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 a 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) $0.2\,m$

Answer 1

When sound waves get reflected from a perfectly reflecting wall, the reflected waves and the incident waves interfere with each other to form an interference pattern. In an interference pattern, the distance between a node and an antinode is equal to a quarter of the wavelength.
$\Rightarrow v = f\lambda$ where $v$ is the velocity of sound in air, $f$ is the frequency, and $\lambda$ is the wavelength of the wave

Complete step by step solution:
When waves are reflected from a perfectly reflecting wall, the reflected waves and incident waves interfere with each and form an interference pattern. So the interference pattern will have node points which are points of zero displacement and antinode points which are points of maximum amplitude. The distance between a node and an antinode is dependent on the wavelength of the wave.
The wavelength of the wave can be calculated using the relation
$\Rightarrow v = f\lambda$
On substituting $v = 330\,m/s$ and $f = 660\,Hz$, we can calculate the wavelength as:
$\Rightarrow \lambda = \dfrac{{330}}{{660}}$
$\Rightarrow 0.5\,m$
For a perfectly reflecting wall, a node is formed at the point of reflection i.e. the wave has zero amplitude at the point of reflection. So the point where the amplitude is maximum is the first antinode of the wave which is situated at a distance of $\lambda /4$ meters away from the wall.
Hence the distance between the wall and the first point where the amplitude is maximum can be calculated as:
$\Rightarrow \dfrac{\lambda }{4} = \dfrac{{0.5}}{4}$
$\Rightarrow 0.125\,m$
Correct answer is option (A).

Note:
To solve such questions, we should talk about the way sound waves interfere and hence form nodes and antinodes. We must also be aware that when nodes are formed at the point of reflection since the incident and the reflected waves cancel each other out at that point which can then help us in breaking the problem down in finding the distance between a node and an antinode.