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.

Explanation

Linear mass density ofstring is
`mu = (m)/(L) = (2.5 xx 10^(-3))/(0.25)` 0.01`
For first overtone (i.e. second harmonik) the frequency of the string
`n_(1) = (2)/(2L)sqrt((T)/(mu))= (2)/(2xx0.25)sqrt((T)/(0.01)) = 40sqrt(T)`
Fundamental frequency of closed organ pipe is given as
`n_(c) = (v)/(4l) = (320)/(4 xx 0.4) = 200 Hz`
Now as the two produces 8 beats per second, we have
`200 - 40sqrt(T) = 8` [caseI]
`40sqrt(T) - 200 = 8`
Now as decreasing the tension decreases the beat frequency, case-I is not permissible. Thus we have
`40sqrt(T) = 208`
`T = [(208)/(40)]^(2)`
= 27.04 N