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(A) The frequency of the first overtone (third harmonic) of a closed organ pipe of length $L = 75\, cm = 0.75\, m$ is given by $f = \frac{3v}{4L}$.Sub

Vedclass · Accepted Answer

$344$

Vedclass · Answer

(A) The frequency of the first overtone (third harmonic) of a closed organ pipe of length $L = 75\, cm = 0.75\, m$ is given by $f = \frac{3v}{4L}$.Substituting the values: $f = \frac{3 	imes 340}{4 	imes 0.75} = \frac{1020}{3} = 340\, Hz$.Since the string resonates with this mode,the frequency of the string $f_s = 340\, Hz$.Given that the string produces $4\, beats/s$ with a tuning fork of frequency $n$,we have $|n - 340| = 4$,which implies $n = 344\, Hz$ or $n = 336\, Hz$.When the tension $T$ is increased,the frequency of the string $f_s$ increases (since $f_s \propto \sqrt{T}$).If $n = 336\, Hz$,increasing $f_s$ would move it further away from $336\, Hz$,increasing the beat frequency.If $n = 344\, Hz$,increasing $f_s$ from $340\, Hz$ towards $344\, Hz$ decreases the difference $|344 - f_s|$,thus reducing the beat frequency to $2\, beats/s$.Therefore,the frequency of the tuning fork is $344\, Hz$.

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