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

Options

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`.