Sum

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.

Concept: Wave Motion

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