AP EAMCET · PHYSICS · Waves and Sound
Two sources \(A\) and \(B\) are producing notes of frequency \(680 \mathrm{~Hz}\). A listener moves from \(A\) to \(B\) with a constant velocity \(v\). if the speed of sound in air is \(340 \mathrm{~ms}^{-1}\), the value of \(y\) so that he hears 10 beats per second is
- 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
(B) \(2.5 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, notes of frequency produced by the sources \(A\) and \(B\) is \(680 \mathrm{~Hz}\). i. e., \(f_A\) and \(f_B=680 \mathrm{~Hz}\) Velocity of listener moves from \(A\) to \(B\) is constant \(=v\), speed of sound, \(v_s=340 \mathrm{~m} / \mathrm{s}\), and beats per second,…
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