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(A) In a resonance tube,the first resonance occurs at length $\ell_1 = \frac{\lambda}{4} - e$ and the second resonance occurs at $\ell_2 = \frac{3\lam

Vedclass · Accepted Answer

$348$

Vedclass · Answer

(A) In a resonance tube,the first resonance occurs at length $\ell_1 = \frac{\lambda}{4} - e$ and the second resonance occurs at $\ell_2 = \frac{3\lambda}{4} - e$,where $e$ is the end correction.Subtracting the two lengths: $\ell_2 - \ell_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2}$.Given $\ell_1 = 17 \, cm$ and $\ell_2 = 51 \, cm$,we have $\frac{\lambda}{2} = 51 - 17 = 34 \, cm$.Therefore,the wavelength $\lambda = 2 	imes 34 \, cm = 68 \, cm = 0.68 \, m$.The speed of sound $v$ is given by $v = f \lambda$,where $f = 512 \, Hz$.$v = 512 	imes 0.68 = 348.16 \, m/s$.Rounding to the nearest integer,the speed of sound is $348 \, m/s$.

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