Question

A Kundt's tube apparatus has a steel rod of length 1.0 m clamped at the centre. It is vibrated in its fundamental mode at a frequency of 2600 Hz. The lycopodium powder dispersed in the tube collects into heaps separated by 6.5 cm. Calculate the speed of sound in steel and in air.

Answer 1

Given:
Length at which steel rod is clamped l = $\frac{1}{2}=0.5\mathrm{m}$
Fundamental mode of frequency f = 2600 Hz
Distance between the two heaps $∆l$ = 6.5 cm = $6.5\times {10}^{-2}\mathrm{m}$

Since Kundt's tube apparatus is a closed organ pipe, its fundamental frequency is given by:

$$\begin{gathered} f=\frac{v_{\mathrm{air}}}{4L} \\ \Rightarrow v_{\mathrm{air}}=f\times 2\times ∆L \\ \Rightarrow v_{air}=2600\times 2\times 6.5\times {10}^{-2}=338\mathrm{m}/\mathrm{s} \\ \left(\mathrm{b}\right)\frac{v_{steel}}{v_{air}}=\frac{2\times l}{∆l} \\ \Rightarrow v_{steel}=\frac{2l}{∆l}\times v_{air} \\ \Rightarrow v_{steel}=\frac{2\times 0.5\times 338}{6.5\times {10}^{-2}} \\ \Rightarrow v_{steel}=5200\mathrm{m}/\mathrm{s} \end{gathered}$$