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
$f_1=\frac{3v}{4L_1}$,
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:
$f_2=\left(\frac{v}{2L_2}\right)$
As per the question,
$$\begin{gathered} f_1=f_2 \\ \left(\frac{3\times v}{4\times 30}\right)=\left(\frac{v}{2L_2}\right) \\ \Rightarrow L_2=20\mathrm{cm} \end{gathered}$$
∴ The length of the pipe P₂ will be 20 cm.