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 = \[\frac{1}{2} = 0 . 5  \text { m }\]

Fundamental mode of frequency f = 2600 Hz
Distance between the two heaps \[∆ l\] = 6.5 cm = \[6 . 5 \times  {10}^{- 2} \text { m }\]

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

\[f = \frac{v_{air}}{4L}\] 

\[ \Rightarrow  v_{air}  = f \times 2 \times  ∆ L\] 

\[ \Rightarrow  v_{air}  = 2600 \times 2 \times 6 . 5 \times  {10}^{- 2}  = 338 \text {  m/s }\] 

\[(b)  \frac{v_{steel}}{v_{air}} = \frac{2 \times l}{∆ l}\] 

\[ \Rightarrow  v_{steel}    = \frac{2l}{∆ l} \times  v_{air} \] 

\[ \Rightarrow    v_{steel}  = \frac{2 \times 0 . 5 \times 338}{6 . 5 \times {10}^{- 2}}\] 

\[ \Rightarrow    v_{steel}  = 5200  \text { m/s }\]