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 m/s) in air from these data.

For the tube open at one end, the resonance frequencies are  $\frac{n\upsilon }{4l}$  , where n  is a positive odd integer. If the tuning 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 

$\frac{n\upsilon }{4l_1}=v \frac{(n+2)\upsilon }{4l_2}=v \frac{(n+4)\upsilon }{4l_3}=v$

Giving  $l_3-l_2=l_2-l_1=\frac{2\upsilon }{4v}=\frac{\upsilon }{2v}$    By the question,  $l_3-l_2=(52.75-31.25)cm=21.50cm$  and  $l_2-l_1=(31.25-9.75)cm=21.50cm$

Thus,  $\frac{\upsilon }{2v}=21.50cm$   $or,\upsilon =2v\times 21.50cm=2\times 800s^{-1}\times 21.50cm=344m{s}^{-1}$