Question

An organ pipe of length L is open at one end and closed at other end. The wavelengths of the three lowest frequencies that can be produced by this pipe are

• A. $2L,L,\frac{L}{2}$
• B. $2L,L,\frac{2L}{3}$
• C. $4L,2L,L$
• D. $4L,\frac{4L}{3},\frac{4L}{5}$

Answer 1

The correct option is D $4L,\frac{4L}{3},\frac{4L}{5}$

As organ pipe is closed:

$\frac{\lambda }{4}+\frac{n\lambda }{2}=l$

$\Rightarrow (2n+1)\frac{\lambda }{4}=l$

$\Rightarrow \lambda =\frac{4l}{2n+1}$

First three lowest wavelengths are $4l,\frac{4l}{3},\frac{4l}{5}$