Q. Sound waves of v=600Hz fall normally on a perfectly reflecting wall. The shortest distance from the wall at which all particles will have maximum amplitude of vibration will be (speed of sound=$300ms^{- 1}$)
A) $\frac{7}{8}m$
B) $\frac{3}{8}m$
C) $\frac{1}{8}m$
D) $\frac{1}{4}m$
Explanation
(c) 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
$\frac{\lambda}{4}$
Therefore, if v be the frequency of note emitted then
$\lambda = \frac{v}{v}$
$\Longrightarrow \ \lambda = \frac{300}{600} = \frac{1}{2}m\ $
Maximum amplitude is obtained at distance
$= \frac{\lambda}{4} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.