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?

Answer 1

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.