Physics - Sound Wave Question with Solution | TestHub

For a pipe open at one end and closed at the other, the boundary conditions require an antinode at the open end and a node at the closed end.

The length of the pipe $L$ is related to the wavelength $\lambda$ by $L=(2n-1)\frac{\lambda }{4}$, where $n=1,2,3,\dots$

The frequencies of the standing waves are given by $f=\frac{v}{\lambda }=(2n-1)\frac{v}{4L}$.

Since $(2n-1)$ represents odd integers, only odd harmonics ($1f_1,3f_1,5f_1,\dots$) are present in a pipe open at one end.

Answer: Odd harmonics