JEE Advanced · Physics · 16. Waves & Sound
Paragraph:
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 engine 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 \(340 \mathrm{~m} / \mathrm{s}\).
Question:
The spread of frequency as observed by the passengers in train \(B\) is
- A \(310 \mathrm{~Hz}\)
- B \(330 \mathrm{~Hz}\)
- C \(350 \mathrm{~Hz}\)
- D \(290 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(A) \(310 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
For the passengers in train \(B\), observer is receding with velocity \(30 \mathrm{~m} / \mathrm{s}\) and source is approaching with velocity \(20 \mathrm{~m} / \mathrm{s}\).
\(\therefore f_1^{\prime}=800\left(\frac{340-30}{340-20}\right)=775 \mathrm{~Hz} \)
\( \text { and } f_2^{\prime}=1120\left(\frac{340-30}{340-20}\right)=1085 \mathrm{~Hz}\)
\(\therefore\) Spread of frequency \(=f_2^{\prime}-f_1^{\prime}=310 \mathrm{~Hz}\)
\(\therefore\) The correct option is (a).
\(\therefore f_1^{\prime}=800\left(\frac{340-30}{340-20}\right)=775 \mathrm{~Hz} \)
\( \text { and } f_2^{\prime}=1120\left(\frac{340-30}{340-20}\right)=1085 \mathrm{~Hz}\)
\(\therefore\) Spread of frequency \(=f_2^{\prime}-f_1^{\prime}=310 \mathrm{~Hz}\)
\(\therefore\) The correct option is (a).
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