derive y=asin(kx-wt)in negative direction
Answer (1)
4th Oct, 2020
Hello,
y = A sin(kx-ωt) travels in direction of positive x axis & has vibrations along x axis.
y = A sin(kx+ωt) travels in direction of negative x axis & has vibrations along y axis.
Both equations represent transverse waves.
A particular waveform travels right along the x axis from left.
Hope it helps
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Comments (2)
but could you please derive them
Reply
4th Oct, 2020
shanmukhaadithyaakkiraju
Consider that , a wave form is traveling right along the x axis from left.
Since the particles are oscillating in simple harmonic motion ,then It will maintain y = A sinθ
When t =0, point P is at x distance right from O has phase difference, then oscillation of particle at P maintain y = A sin
Phase difference at λ is 2 π ; at x is (2π/ λ)x
So, y = A sin (2π/ λ)x
Phase velocity - v ; time - t
Wave travels right with v at t then phase lag between particle at origin O and particle of right go on increasing since wave proceeds away from O towards right.
The equation of motion of particle at right is :
y = A sin [( 2π/ λ ) x - ωt] = A sin (kx - ωt)
For left is :
y = A sin ( kx + ωt)
Hope it helps
Since the particles are oscillating in simple harmonic motion ,then It will maintain y = A sinθ
When t =0, point P is at x distance right from O has phase difference, then oscillation of particle at P maintain y = A sin
Phase difference at λ is 2 π ; at x is (2π/ λ)x
So, y = A sin (2π/ λ)x
Phase velocity - v ; time - t
Wave travels right with v at t then phase lag between particle at origin O and particle of right go on increasing since wave proceeds away from O towards right.
The equation of motion of particle at right is :
y = A sin [( 2π/ λ ) x - ωt] = A sin (kx - ωt)
For left is :
y = A sin ( kx + ωt)
Hope it helps
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