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 wires will lead to the occurrence of 6 beats

Question

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 wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

TestHub · Accepted Answer

Beats per second when both the wires vibrate simultaneously

n₁± n₂= 6

or $\frac{1}{2 l} \sqrt{\frac{T}{m}} \pm \frac{1}{2 l} \sqrt{\frac{T^{'}}{m}} = 6$

or $\frac{1}{2 l} \sqrt{\frac{T^{'}}{m}} - \frac{1}{2 l} \sqrt{\frac{T}{m}} = 6$

or $\frac{1}{2 l} \sqrt{\frac{T^{'}}{m}} - 600 = 6$ ...(i)

Given that fundamental frequency

$\begin{matrix} \frac{1}{2 l} & \sqrt{\frac{T}{m}} = 600 \end{matrix}$

Dividing Eq.(i) by Eq. (ii), we get

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

Or $\sqrt{\frac{T^{'}}{T}} = 1.01$

Or $\frac{T^{'}}{T} = 1.02 %$

Or $T' = T (1.02)$

Increase in tension

$Δ T' = (T \times 1.02) - T$

$= 0.02 T$

Hence, $Δ T' = 0.02$

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