In an experiment to measure the speed of soun | vedclass

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(B) For two successive resonances at lengths $L_1$ and $L_2$,the half-wavelength is given by $\lambda/2 = L_2 - L_1$.$\lambda/2 = 83.9 	ext{ cm} - 50

Vedclass · Accepted Answer

$A, C$

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(B) For two successive resonances at lengths $L_1$ and $L_2$,the half-wavelength is given by $\lambda/2 = L_2 - L_1$.$\lambda/2 = 83.9 	ext{ cm} - 50.7 	ext{ cm} = 33.2 	ext{ cm}$.Therefore,the wavelength $\lambda = 2 	imes 33.2 	ext{ cm} = 66.4 	ext{ cm}$. (Statement $C$ is true).Speed of sound $v = f \lambda = 500 	ext{ Hz} 	imes 0.664 	ext{ m} = 332 	ext{ m s}^{-1}$. (Statement $A$ is true).For the first resonance,$L_1 + e = \lambda/4$,where $e$ is the end correction.$50.7 	ext{ cm} + e = 66.4 	ext{ cm} / 4 = 16.6 	ext{ cm}$.$e = 16.6 	ext{ cm} - 50.7 	ext{ cm} = -34.1 	ext{ cm}$.Note: The problem setup implies $L_1$ and $L_2$ are the $2^{	ext{nd}}$ and $3^{	ext{rd}}$ harmonics (or higher) because $L_1$ is quite large. The end correction $e$ is typically small and positive. Given the options,$A$ and $C$ are correct.

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