Physics - Sound Wave Question with Solution | TestHub

The wall acts like a rigid boundary and reflects this wave and sends it back towards the open end. At the open an antinode is formed and a node is formed at the wall. The distance between antinode and node is

$\frac{\mathrm{\lambda }}{4}$

Therefore, if v be the frequency of note emitted then

$\mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{v}}$

$⟹\mathrm{ }\mathrm{\lambda }=\frac{300}{600}=\frac{1}{2} \mathrm{m}\mathrm{ }$

Maximum amplitude is obtained at a distance

$=\frac{\mathrm{\lambda }}{4}=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8} \mathrm{m}$