In the Kundt's tube experiment (shown in fig. | vedclass

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(B) In the rod clamped at the end,the fundamental mode of vibration corresponds to a quarter wavelength,so $L = \frac{\lambda_r}{4}$,which gives $\lam

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$1100$

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(B) In the rod clamped at the end,the fundamental mode of vibration corresponds to a quarter wavelength,so $L = \frac{\lambda_r}{4}$,which gives $\lambda_r = 4L$. The frequency is $f = \frac{v_r}{\lambda_r} = \frac{v_r}{4L}$.In the air column,the distance between successive nodes (powder heaps) is $\frac{\lambda_a}{2} = 0.6\ m$,so $\lambda_a = 1.2\ m$. The frequency of the air column is $f = \frac{v_a}{\lambda_a} = \frac{330}{1.2} = 275\ Hz$.Since the rod and air column vibrate at the same frequency,we have $\frac{v_r}{4L} = 275$.Substituting $L = 1\ m$,we get $v_r = 275 	imes 4 = 1100\ m/s$.

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