Sound waves of ν = 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 )

Step-by-step solution

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 λ 4 Therefore, if v be the frequency of note emitted then λ = v v ⟹   λ = 300 600 = 1 2   m   Maximum amplitude is obtained at a distance = λ 4 = 1 2 × 1 4 = 1 8   m