Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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The Separation Between a Node and the Next Antinode in a Vibrating Air Column is 25 Cm. If the Speed of Sound - Physics

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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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 39 | Page 355
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