AP EAMCET · PHYSICS · Waves and Sound
Two sources \(A\) and \(B\) are sending notes of frequency \(680 \mathrm{~Hz}\). A listener moves from \(A\) and \(B\) with a constant velocity \(u\). If the speed of sound in air is \(340 \mathrm{~ms}^{-1}\), what must be the value of \(u\) so that he hears 10 beats per second?
- A \(2.0 \mathrm{~ms}^{-1}\)
- B \(2.5 \mathrm{~ms}^{-1}\)
- C \(3.0 \mathrm{~ms}^{-1}\)
- D \(3.5 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(3.0 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Listener go from \(A \rightarrow B\) with velocity \((u)\) let the apparent frequency of sound from source \(A\) by listener or \(\begin{aligned} n^{\prime} & =n\left(\frac{v-v_o}{v+v_s}\right) \\ n^{\prime} & =680\left(\frac{340-u}{340+0}\right) \end{aligned}\) The apparent…
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