Sound waves of `v=60 Hz` 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 (Given, speed of sound `=300 ms^(-1)`)

Options

A. `7/8 m`
B. `3/8 m`
C. `1/8 m`
D. `1/4 m`

Answer

C

Explanation

The wall acts like a rigid boundary and reflects this wave and sends it back towards the open end. At the open end an antinode is formed and a node is formed at the wall. The distance between antinode and node is `lambda/4`. Therefore, if v be the frequency of note emitted, then
`lambda=v/v=300/600=1/2 m`
Maximum amplitude is obtained at distance
`=lambda/4=1/2xx1/4=1/8 m`