Question

Sound waves of $\nu=600 \, Hz$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which all particles will have a maximum amplitude of vibration will be (speed of sound= $300 \, m \, s^{- 1}$ )

• A. $\frac{7}{8}m$
• B. $\frac{3}{8} \, m$
• C. $\frac{1}{8} \, m$
• D. $\frac{1}{4} \, m$

Answer 1

Answer: (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 a distance
$=\frac{\lambda }{4}=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}m$