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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.
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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 :
\[f = \frac{v_{air}}{4L}\]
\[ \Rightarrow v_{air} = f \times 2 \times ∆ L\]
\[ \Rightarrow v_{air} = 2600 \times 2 \times 6 . 5 \times {10}^{- 2} = 338 \text { m/s }\]
\[(b) \frac{v_{steel}}{v_{air}} = \frac{2 \times l}{∆ l}\]
\[ \Rightarrow v_{steel} = \frac{2l}{∆ l} \times v_{air} \]
\[ \Rightarrow v_{steel} = \frac{2 \times 0 . 5 \times 338}{6 . 5 \times {10}^{- 2}}\]
\[ \Rightarrow v_{steel} = 5200 \text { m/s }\]
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