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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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Verified Answer
The correct answer is:
$0.02$
When both the wires vibrate simultaneously, beats per second,
$$
\begin{array}{c}
n_{1} \pm n_{2}=6 \\
\text { or } \frac{1}{2 l} \sqrt{\frac{T}{m}} \pm \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}=6 \\
\Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-\frac{1}{2 l} \sqrt{\frac{T}{m}}=6 \\
\Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-600=6 \Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}=606...(i)
\end{array}
$$
Given that fundamental frequency
$$
\frac{1}{2 l} \sqrt{\frac{T}{m}}=600...(ii)
$$
Dividing Eq. (i) by(ii),
$$
\begin{array}{l}
\frac{\frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}=\frac{606}{600} \\
\Rightarrow \sqrt{\frac{T^{\prime}}{T}}=(1.01) \Rightarrow \frac{T^{\prime}}{T}=(1.02) \% \\
\Rightarrow T^{\prime}=T(1.02)
\end{array}
$$
Increase in tension
$$
\begin{array}{l}
\Delta T^{\prime}=T \times 1.02-T=(0.02 T) \\
\therefore \Delta T=0.02
\end{array}
$$
$$
\begin{array}{c}
n_{1} \pm n_{2}=6 \\
\text { or } \frac{1}{2 l} \sqrt{\frac{T}{m}} \pm \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}=6 \\
\Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-\frac{1}{2 l} \sqrt{\frac{T}{m}}=6 \\
\Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-600=6 \Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}=606...(i)
\end{array}
$$
Given that fundamental frequency
$$
\frac{1}{2 l} \sqrt{\frac{T}{m}}=600...(ii)
$$
Dividing Eq. (i) by(ii),
$$
\begin{array}{l}
\frac{\frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}=\frac{606}{600} \\
\Rightarrow \sqrt{\frac{T^{\prime}}{T}}=(1.01) \Rightarrow \frac{T^{\prime}}{T}=(1.02) \% \\
\Rightarrow T^{\prime}=T(1.02)
\end{array}
$$
Increase in tension
$$
\begin{array}{l}
\Delta T^{\prime}=T \times 1.02-T=(0.02 T) \\
\therefore \Delta T=0.02
\end{array}
$$
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