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Question
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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Solution
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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