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.

#### Solution

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 :

\[f = \frac{v_{air}}{4L}\]

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

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

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

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

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

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