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Travelling Waves - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • General equation of travelling is considered one of the most asked concept.

  • 24 Questions around this concept.

Solve by difficulty

The displacement y of a particle in a medium can be expressed as:

y=10^{-6}\sin \left ( 100t +20x+\pi /4 \right )m where t is in second and x in meter. The speed of the wave is:

Concepts Covered - 1

General equation of travelling

The function f(x,t) represents the displacements y of the particle at t=0 and x=x'

y=f(x=x',t=0)=A\sin(kx')

k- propagation constant or angular wave number

A- Amplitude

For a given time, between position x=0 to x=x' the phase changes from 0 to kx'

similarly, x=0 to x=\lambda the phase changes from 0 to 2\pi

\\ x'\rightarrow kx' \\ \lambda \rightarrow 2\pi

k=\frac{2\pi }{\lambda } \\ \text{kx' represents phase of wave at x=x'}

The disturbance travels on the strings towards along positive x-axis with a constant speed 'v'. Thus, the displacement produced at the left end at time 't=0', reaches the point 'x' at time 't=(x-x')/v'.

As wave shape remains same for progressive wave, particle's displacement at x=x', t=0 and x=x'+vt, t=t are same

i.e., y=f(x=x',t=0) is same as y=f(x,t)

Let's now write the equation in terms of the stationary coordinate x, where x'  =  x − vt 

y=f(x',0)= f(x,t) \\ \therefore y(x,t)=A\sin(kx') =A \sin(k(x-vt))

 

y(x,t)=A \sin(k(x-vt))= A \sin (\frac{2\pi x}{\lambda }-\frac{2\pi vt}{\lambda })=A \sin(kx-\omega t)

v=f\lambda=\frac{\lambda }{T} \\ T=\frac{2\pi }{\omega }

y(x,t)=A \sin(k(x-vt))= A \sin (\frac{2\pi x}{\lambda }-\frac{2\pi t}{T })=A \sin(kx-\omega t)

For wave travelling along negative x-axis,

y(x,t)=A \sin(k(x+vt))= A \sin (\frac{2\pi x}{\lambda }+\frac{2\pi t}{T })=A \sin(kx+\omega t)

GENERAL EQUATION OF TRAVELLING WAVE

y(x,t)=A \sin(k(x\pm vt)+\phi )= A \sin (\frac{2\pi x}{\lambda }\pm \frac{2\pi t}{T }+\phi )=A \sin(kx\pm\omega t+\phi )

 

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General equation of travelling

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