An organ pipe P_1,closed at one end,is vibrat | vedclass

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(A) Let $v$ be the speed of sound in air.For a pipe $P_1$ closed at one end,the frequency of the first harmonic (fundamental frequency) is given by $f

Vedclass · Accepted Answer

$1/6$

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(A) Let $v$ be the speed of sound in air.For a pipe $P_1$ closed at one end,the frequency of the first harmonic (fundamental frequency) is given by $f_1 = \frac{v}{4l_1}$.For a pipe $P_2$ open at both ends,the frequency of the third harmonic is given by $f_2 = \frac{3v}{2l_2}$.Since both pipes are in resonance with the same tuning fork,their frequencies are equal: $f_1 = f_2$.Therefore,$\frac{v}{4l_1} = \frac{3v}{2l_2}$.Simplifying the equation: $\frac{1}{4l_1} = \frac{3}{2l_2}$.Rearranging for the ratio of lengths: $\frac{l_1}{l_2} = \frac{2}{12} = \frac{1}{6}$.Thus,the ratio of the lengths of $P_1$ and $P_2$ is $1/6 \approx 0.167$.

An organ pipe $P_1$,closed at one end,is vibrating in its first harmonic. Another pipe $P_2$,open at both ends,is vibrating in its third harmonic. Both are in resonance with a given tuning fork. The ratio of the lengths of $P_1$ and $P_2$ is:

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