A vibrating string of certain length `l` under a tension `T` resonates with a maode corresponding to the first overtone (third harmonic) of an air column of length `75cm` inside a tube closed at one end. The string also gereates `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 ti be `340 m//s`, the frequency `n` of the tuning fork in `Hz` is

Options

A. (a) `344`
B. (b) `336`
C. ( c ) `117.3`
D. ( d ) `109.3`

Answer

A

Explanation

(a) The frequency `(v) produced by the air column is given by
`lambda xx v = v` rArr `v = (v)/(lambda)`
Also, `(3lambda)/(4) = l = 75cm = 0.75m`
:. `lambda = (4 xx 0.75)/(3)` rArr `v = (340 xx 3)/(4 xx 0.75) = 340 Hz`
:. The `frequency of vibrating string = 340`. since this string produced `4beats//sec` with a tuning fork of frequency `n` therefore `n = 340 + 4` or `n = 340 - 4`. With increase in tension, the frequency produced by string increases. As the beats//sec decreases therefore `n = 340 + 4 = 344 Hz`.