A string `25 cm` long and having a mass of `2.5 g` is under tension. A pipe closed at one end is `40 cm` 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 tencion in the string decrease the beat frequency. If the speed of sound in air is `320 m//s`, find the tension in the string.

Answer

`27.0400 N`

Explanation

Frequency of first over tone of string `= (V)/(t)`
`f_(1) = (sqrt((T)/(mu)))/(l_(s))`
fundanebtak frequency of close pipe
`f_(2) = (v)/(4l_(p)) = (320)/(4 xx 0.4) = 200 Hz`
`f_(1)` and `f_(2)` produce `8` beats so `f_(1) = 200 + 8` or `200 - 8` on decreasing tension `f_(1)` will decrease and beats decrease so it must be `200 + 8`
`208 = (sqrt((T)/(((2.5xx10^(-3))/(25xx10^(-2))))))/(25xx10^(-2))`
`T = (208 xx 25 xx 10^(-3))^(2) = 27.04N`