TS EAMCET · Physics · Waves and Sound
An engine sounding a whistle of frequency \(2000 \mathrm{~Hz}\) is receding from the stationary observer at \(72 \mathrm{~km} / \mathrm{h}\). What is the apparent frequency of the observer? The velocity of sound in air is \(340 \mathrm{~m} / \mathrm{s}\).
- A \(1889 \mathrm{~Hz}\)
- B \(2889 \mathrm{~Hz}\)
- C \(3889 \mathrm{~Hz}\)
- D \(4889 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(A) \(1889 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
Here a source is moving away from a stationary observer. So, frequency recorded at source will be lower than true frequency. Apparent frequency at observer, \(f^{\prime}=f\left(\frac{v}{v+v_s}\right)\) where, \(v=\) speed of sound \(=340 \mathrm{~m} / \mathrm{s}\) and…
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