Two tuning forks `A and B` are sounded together and `8 beats//s` are heard . `A` is in resonance with a column of air `32 cm` long in a pipe closed at one end and `B` is increased by one `cm`. Calculate the frequency of fork .

Answer

`264 Hz , 256 Hz`

Explanation

`n_(A) - n_(B) = 8`
Also
`n_(A) = (v)/( 4 (0.32)) , n_(B) = (v)/( 4 (0.33))`
`(v)/( 4 (0.32)) - (v)/( 4(0.33) = 8`
`v = 338 m//s`
`:. n_(A) = (v)/( 4 xx 0.32) = (338)/(4 xx 0.32) = 264 Hz`.
`n_(B) = n_(A) - 8 = 256 Hz`.