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

Frequency of fundamental mode of closed pipe,
`f_(1) = (nu)/(4l) = 200 H_(Z)`
Decreasing the tension in the string decrease the beat frequenct.
Hence, the first overtone frequency of the string should be `208 H_(Z)` (not `192 H_(Z)`
:. `208 = (1)/(l) sqrt((T)/(mu))`
:. `T = mu. l^(2) (208)`
`= (2.5 xx 10^(-3)/(0.25)) (0.25)^(2) (208)^(2)`
`= 27.04 N`