The first overtone frequency of a closed organ pipe *P*_{1} is equal to the fundamental frequency of a open organ pipe *P*_{2}. If the length of the pipe *P*_{1} is 30 cm, what will be the length of *P*_{2}?

#### Solution

Given:

Length of closed organ pipe *L*_{1} = 30 cm

Length of open organ pipe *L*_{2} = ?

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 *P*_{1} 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 P_{2} will be 20 cm.