COMEDK · Physics · 16. Waves and Sound
A bat emitting an ultrasonic wave of frequency \(4.5 \times 10^4 \mathrm{~Hz}\) at speed of \(6 \mathrm{~m} / \mathrm{s}\) between two parallel walls. The two frequencies heard by the bat will be
- A \(4.67 \times 10^4 \mathrm{~Hz}, 4.34 \times 10^4 \mathrm{~Hz}\)
- B \(4.34 \times 10^4 \mathrm{~Hz}, 4.67 \times 10^4 \mathrm{~Hz}\)
- C \(4.5 \times 10^4 \mathrm{~Hz}, 5.4 \times 10^4 \mathrm{~Hz}\)
- D \(4.67 \times 10^3 \mathrm{~Hz}, 4.34 \times 10^4 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(A) \(4.67 \times 10^4 \mathrm{~Hz}, 4.34 \times 10^4 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
The frequency of the source is \(f_0 = 4.5 \times 10^4 \text{ Hz}\). The speed of the bat is \(v_b = 6 \text{ m/s}\). The speed of sound in air is taken as \(v = 330 \text{ m/s}\). When the bat moves towards one wall, the wall acts as a stationary observer receiving the sound…
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