KCET · Physics · Waves and Sound
An engine moving towards a wall with a velocity \(50 \mathrm{~ms}^{-1}\) emits a note of \(1.2 \mathrm{kHz}\). Speed of sound in air is \(350 \mathrm{~ms}^{-1}\). The frequency of the note after reflection from the wall as heard by the driver of the engine is
- A \(2.4 \mathrm{kHz}\)
- B \(0.24 \mathrm{kHz}\)
- C \(1.6 \mathrm{kHz}\)
- D \(1.2 \mathrm{kHz}\)
Answer & Solution
Correct Answer
(C) \(1.6 \mathrm{kHz}\)
Step-by-step Solution
Detailed explanation
The reflected sound appears to propagate in a direction opposite to that of moving engine. Thus, the source and the observer can be presumed to approach each other with same velocity
\(v^{\prime} =\frac{v\left(v+v_{0}\right)}{\left(v-v_{s}\right)} \)
\(=v\left(\frac{v+v_{s}}{v-v_{s}}\right) \quad\left[\because v_{0}=v_{s}\right] \)
\(\Rightarrow v^{\prime} =1.2\left(\frac{350+50}{350-50}\right) \)
\(=\frac{1.2 \times 400}{300}=1.6 \mathrm{kHz}\)
\(v^{\prime} =\frac{v\left(v+v_{0}\right)}{\left(v-v_{s}\right)} \)
\(=v\left(\frac{v+v_{s}}{v-v_{s}}\right) \quad\left[\because v_{0}=v_{s}\right] \)
\(\Rightarrow v^{\prime} =1.2\left(\frac{350+50}{350-50}\right) \)
\(=\frac{1.2 \times 400}{300}=1.6 \mathrm{kHz}\)
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