A string `25cm` long and having a mass of `2.5 gm` is under tension. A pipe closed at one end is `40cm` long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, `8` beats per second are heard. It is observed that decreasing the tension in the string decreases beat frequency. if the speed of sound in air is `320m//s`, find the tension in the string.

Answer

B, D

Explanation

,

First Overtone
`Mass of string per unit length = (2.5 xx 10^(-3))/(0.25) = 0.01 kg//m`
:. Frequency, `v_(s) = sqrt((T)/(m)) = (1)/(0.25)sqrt((T)/(0.01))` …(i)
Fundamental frequency
:. `(lambda)/(4) = 0.4` rArr `lambda = 1.6m`
:. `v_(T) = (c)/(lambda_(T) = (320)/(1.6) = 200 Hz` ...(ii)
Given that `8 beats//second are heard. the beat frequency decreases with the decreasing tension. this means that beat frequency decreases with decreasing `v_(s)`. So beat frequency is given by the expression.
`v = v_(s) - v_(T)`
:. `8 = (1)/(0.25)sqrt((T)/(0.01)) - 200` rArr `T = 27.04 N`