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The First Overtone Frequency of a Closed Organ Pipe P1 is Equal to the Fundamental Frequency of a Open Organ Pipe P2.

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प्रश्न

The first overtone frequency of a closed organ pipe P1 is equal to the fundamental frequency of a open organ pipe P2. If the length of the pipe P1 is 30 cm, what will be the length of P2?

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उत्तर

Given:
Length of closed organ pipe L1 = 30 cm
Length of open organ pipe L2 = ?
Let \[f_1\] and \[f_2\] be the frequencies of the closed and open organ pipes, respectively.
The first overtone frequency of a closed organ pipe P1 is given by 

\[f_1  = \frac{3v}{4 L_1}\]

where v is the speed of sound in air.
On substituting the respective values, we get :

\[f_1  = \frac{3v}{4 \times 30}\]

Fundamental frequency of an open organ pipe is given by:

\[f_2  = \left( \frac{v}{2 L_2} \right)\]

As per the question,

\[f_1  =  f_2 \] 

\[  \left( \frac{3 \times v}{4 \times 30} \right) = \left( \frac{v}{2 L_2} \right)\] 

\[     \Rightarrow    L_2  = 20  \text { cm }\]

∴ The length of the pipe P2 will be 20 cm.

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पाठ 16: Sound Waves - Exercise [पृष्ठ ३५५]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 16 Sound Waves
Exercise | Q 42 | पृष्ठ ३५५

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