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

A

Explanation

`(1)/(2L)sqrt((T)/(m))=(3v)/(4l)`
where `v = 340ms^(-1),l=75 cm= 0.75m`
Now according to given condition
`n-(1)/(2L)sqrt((T)/(m))=4` so `n=(1)/(2L)sqrt((T)/(m))+4=((3v)/(4l)+4)`
`=(3)/(4)xx(340)/(0.75)+4 = 344 Hz`.