A vibrating string of certain length l under | vedclass

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(A) The frequency of the first overtone (third harmonic) of a closed organ pipe is given by $f = \frac{3v}{4L}$.Given $v = 340 \ m/s$ and $L = 0.75 \

Vedclass · Accepted Answer

$344$

Vedclass · Answer

(A) The frequency of the first overtone (third harmonic) of a closed organ pipe is given by $f = \frac{3v}{4L}$.Given $v = 340 \ m/s$ and $L = 0.75 \ m$,we have $f = \frac{3 	imes 340}{4 	imes 0.75} = \frac{1020}{3} = 340 \ Hz$.The string resonates at $340 \ Hz$,so the frequency of the string $f_s = 340 \ Hz$.Beats produced with a tuning fork of frequency $n$ is $|f_s - n| = 4$,so $n = 340 \pm 4$,which means $n = 344 \ Hz$ or $336 \ Hz$.When tension $T$ is increased,the frequency of the string $f_s$ increases $(f_s \propto \sqrt{T})$.If $n = 344 \ Hz$,the beat frequency $|344 - f_s|$ decreases from $4$ to $2$ as $f_s$ increases towards $344 \ Hz$.If $n = 336 \ Hz$,the beat frequency $|f_s - 336|$ would increase as $f_s$ increases from $340 \ Hz$.Therefore,the correct frequency $n$ is $344 \ Hz$.

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