Question

The fundamental frequency of an open organ pipe is equal to the second overtone of a closed organ pipe. If the length of closed organ pipe is $40\text{cm}$, the length of open organ pipe is

• A. $13.3\text{cm}$
• B. $8\text{cm}$
• C. $12.5\text{cm}$
• D. $16\text{cm}$

Answer 1

The correct option is D $16\text{cm}$

Fundamental frequency of open organ pipe is,
$f_0=\frac{v}{2l_1}$

Second overtone of closed organ pipe is,
$f_2=\frac{5v}{4l_2}$

From the data given in the question,
$f_0=f_2$

$\Rightarrow \frac{v}{2l_1}=\frac{5v}{4l_2}$

$\Rightarrow l_1=\frac{2}{5}{l}_2$

Given, $l_2=40\text{cm}$

$\Rightarrow l_1=\frac{80}{5}=16\text{cm}$