Physics - Sound Wave Question with Solution | TestHub

If we adjust the length of air-column in closed organ pipe as such its any natural frequency equals to the frequency of tuning fork, then the amplitude of forced vibrations of air-column increases very much. This is the state of resonance. At first resonance $l=\frac{\lambda }{4}$ So, frequency of tuning fork $f=\frac{v}{\lambda }=\frac{v}{4l}$Given, $l=20\mathrm{cm}=0.2\mathrm{m},v=332\mathrm{m}/\mathrm{s}$ Hence, $f=\frac{0.332}{4\times 0.2}=415\mathrm{Hz}$