Question

Sound waves of frequency 660 Hz falls normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particles have maximum amplitude of vibration is (velocity of sound in air = 330 m$s^{-1}$)

• A. 0.125 m
• B. 0.5 m
• C. 0.25 m
• D. 2 m

Answer 1

The correct option is A 0.125 m

Frequency of sound waves $\nu =660Hz$ and velocity $v=330m/s$

Now wavelength of the sound waves $\lambda =\frac{v}{\nu }=\frac{330}{660}=0.5m$

Stationary waves will be formed due to the interference between the incident and the reflected wave and the displacement node is formed at the reflecting surface.

Thus the shortest distance between the wall (node) and the point where the air particles have maximum amplitude of vibration (antinode) $d=\frac{\lambda }{4}=\frac{0.5}{4}=0.125m$