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(A) For a closed resonance tube,the resonance lengths are given by $l_1 = \frac{\lambda}{4}$ and $l_2 = \frac{3\lambda}{4}$.The difference between the

Vedclass · Accepted Answer

$500$

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(A) For a closed resonance tube,the resonance lengths are given by $l_1 = \frac{\lambda}{4}$ and $l_2 = \frac{3\lambda}{4}$.The difference between the two resonance lengths is $l_2 - l_1 = \frac{\lambda}{2}$.Given $l_1 = 16 \ cm = 0.16 \ m$ and $l_2 = 49 \ cm = 0.49 \ m$.Thus,$\frac{\lambda}{2} = 0.49 \ m - 0.16 \ m = 0.33 \ m$,which implies $\lambda = 0.66 \ m$.Using the wave equation $v = n\lambda$,where $v$ is the velocity of sound and $n$ is the frequency:$n = \frac{v}{\lambda} = \frac{330 \ m/s}{0.66 \ m} = 500 \ Hz$.

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

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