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Two identical wires are stretched by same tension of 100 N and each wire emits note of frequency 400 Hz. If the tension of any wire is increased by 2 N then beat frequency heard is

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1 Hz

2 Hz

3 Hz

4 Hz

answer is D.

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Detailed Solution

$\frac{Δf}{f} = \frac{1}{2} \frac{ΔT}{T} ∴ \frac{Δf}{400} = \frac{1}{2} (\frac{2}{100}) ∴ Δf = 4 Hz$

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