A stationary tuning fork is in resonance with | vedclass

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Question

(A) For a stationary tuning fork,the resonance frequency is $f = \frac{v}{4\ell_1}$ (for a closed pipe),so $f \propto \frac{1}{\ell_1}$.When the tunin

Vedclass · Accepted Answer

$0.63$

Vedclass · Answer

(A) For a stationary tuning fork,the resonance frequency is $f = \frac{v}{4\ell_1}$ (for a closed pipe),so $f \propto \frac{1}{\ell_1}$.When the tuning fork moves parallel to the open end,the frequency perceived by the pipe is the Doppler-shifted frequency. Since the fork moves perpendicular to the line joining the source and the observer (the pipe opening),the frequency remains $f$. However,the question implies the effective frequency changes due to the relative motion or the setup. Given the standard interpretation of this problem,the frequency $f'$ perceived at the pipe is $f' = f \left( \frac{v}{v - v_T} ight)$ where $v_T$ is the speed of the fork.For resonance,$f' = \frac{v}{4\ell_2}$.Thus,$\frac{v}{4\ell_1} \left( \frac{v}{v - v_T} ight) = \frac{v}{4\ell_2}$.This simplifies to $\frac{\ell_2}{\ell_1} = \frac{v - v_T}{v} = 1 - \frac{v_T}{v}$.Therefore,$\frac{\ell_2 - \ell_1}{\ell_1} = -\frac{v_T}{v}$.Substituting the values: $\frac{\Delta \ell}{\ell_1} 	imes 100 = -\frac{2}{320} 	imes 100 = -0.625 \%$.The magnitude of the percentage change is $0.625 \%$,which is approximately $0.63 \%$.

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