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 bythe 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 end , the resonance frequencies are `(nv)/(4l)`, where n is a positive odd integer. If the turning fork has a frequency v and `l_1,l_2,l_3` are the successive lengths of the tube in resonance with it, we have
`(nv)/(4l_1)=v`
`((n+2)v)/(4l_2)=v`
`((n+4)v)/(4l_2)=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 =21.50 cm and `l_2-l_1` = (31.25-9.57) cm=21.50 cm
Thus, `v/(2v)` =21.50 cm
or v=2v x 21.50 cm = `2xx800s^(-1)xx21.50` cm =344 m/s