~m} / \mathrm{s}$$ respectively in the same direction on the same straight track, with $$B$$ ahead of $$A$$. The engines are at the front ends. The engines of train A blows a long whistle.

Assume that the sound of the whistle is composed of components varying in frequency from $$f_{1}=800 \mathrm{~Hz}$$ to $$f_{2}=1120 \mathrm{~Hz}$$, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus $$320 \mathrm{~Hz}$$. The speed of sound in still air is $$340 \mathrm{~m} / \mathrm{s}$$.

Two trains $$A$$ and $$B$$ are moving with speeds $$20 \mathrm{~m} / \mathrm{s}$$ and $$30 \mathrm{~m} / \mathrm{s}$$ respectively in the same direction on the same straight track, with $$B$$ ahead of $$A$$. The engines are at the front ends. The engines of train A blows a long whistle.

Assume that the sound of the whistle is composed of components varying in frequency from $$f_{1}=800 \mathrm{~Hz}$$ to $$f_{2}=1120 \mathrm{~Hz}$$, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus $$320 \mathrm{~Hz}$$. The speed of sound in still air is $$340 \mathrm{~m} / \mathrm{s}$$.

Two trains $$A$$ and $$B$$ are moving with speeds $$20 \mathrm{~m} / \mathrm{s}$$ and $$30 \mathrm{~m} / \mathrm{s}$$ respectively in the same direction on the same straight track, with $$B$$ ahead of $$A$$. The engines are at the front ends. The engines of train A blows a long whistle.

Assume that the sound of the whistle is composed of components varying in frequency from $$f_{1}=800 \mathrm{~Hz}$$ to $$f_{2}=1120 \mathrm{~Hz}$$, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus $$320 \mathrm{~Hz}$$. The speed of sound in still air is $$340 \mathrm{~m} / \mathrm{s}$$.

The spread of frequency as observed by the passengers in train B is
310 Hz
330 Hz
350 Hz
290 Hz

Explanation

Frequency $$(f') = {f_1}\left( {{{v - {v_0}} \over {v - {v_s}}}} \right) = 800\left[ {{{340 - 30} \over {340 - 20}}} \right]$$

$$=800 \times \frac{31}{32}$$ ..... (i)

where, $$\left(\frac{v-v_{0}}{v-v_{\mathrm{S}}}\right)=\frac{31}{32}$$ ..... (ii)

Frequency $$\left(f^{\prime}\right)=f_{2}\left(\frac{v-v_{0}}{v-v_{\mathrm{S}}}\right)=1120 \times \frac{31}{32}$$ ..... (iii)

$$\therefore \quad f^{\prime}-f^{\prime}=(1120-800) \frac{31}{32}=320 \times \frac{31}{32}$$

$$=310 \mathrm{~Hz}$$

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