JEE Advanced · Physics · 16. Waves & Sound
A source (S) of sound has frequency \(240 \mathrm{~Hz}\). When the observer \((\mathrm{O})\) and the source move towards each other at a speed \(v\) with respect to the ground (as shown in Case 1 in the figure), the observer measures the frequency of the sound to be \(288 \mathrm{~Hz}\). However, when the observer and the source move away from each other at the same speed \(v\) with respect to the ground (as shown in Case 2 in the figure), the observer measures the frequency of sound to be \(n \mathrm{~Hz}\). The value of \(n\) is ______

- A 100
- B 150
- C 200
- D 250
Answer & Solution
Correct Answer
(C) 200
Step-by-step Solution
Detailed explanation
Frequency received by observer \(f_0=\left(\frac{C \pm V_0}{C \pm V_S}\right) f_{\mathrm{s}}\), C is speed of sound
Case-1:
\(\begin{aligned} & \mathrm{f}_1=\left(\frac{\mathrm{C}+\mathrm{V}}{\mathrm{C}-\mathrm{V}}\right) \mathrm{f}_{\mathrm{s}} \\ & 288=\left(\frac{\mathrm{C}+\mathrm{V}}{\mathrm{C}-\mathrm{V}}\right) 240\end{aligned}\)
Case-2:
\(f_2=\left(\frac{C-V}{C+V}\right) f_S\)
\(\mathrm{n}=\left(\frac{\mathrm{C}-\mathrm{V}}{\mathrm{C}+\mathrm{V}}\right) 240\)
multiply the two equations, we get.
\(\begin{aligned} & (288)(\mathrm{n})=(240)(240) \\ & \mathrm{n}=200\end{aligned}\)
Case-1:
\(\begin{aligned} & \mathrm{f}_1=\left(\frac{\mathrm{C}+\mathrm{V}}{\mathrm{C}-\mathrm{V}}\right) \mathrm{f}_{\mathrm{s}} \\ & 288=\left(\frac{\mathrm{C}+\mathrm{V}}{\mathrm{C}-\mathrm{V}}\right) 240\end{aligned}\)
Case-2:
\(f_2=\left(\frac{C-V}{C+V}\right) f_S\)
\(\mathrm{n}=\left(\frac{\mathrm{C}-\mathrm{V}}{\mathrm{C}+\mathrm{V}}\right) 240\)
multiply the two equations, we get.
\(\begin{aligned} & (288)(\mathrm{n})=(240)(240) \\ & \mathrm{n}=200\end{aligned}\)
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