Explain the superposition principle for waves forming in calm water.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The figure $(a)$ shows two needles oscillating with equal phase,representing two coherent sources. Consider two needles moving periodically up and down in an identical fashion in a trough of water as shown in the figure; they produce two water waves.
At any particular point,the phase difference between the displacements produced by each of the waves does not change with time,so both sources are called coherent sources.
Figure $(b)$ shows the position of crests (solid circles) and troughs (dashed circles) at a given instant of time.
Now consider a point $P$ as shown in figure $(a)$ for which $S_{1}P = S_{2}P$.
Since waves from $S_{1}$ and $S_{2}$ will take the same time to travel to point $P$,they arrive with the same phase.
The displacement produced by the source $S_{1}$ at point $P$ is given by $y_{1} = a \cos \omega t$,and the displacement produced by the source $S_{2}$ at point $P$ is given by $y_{2} = a \cos \omega t$,where $a$ is the amplitude.
According to the superposition principle,the resultant displacement at $P$ is:
$y = y_{1} + y_{2} = a \cos \omega t + a \cos \omega t$
$\therefore y = 2a \cos \omega t$

Explore More

Similar Questions

State the principle of superposition and explain it.

Two waves have equations $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$. If the frequency and amplitude of the resultant wave remain equal to those of the superimposing waves,then the phase difference between them is:

Two small identical speakers are connected in phase to the same source. The speakers are $3 \,m$ apart and at ear level. An observer stands at $P, 4 \,m$ in front of one speaker as shown. The sound she hears is least intense when the wavelength is $\lambda_1$ and most intense when the wavelength is $\lambda_2$. Then,the possible values of $\lambda_1$ and $\lambda_2$ are

Two sources of sound $A$ and $B$ produce waves of $350 Hz$,and they vibrate in the same phase. $A$ particle $P$ is vibrating under the influence of these two waves. If the amplitudes at point $P$ produced by the two waves are $0.3 mm$ and $0.4 mm$,what will be the resultant amplitude of the point $P$ when $AP - BP = 25 cm$ and the velocity of sound is $350 m/s$?

Difficult
View Solution

There is a destructive interference between two waves of wavelength $\lambda$ coming from two different paths at a point. To get maximum sound or constructive interference at that point,the path of one wave is to be increased by

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo