MCQ
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
    $\frac{1}{2}\left( {{2l_{2  }}-{l_1}} \right)$
  • $\frac{{{l_2} - 3{l_1}}}{2}$
  • D
    $\frac{{{l_2} - {l_1}}}{2}$

Answer

Correct option: 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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