A tuning fork is vibrating at 250, {Hz}. The | vedclass

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(B) For the shortest closed organ pipe,the fundamental frequency corresponds to the first harmonic,where the length of the pipe $\ell$ is equal to one

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$34$

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(B) For the shortest closed organ pipe,the fundamental frequency corresponds to the first harmonic,where the length of the pipe $\ell$ is equal to one-fourth of the wavelength $\lambda$.$\ell = \frac{\lambda}{4} \Rightarrow \lambda = 4\ell$The frequency $f$ is given by the relation $f = \frac{v}{\lambda}$,where $v$ is the speed of sound.Substituting $\lambda = 4\ell$ into the frequency formula:$f = \frac{v}{4\ell}$Given $f = 250\, {Hz}$ and $v = 340\, {ms}^{-1}$,we solve for $\ell$:$250 = \frac{340}{4\ell}$$4\ell = \frac{340}{250} = 1.36\, {m}$$\ell = \frac{1.36}{4} = 0.34\, {m}$Converting to centimeters:$\ell = 0.34 	imes 100 = 34\, {cm}$

$A$ tuning fork is vibrating at $250\, {Hz}$. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ..... ${cm}$.
(Take speed of sound in air as $340\, {ms}^{-1}$)

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$A$ tuning fork is vibrating at $250\, {Hz}$. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ..... ${cm}$.(Take speed of sound in air as $340\, {ms}^{-1}$)