In a resonance tube experiment a student using 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 changed. Successive resonances are observed at lengths 9.75 cm, 31.25 cm and 52.75 cm. the approximate speed of sound in air from this data will be?

For the tube open at one end, the resonance frequencies are $\frac{\mathrm{nv}}{4\mathrm{\ell }}$, where n is a positive odd integer. If the tuning fork has a frequency f and f₁, f₂, f₃ are the successive lengths of the tube in resonance with it, we have
$\frac{\mathrm{nv}}{4{\mathrm{\ell }}_1}={\mathrm{f}}_1;\frac{(\mathrm{n}+2)\mathrm{v}}{4{\mathrm{\ell }}_2}={\mathrm{f}}_2;\frac{(\mathrm{n}+4)\mathrm{v}}{4{\mathrm{\ell }}_3}={\mathrm{f}}_3$
giving ${\mathrm{f}}_3-{\mathrm{f}}_2={\mathrm{f}}_2-{\mathrm{f}}_1=\frac{2\mathrm{v}}{4\mathrm{v}}=\frac{\mathrm{v}}{2\mathrm{v}}$
By question,${\mathrm{f}}_3-{\mathrm{f}}_2=52.75-31.25=21.5\mathrm{cm}$
and ${\mathrm{f}}_2-{\mathrm{f}}_1=31.25-9.75=21.5\mathrm{cm}$
Thus $\frac{\mathrm{v}}{2f}=21.50$
$\mathrm{v}=2\mathrm{f}\times 21.50=2\times 800\times 0.2150=344 \mathrm{m}/\mathrm{s}$