Q60 / 66 · JEE Advanced 2007 (Paper 2) · Physics
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
- Correct answerA$310 \mathrm{~Hz}$
- B$330 \mathrm{~Hz}$
- C$350 \mathrm{~Hz}$
- D$290 \mathrm{~Hz}$
Solution
For the passengers in train $B$, observer is receding with velocity $30 \mathrm{~m} / \mathrm{s}$ and source is approaching with velocity $20 \mathrm{~m} / \mathrm{s}$.
$$
\begin{array}{ll}
\therefore & f_1^{\prime}=800\left(\frac{340-30}{340-20}\right)=775 \mathrm{~Hz} \\
\text { and } & f_2^{\prime}=1120\left(\frac{340-30}{340-20}\right)=1085 \mathrm{~Hz}
\end{array}
$$
$\therefore$ Spread of frequency $=f_2^{\prime}-f_1^{\prime}=310 \mathrm{~Hz}$
$\therefore$ The correct option is (a).
Hence, the correct answer is option: 'A'.