1. Tardigrade
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  4. In a resonance tube, the first resonance with a tuning fork occurs at 16 cm and second at 49 cm. If the velocity of sound is 330 m / s, the frequency of tunning fork is

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Q. In a resonance tube, the first resonance with a tuning fork occurs at $16\, cm$ and second at $49\, cm$. If the velocity of sound is $330\, m / s$, the frequency of tunning fork is

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Solution:

For closed pipe $l_{1}=\frac{v}{4 n},$
$l_{2}=\frac{3 v}{4 n}$
$\Rightarrow n=\frac{v}{2\left(l_{2}-l_{1}\right)}$
$=\frac{330}{2 \times(0.49-0.16)}=500\, Hz$