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A pipe $20 \mathrm{~cm}$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430 \mathrm{~Hz}$ source? Will the same source be in resonance with the pipe, if both ends are open. Speed of sound $=340 \mathrm{~m} / \mathrm{s}$
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Given, $L=20 \mathrm{~cm}=0.2 \mathrm{~m}$,
Natural frequency of the pipe
$$
\begin{aligned}
&v_{\mathrm{n}}=430 \mathrm{~Hz}, v=340 \mathrm{~m} / \mathrm{s} \\
&v_n=(2 n-1) \frac{v}{4 L} \Rightarrow 430=(2 n-1) \frac{340}{4 \times 0.2} \\
&\Rightarrow 2 n-1=\frac{430 \times 4 \times 0.2}{340}=1.02 \\
&\Rightarrow 2 n=2.02 \Rightarrow n=1.01
\end{aligned}
$$
$\therefore \quad$ It is 1 st normal mode of vibration.
In a pipe, with both ends open
$$
v_n=n \times \frac{v}{2 L}=\frac{n \times 340}{2 \times 0.2}=430
$$
$$
\therefore \quad n=\frac{430 \times 2 \times 0.2}{340}=0.5
$$
$\because n$ should be an integer, therefore it is not in resonance with the source.
Natural frequency of the pipe
$$
\begin{aligned}
&v_{\mathrm{n}}=430 \mathrm{~Hz}, v=340 \mathrm{~m} / \mathrm{s} \\
&v_n=(2 n-1) \frac{v}{4 L} \Rightarrow 430=(2 n-1) \frac{340}{4 \times 0.2} \\
&\Rightarrow 2 n-1=\frac{430 \times 4 \times 0.2}{340}=1.02 \\
&\Rightarrow 2 n=2.02 \Rightarrow n=1.01
\end{aligned}
$$
$\therefore \quad$ It is 1 st normal mode of vibration.
In a pipe, with both ends open
$$
v_n=n \times \frac{v}{2 L}=\frac{n \times 340}{2 \times 0.2}=430
$$
$$
\therefore \quad n=\frac{430 \times 2 \times 0.2}{340}=0.5
$$
$\because n$ should be an integer, therefore it is not in resonance with the source.
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