Question

Sound waves of frequency 660 Hz fall 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 is 330 m/s)
A. 0.125 m
B. 0.5 m
C. 0.25 m
D. 2 m

Answer 1

Hint: The velocity of sound is equal to the product of the wavelength of the sound waves and the frequency of the sound waves. The standing waves are formed after reflection of the sound waves from the wall. The first maxima is formed at a distance equal to one fourth of the wavelength of the sound waves.
Formula used:
The relation between velocity of the sound, the frequency and the wavelength of the sound is given as
$v = \nu \lambda $

Complete answer:
We are given sound waves which have frequency given as
$\nu = 660Hz$
We are given the velocity of the sound to be
$v = 330m/s$
Using the relation between velocity of the sound, the frequency and the wavelength of the sound, we can calculate the wavelength of the sound in the following way.
$\lambda = \dfrac{v}{\nu }$
Inserting the known values, we get
$\lambda = \dfrac{{330}}{{660}} = 0.5m$
When sound waves are made incident on the reflecting wall then sound waves get reflected from the wall. The incident waves and the reflected waves interfere with each other forming standing waves. The shortest distance from the wall at which the air particles have maximum amplitude of vibration is at a distance of one fourth the wavelength of the sound waves. It is given as
$\dfrac{\lambda }{4} = \dfrac{{0.5}}{4} = 0.125m$

So, the correct answer is “Option A”.

Note:
Since sound waves require a medium to travel. The particles of the medium vibrate when sound waves pass through them. At the maximum position, the particles of the medium have maximum displacement from their mean position while at mean position, there is zero displacement from the mean position and this is called a node.