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(B) The condition for the first resonance in a closed organ pipe is given by: $\frac{V}{4(\ell + e)} = f$,where $V$ is the speed of sound,$\ell$ is th

Vedclass · Accepted Answer

$15.2$

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(B) The condition for the first resonance in a closed organ pipe is given by: $\frac{V}{4(\ell + e)} = f$,where $V$ is the speed of sound,$\ell$ is the length of the air column,$e$ is the end correction,and $f$ is the frequency.First,calculate the end correction $e$. For a tube of radius $r$,$e = 0.6r$. Given diameter $d = 4 \ cm$,so $r = 2 \ cm$. Thus,$e = 0.6 	imes 2 = 1.2 \ cm$.Rearranging the formula for $\ell$: $\ell = \frac{V}{4f} - e$.Substitute the given values: $V = 336 \ m/s = 33600 \ cm/s$,$f = 512 \ Hz$,and $e = 1.2 \ cm$.$\ell = \frac{33600}{4 	imes 512} - 1.2 = \frac{33600}{2048} - 1.2 = 16.406 - 1.2 = 15.206 \ cm$.Rounding to the nearest value,the reading is $15.2 \ cm$.

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