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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 $75 \mathrm{~cm}$ inside a tube closed at one end. The string also generates 4 beat/s 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 $340 \mathrm{~m} / \mathrm{s}$, the frequency $n$ of the tuning fork (in $\mathrm{Hz}$ ) is
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Verified Answer
The correct answer is:
344
344
With increase in tension, frequency of vibrating string will increase. Since number of beats are decreasing. Therefore, frequency of vibrating string or third harmonic frequency of closed pipe should be less than the frequency of tuning fork by 4 .
$\therefore$ Frequency of tuning fork $=$ Third harmonic frequency of closed pipe $+4$
$$
=3\left(\frac{v}{4 l}\right)+4=3\left(\frac{340}{4 \times 0.75}\right)+4=344 \mathrm{~Hz} .
$$
$\therefore$ correct option is (a).
$\therefore$ Frequency of tuning fork $=$ Third harmonic frequency of closed pipe $+4$
$$
=3\left(\frac{v}{4 l}\right)+4=3\left(\frac{340}{4 \times 0.75}\right)+4=344 \mathrm{~Hz} .
$$
$\therefore$ correct option is (a).
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