A tuning fork vibrating at frequency 800 Hz produces resonance in a resonance column tube. The upper end is open and the lower end is closed by the water surface which can be varied. Successive resonances are observed at lengths 9.75 cm, 31.25 cm and 52.75 cm. Calculate the speed of sound in air from these data.

Explanation

For the tube open at one the resoN/Asnce frequencies are `(nv)/(4l)` when n is a positive odd interger. If the tuning fork has a frequency v and `l_1, l_2, l_3` are the successive lengths of the tube in resoN/Ance with it, we have
`(nv)/(4l_1)=v`
`((n+2)v)/(4l_2)=v`
`(n+4)v)/(4l_3)=v`
`giving l_3-l_2=l_2-l_1=(2v)/(4v)=v/(2v)`
By the question l_3-l_2=(52.75-31.25)cm=21.50cm` and `l_2-l_1=(31.25-9.75)m=21.50cm`.
`Thus, v/(2v)=21.50cm`
`or v=2vxx21.50cm=2xx800s^-1xx21.50cm=344ms^-1`