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Two Successive Resonance Frequencies in an Open Organ Pipe Are 1944 Hz and 2592 Hz. Find the Length of the Tube. - Physics

Sum

Two successive resonance frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air is 324 ms−1.

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Solution

Given:
Velocity of sound in air v = 324 ms−1
Let l be the length of the resonating column.
Then, the frequencies of the two successive resonances will be \[\frac{(n + 2)v}{4I}  \text { and } \frac{nv}{4I}\]

As per the question,

\[\frac{\left( n + 2 \right)v}{4l}\]= 2592

\[\frac{nv}{4l}\]= 1944

So ,

\[\frac{(n + 2)v}{4l} - \frac{nv}{4I} = 2592 - 1944 = 648\] 

\[ \Rightarrow \frac{2v}{4l} = 648\] 

\[ \Rightarrow l = \frac{2 \times 324 \times 100}{4 \times 648}\text { cm } = 25  \text { cm }\]

Hence, the length of the tube is 25 cm.

Concept: Speed of Wave Motion
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 47 | Page 355
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