Two open pipes of length $L$ are vibrated simultaneously. If length of one of the pipes is reduced by $y$ , then the number of beats heard per second will nearly be $($ if the velocity of sound is $v$ and $y < L)$

\frac{v y}{2 L}

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\frac{2 L^{2}}{V y}

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\frac{v y}{2 L^{2}}

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\frac{v y}{L^{2}}

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Solution

The correct option is A $\frac{v y}{2 L^{2}}$
In an open pipe, there will be antinodes at the ends of the pipe. The fundamental frequency $($ or the lowest frequency $)$ corresponds to the largest wavelength.
If the length of the pipe is $L$ , $\frac{λ}{2} = L$ . Frequency is related to wavelength as $f = \frac{v}{λ}$ .
Therefore, $f_{1} = \frac{v}{2 L}$ and $f_{2} = \frac{v}{2 (L - y)}$ . Beats per second is

$Δ f = f_{2} - f_{1} = \frac{v}{2} (\frac{1}{L - y} - \frac{1}{L}) = \frac{v}{2} (\frac{y}{L (L - y)}) \approx \frac{v y}{2 L^{2}}$ .
Option "C" is correct.

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Two open pipes of length L are vibrated simultaneously. If length of one of the pipes is reduced by y, then the number of beats heard per second will nearly be (if the velocity of sound is v and y<L)

Two open pipes of length $L$ are vibrated simultaneously. If length of one of the pipes is reduced by $y$ , then the number of beats heard per second will nearly be $($ if the velocity of sound is $v$ and $y < L)$