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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

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प्रश्न

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.

योग
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उत्तर

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.

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Speed of Wave Motion
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अध्याय 16: Sound Waves - Exercise [पृष्ठ ३५५]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 16 Sound Waves
Exercise | Q 47 | पृष्ठ ३५५

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