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A vibrating string of certain length $l$ under a tension $T$ resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length $75\, cm$ inside a tube closed at one end. The string also generates $4\, beats$ per second when excited along with a tuning fork of frequency $n$. Now when the tension of the string is slightly increased the number of beats reduces to $2\, per second$. Assuming the velocity of sound in air to be $340\, m/s$, the frequency $n$ of the tuning fork in $Hz$ is
$344$
$336$
$117.3$
$109.3$
Solution
${{\rm{n}}_1} = \frac{{3{\rm{V}}}}{{4l}} = \frac{{3 \times 340}}{{4 \times (3/4)}} = 340{\mkern 1mu} {\rm{Hz}}$
for $4$ beats, $n$ is $344 \mathrm{\,Hz}$ or $336 \mathrm{\,Hz}$
If tension in increased $\mathrm{n}_{1}$ should be increased and beats reduces to $2$ per second. So $n$ should be $344 \mathrm{\,Hz}.$