Question

Sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particle has maximum amplitude of vibration is (velocity of sound in air is 330 m/s)

• 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

Here, sound waves fall normally on a perfectly reflecting wall, hence the incident and reflected wave will form the stationary wave-form. Since there is no displacement of particles at wall hence this point is node.

As the separation between two nodes is $\frac{\lambda }{2}m$, the antinode i.e. the point of displacement of particles is at a distance

$\frac{\frac{\lambda }{2}}{2}=\frac{\lambda }{4}m$ from wall.

Now, we know

$\lambda =\frac{v}{n}$

$\lambda =\frac{330}{660}$ ..............................(given)

$\lambda =\frac{1}{2}m$

Hence, the distance of antinode from wall is

$\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}m$

Therefore the shortest distance from the wall at which the air particle has maximum amplitude of vibration(anti-node) is $\frac{1}{8}m$ i.e. 0.125 m.