A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of $2 \mathrm{~ms}^{-1}$ in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is $320 \mathrm{~ms}^{-1}$, the smallest value of the percentage change required in the length of the pipe is
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Solution
Sol. $\mathrm{f} \propto \frac{1}{\ell_1} \Rightarrow \mathrm{f}=\frac{\mathrm{K}}{\ell_1}$
( $\ell_1 \Rightarrow$ initial length of pipe)
$$
\left(\frac{\mathrm{V}}{\mathrm{~V}-\mathrm{V}_{\mathrm{T}}}\right) \mathrm{f}=\frac{\mathrm{k}}{\ell_2}\left\{\mathrm{~V}_{\mathrm{T}} \text { Speed of tuning fork, } \ell_2 \rightarrow \text { new length of pipe }\right\}
$$
(1) ÷ (2)
$$
\begin{aligned}
& \frac{\mathrm{V}-\mathrm{V}_{\mathrm{T}}}{\mathrm{~V}}=\frac{\ell_2}{\ell_1} \\
& \frac{\ell_2}{\ell_1}-1=\frac{\mathrm{V}-\mathrm{V}_{\mathrm{T}}}{\mathrm{~V}}-1 \\
& \frac{\ell_2-\ell_1}{\ell_1}=\frac{-\mathrm{V}_{\mathrm{T}}}{\mathrm{~V}} \\
& \frac{\ell_2-\ell_1}{\ell_1} \times 100=\frac{-2}{320} \times 100=-0.625
\end{aligned}
$$
Therefore smallest value of percentage change required in the length of pipe is 0.625
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