Question

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

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

Answer 1

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

If organ pipe is closed, expression for the length,

$\Rightarrow L=(2n-1)\frac{\lambda }{4}$

$\Rightarrow \lambda =\frac{4L}{(2n-1)}\dots \left(i\right)$

Substitute different $n$ values for finding wavelengths of the three lowest resonating frequencies,

$\Rightarrow {\lambda }_1=\frac{4l}{\left(2\right(1)-1)}=4L$

$\Rightarrow {\lambda }_2=\frac{4L}{\left(2\right(2)-1)}=\frac{4L}{3}$

$\Rightarrow {\lambda }_3=\frac{4L}{\left(2\right(3)-1)}=\frac{4L}{5}$

Final answer :$\left(d\right)$