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//s` 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 `340 m//s` , the frequency `n` the tuning fork in `H_(Z)` is (a) `344` (b) `336` (c ) `117.3` (d) `109.3`

A vibrating string of certain length `l` under a tension `T` resonates

1.	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//s` 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 `340 m//s` , the frequency `n` the tuning fork in `H_(Z)` is (a) `344` (b) `336` (c ) `117.3` (d) `109.3`
Answer» 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 or closed pipe should be less than the frequency of tuning fork by `4` . `:.` Frequency of tuning fork = third harmonic frequency of closed pipe `+ 4` = `3 ((upsilon)/(4l)) + 4 = 3 ((340)/(4xx0.75)) + 4` = `344 H_(Z)` `:.` Correct option is (a).

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