Question

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.

Answer 1

For the tube open at one end, the resonance frequencies are $\frac{nv}{4l}$, 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
$\frac{nv}{4l_1}=v$
$\frac{(n+2)v}{4l_2}=v$
$\frac{(n+4)v}{4l_2}=v$
giving $l_3-l_2=l_2-l_1=\frac{2v}{4v}=\frac{v}{2v}$
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, $\frac{v}{2v}=21.50$cm
or $v=2v\times 21.50cm=2\times 800s^{-1}\times 21.50cm=344m/s$.