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Question
The separation between a node and the next antinode in a vibrating air column is 25 cm. If the speed of sound in air is 340 m s−1, find the frequency of vibration of the air column.
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Solution
Given:
Separation between the node and anti-node = 25 cm
Speed of sound in air v = 340 ms−1
Frequency of vibration of the air column f = ?
The distance between two nodes or anti-nodes is λ.
We have :
\[\frac{\lambda}{4} = 25 \text { cm }\]
\[ \Rightarrow \lambda = 100 \text { cm } = 1 \text { m }\]
Also ,
\[v = f\lambda\]
\[\Rightarrow f = \frac{v}{\lambda} = \frac{340}{1} = 340 \text { Hz }\]
Hence, the frequency of vibration of the air column is 340 Hz.
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