If $l_1$ and $l_2$ are the lengths of air column for the first and second resonance when a tuning fork of frequency $n$ is sounded on a resonance tube, then the distance of the displacement antinode from the top end of the resonance tube is:
A$2\left( {{l_2} - {l_1}} \right)$
B$\frac{1}{2}\left( {{2l_{2 }}-{l_1}} \right)$
C$\frac{{{l_2} - 3{l_1}}}{2}$
D$\frac{{{l_2} - {l_1}}}{2}$
Medium
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C$\frac{{{l_2} - 3{l_1}}}{2}$
c If $l_{1}=\frac{\lambda}{4},$ and $l_{2}=\frac{3 \lambda}{4},$ and according to the image, displacement anti node from the top should be zero,
therefore,
distance $=\left(l_{2}-3 l_{1}\right) / 2=0$
answer $C$ is correct.
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