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Q.

Two identical sonometer wires have a fundamental frequency of 1000 Hz when kept under the same tension. What fractional increase in the tension of one wire nearly would cause an occurrence of 5 beats per second, when both wires vibrate together?

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a

$0.02$

b

$0.04$

c

0.01

d

$0.08$

answer is C.

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Detailed Solution

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The fundamental frequency of wire is given as
$n_{0} = \frac{1}{2 L} \sqrt{\frac{T}{μ}}$
On increasing tension by $Δ T$ the new frequency will be
$n_{0}^{'} = \frac{1}{2 L} \sqrt{\frac{T + Δ T}{μ}} = n_{0} \sqrt{1 + \frac{Δ T}{T}} \approx n_{0} (1 + \frac{Δ T}{2 T}) \frac{n_{0}^{'} - n_{0}}{n_{0}} = \frac{1}{2} \frac{Δ T}{T} O r \frac{Δ T}{T} = 2 \times \frac{5}{1000} = 0.01$

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