MHT CET · Physics · Waves and Sound
An obstacle is moving towards the source with velocity 'v'. The sound is reflected from the obstacle. If ' \(\mathrm{c}^{\prime}\) is the speed of sound and \({ }\lambda^{\prime}\) ' is the wavelength, then the
wavelength of the reflected wave \(\left(\lambda_{r}\right)\) is
- A \(\lambda_{r}=\left(\frac{c-v}{c+v}\right) \lambda\)
- B \(\lambda_{r}=\left(\frac{c-v}{c}\right) \lambda\)
- C \(\lambda_{r}=\left(\frac{c+v}{c-v}\right) \lambda\)
- D \(\lambda_{r}=\left(\frac{c+v}{c}\right) \lambda\)
Answer & Solution
Correct Answer
(A) \(\lambda_{r}=\left(\frac{c-v}{c+v}\right) \lambda\)
Step-by-step Solution
Detailed explanation
The frequency of reflected sound wave is \(f_{r}=f\left(\frac{c+v}{c-v}\right)\)
\(\because\) No change in velocity occurs due to reflection of sound wave.
Hence,
\(\begin{array}{l}
\frac{c}{\lambda_{r}}=\frac{c}{\lambda}\left(\frac{c+v}{c-v}\right) \Rightarrow \frac{1}{\lambda_{r}}=\frac{1}{\lambda}\left(\frac{c+v}{c-v}\right) \\
\lambda_{r}=\frac{c-v}{c+v} \lambda
\end{array}\)
\(\because\) No change in velocity occurs due to reflection of sound wave.
Hence,
\(\begin{array}{l}
\frac{c}{\lambda_{r}}=\frac{c}{\lambda}\left(\frac{c+v}{c-v}\right) \Rightarrow \frac{1}{\lambda_{r}}=\frac{1}{\lambda}\left(\frac{c+v}{c-v}\right) \\
\lambda_{r}=\frac{c-v}{c+v} \lambda
\end{array}\)
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