Two identical wires are stretched by the same tension of 101 N and each emits a note of frequency 202 Hz. If the tension in one wire is increased by 1 N, then the beat frequency is:

Two identical wires are stretched by the same tension of 101 N and each emits a note of frequency 202 Hz. If the tension in one wire is increased by 1 N, then the beat frequency is:

A
2 Hz
B
$\frac{1}{2} Hz$
C
1 Hz
D
none of these

Solution:

$f \propto \sqrt{T}$

$f = \frac{1}{2 l} \sqrt{\frac{T}{μ}} \Rightarrow \frac{∆ f}{f} = \frac{1}{2} \frac{∆ T}{T}$

$\Rightarrow ∆ f = \frac{202}{2} \times \frac{1}{101} = 1$

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