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 occurrence of 6 beats per second when both wires vibrate simultaneously?

• Correct answer
  A

  0.01
• B

  0.02
• C

  0.03
• D

  0.04

Solution

Beats per second when both the wires vibrate simultaneously. $n_1 \pm n_2=6$ $\begin{aligned} \text { or } & \frac{1}{2 l} \sqrt{\frac{T}{m}} \pm \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}} & =6 \\ \text { or } & \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-\frac{1}{2 l} \sqrt{\frac{T}{m}} & =6 \\ \text { or } & \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-600 & =6\end{aligned}$
Given that fundamental frequency
Dividing Eq. (i) by Eq. (ii), we get $\frac{\frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}=\frac{606}{600}$ $\begin{array}{ll}\text { or } & \sqrt{\frac{T^{\prime}}{T}}=(1.01) \\ \text { or } & \frac{T^{\prime}}{T}=(1.02) \% \\ \text { or } & T^{\prime}=T(1.02)\end{array}$ Increase in tension $\begin{aligned} \Delta T^{\prime} & =T \times 1.02-T \\ & =(0.02 T) \\ \text{Hence,} \quad \Delta T^{\prime}=0.02 \end{aligned}$

Hence, the correct answer is option: 'A'.