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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?
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The correct answer is:
0.01
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
$\begin{aligned}
\Delta T^{\prime} & =T \times 1.02-T \\
& =(0.02 T) \\
\text{Hence,} \quad \Delta T^{\prime}=0.02
\end{aligned}$
$n_1 \pm n_2=6$
$\begin{aligned} \text { or } & \frac{1}{2 l} \sqrt{\frac{T}{m}} \pm \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}} & =6 \\ \text { or } & \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-\frac{1}{2 l} \sqrt{\frac{T}{m}} & =6 \\ \text { or } & \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-600 & =6\end{aligned}$
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
$\begin{aligned}
\Delta T^{\prime} & =T \times 1.02-T \\
& =(0.02 T) \\
\text{Hence,} \quad \Delta T^{\prime}=0.02
\end{aligned}$
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