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 `75cm` 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 `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. 120 Hz
B. 336 Hz
C. 240 Hz
D. 344 Hz

Answer

D

Explanation

For a pipe closed at one end, the frequency of the third harmonic is given by
`f = (3v)/(4L) = (3 xx 340)/(4 xx 0.75) = 340 ` Hz
` :. ` Frequency of the string = 340 Hz
There are 4 beats/s, produced with a tuning fork of frequency n ,
` :. n = 340 + 4 = 344 Hz " or " n = 340 - 4 = 336 ` Hz
For a vibrating string ` n propto sqrtT`
` :. ` If the tension is slightly increased, its frequency will increase.
Case (1) : If ` n = 336` Hz, then if the frequency of the string increases say becomes , 342 , then the no. of beats/s will increase to 6 ( it will increase). This is not possible because the number of beats is decreased to 2.
Case (2) : But if ` n = 344` Hz and if the frequency of the string becomes 342 , then no . of beats/s will be 2. This is possible .
` :. ` The frequency of the fork ` = 344 ` hz