TS EAMCET · Physics · Waves and Sound
The frequency of sound heard by an observer moving towards a stationary source with certain speed is \(n_1\) and if the observer moves away from the same source with same speed, the frequency of sound heard by the observer is \(\mathrm{n}_2\). If the speed of sound in air is \(340 \mathrm{~ms}^{-1}\) and \(\mathrm{n}_1: \mathrm{n}_2=71: 65\), then speed of observer is
- A \(36 \text { kmph }\)
- B \(27 \text { kmph }\)
- C \(15 \text { kmph }\)
- D \(54 \text { kmph }\)
Answer & Solution
Correct Answer
(D) \(54 \text { kmph }\)
Step-by-step Solution
Detailed explanation
\(n_1 = n \left( \frac{v + v_o}{v} \right)\) \(n_2 = n \left( \frac{v - v_o}{v} \right)\) \(\frac{n_1}{n_2} = \frac{v + v_o}{v - v_o} = \frac{71}{65}\) \(65(v + v_o) = 71(v - v_o)\) \(65v + 65v_o = 71v - 71v_o\) \(136v_o = 6v\) \(v_o = \frac{6 \times 340}{136} = 15 \text{ m/s}\)…
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