Question

An open organ pipe of length $L$ resonates at fundamental frequency with closed organ pipe. Find the length of closed organ pipe.

• A. $2L$
• B. $L\sqrt{3/2}$
• C. $\frac{L}{2}$
• D. $L\sqrt{2}$

Answer 1

The correct option is C $\frac{L}{2}$

Let $f_0$ and $f_C$ be the fundamental frequencies of open and closed organ pipe.

$L$ and $L_C$ be the length of an open and closed organ pipe.
consider closed pipe, $L_C=\frac{{\lambda }_1}{4}$
$\Rightarrow {\lambda }_1=4L_C$

Consider open pipe, $L=\frac{{\lambda }_2}{2}\Rightarrow {\lambda }_2=2L$

Wavelength can be expressed as,
$\Rightarrow \lambda =\frac{v}{f}\Rightarrow f=\frac{v}{\lambda }$

Here,
$f_O=f_C$

$\Rightarrow \frac{v}{{\lambda }_1}=\frac{v}{{\lambda }_2}$

$\Rightarrow \frac{v}{2L}=\frac{v}{4L_C}$

$\Rightarrow L_C=\frac{L}{2}$

Final Answer: (b)