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Two identical wires have a fundamental frequency $f_0$ when kept under the same tension $T$. If the tension of one wire is increased by $\Delta T$, then the $N$ beats occur when both wires oscillate simultaneously.
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Verified Answer
The correct answer is:
$\frac{\Delta T}{T}=\left(\frac{f_0+N}{f_0}\right)^2-1$
Fundamental frequency of the wire,
$$
f_0=\frac{v}{2 L}, \text { where } v=\sqrt{\frac{T}{\mu}}
$$
$\begin{array}{ll}\Rightarrow & f_0=\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \\ \Rightarrow & f_0 \propto \sqrt{T}\end{array}$
Dividing Eq. (ii) by Eq. (i), we get
$$
\begin{aligned}
\frac{\left(f_0+N\right)^2}{f_0^2} & =\frac{T+\Delta T}{T} \\
\Rightarrow \quad \frac{\Delta T}{T} & =\frac{\left(f_0+N\right)^2}{f_0^2}-1=\left(\frac{f_0+N}{f_0}\right)^2-1
\end{aligned}
$$
$$
f_0=\frac{v}{2 L}, \text { where } v=\sqrt{\frac{T}{\mu}}
$$
$\begin{array}{ll}\Rightarrow & f_0=\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \\ \Rightarrow & f_0 \propto \sqrt{T}\end{array}$
Dividing Eq. (ii) by Eq. (i), we get
$$
\begin{aligned}
\frac{\left(f_0+N\right)^2}{f_0^2} & =\frac{T+\Delta T}{T} \\
\Rightarrow \quad \frac{\Delta T}{T} & =\frac{\left(f_0+N\right)^2}{f_0^2}-1=\left(\frac{f_0+N}{f_0}\right)^2-1
\end{aligned}
$$
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