Two identical plain 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 occurrence of 6 beats per second when both wires vibrate simultaneously

Question

Dr. Abubacker · Accepted Answer

(b) $n_{1} - n_{2} = 6$
$\Rightarrow \frac{1}{2l}\sqrt{\frac{T^{'}}{m}} - \frac{1}{2l}\sqrt{\frac{T}{m}} = 6$
$\Rightarrow \frac{1}{2l}\sqrt{\frac{T^{'}}{m}} - 600 = 6$
$\frac{1}{2l}\sqrt{\frac{T^{'}}{m}} = 606 =$ Fundamental frequency …(i)
Given,
$\frac{1}{2l}\sqrt{\frac{T}{m}} = 600$ …(ii)
Dividing Equation (i) by Equation (ii), we get
$\frac{\frac{1}{2l}\sqrt{\frac{T'}{m}}}{\frac{1}{2l}\sqrt{\frac{T}{m}}} = \frac{606}{600}$
$\Rightarrow \sqrt{\frac{T^{'}}{T}} = (1.01) \Rightarrow \frac{T^{'}}{T} = (1.02)$
$\Rightarrow T^{'} = T(1.02)$
Increase in tension
$\mathrm{\Delta}T^{'} = T 	imes 1.02 - T = (0.02T)$
Hence, $\frac{\mathrm{\Delta}T^{'}}{T} = 0.02$

Q. Two identical plain 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 occurrence of 6 beats per second when both wires vibrate simultaneously

Explanation

Reviewed by Dr. Abubacker

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