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VITEEE 2009 Paper

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Question : 19 of 120

Marks: +1, -0

Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurence of 6 beats per second when both wires vibrate simultaneously ?

0.01
0.02
0.03
0.04

Solution:

When both the wires vibrate simultaneously, beats per second,

n_{1} \pm n_{2} = 6

or

\frac{1}{2l} \sqrt{\frac{T}{m}} \pm \frac{1}{2l} \sqrt{\frac{T'}{m}} = 6 \Rightarrow \frac{1}{2l} \sqrt{\frac{T'}{m}} - \frac{1}{2l} \sqrt{\frac{T}{m}} = 6

⇒

\frac{1}{2l} \sqrt{\frac{T'}{m}} - 600 = 6 \Rightarrow \frac{1}{2l} \sqrt{\frac{T'}{m}} = 606

...(i)

Given that fundamental frequency

\frac{1}{2l} \sqrt{\frac{T}{m}} = 600

...(ii)

Dividing Eq.(i) by (ii),

\frac{ \frac{1}{2l} \sqrt{\frac{T'}{m}} }{ \frac{1}{2l} \sqrt{\frac{T}{m}} } = \frac{606}{600}

⇒

\sqrt{\frac{T'}{T}} = (1.01) \Rightarrow \frac{T'}{T} = (1.02)\% \Rightarrow T' = T(1.02)

Increase in tension

ΔT' = T×1.02 - T = (0.02T)

∴ ΔT = 0.02

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