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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 source of $1237.5$ $\mathrm{Hz}$ ? (sound velocity in air $=330 \mathrm{~ms}^{-1}$ )
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Verified Answer
As given that,
Length of pipe, $l=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$
$$
v=1237.5 \mathrm{~Hz}, v=330 \mathrm{~m} / \mathrm{s}
$$
$l=\frac{\lambda}{4}$ or $\lambda=4 l$
$$
\begin{aligned}
&\left(v_{\text {funda }}\right)_1=\frac{v}{4 l}=\frac{330}{4 \times 20 \times 10^{-2}} \\
&\left(v_{\text {funda }}\right)_1=\frac{330 \times 100}{80}=412.5 \mathrm{~Hz} \\
&\frac{\left(v_{\text {given }}\right)}{\left(v_{\text {funda }}\right)_1}=\frac{1237.5}{412.5}=\frac{3}{1}
\end{aligned}
$$
The frequency in Ist, IInd, IIIrd .... harmonic are in the ratio $1: 2: 3: 4 \ldots \ldots$ at one end open organ pipe.
Length of pipe, $l=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$
$$
v=1237.5 \mathrm{~Hz}, v=330 \mathrm{~m} / \mathrm{s}
$$
$l=\frac{\lambda}{4}$ or $\lambda=4 l$
$$
\begin{aligned}
&\left(v_{\text {funda }}\right)_1=\frac{v}{4 l}=\frac{330}{4 \times 20 \times 10^{-2}} \\
&\left(v_{\text {funda }}\right)_1=\frac{330 \times 100}{80}=412.5 \mathrm{~Hz} \\
&\frac{\left(v_{\text {given }}\right)}{\left(v_{\text {funda }}\right)_1}=\frac{1237.5}{412.5}=\frac{3}{1}
\end{aligned}
$$
The frequency in Ist, IInd, IIIrd .... harmonic are in the ratio $1: 2: 3: 4 \ldots \ldots$ at one end open organ pipe.
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