A travelling wave y=A sin (k x-ω t+θ) passes from a heavier string to a lighter string. The reflected wave has amplitude 0.5 A. The junction of the strings is at x=0. The equation of the reflected wave is:

Question

A travelling wave y=A sin (k x-ω t+θ) passes from a heavier string to a lighter string. The reflected wave has amplitude 0.5 A. The junction of the strings is at x=0. The equation of the reflected wave is: (A) y'=0.5 A sin (k x+ω t+θ) (B) y'=-0.5 A sin (k x+ω t+θ) (C) y'=-0.5 A sin (ω t-k x-θ) (D) y'=-0.5 A sin (k x+ω t-θ)

Tardigrade Admin · Accepted Answer

As wave has been reflected from a rarer medium, therefore there is no change in phase. Hence equation for the opposite direction can be written as y =0.5 A sin (-k x-ω t+θ) =-0.5 A sin (k x+ω t-θ)

Q. A travelling wave $y = A sin (k x - ω t + θ)$ passes from a heavier string to a lighter string. The reflected wave has amplitude $0.5 A$ . The junction of the strings is at $x = 0$ . The equation of the reflected wave is:

$y^{'} = 0.5 A sin (k x + ω t + θ)$

$y^{'} = - 0.5 A sin (k x + ω t + θ)$

$y^{'} = - 0.5 A sin (ω t - k x - θ)$

$y^{'} = - 0.5 A sin (k x + ω t - θ)$

As wave has been reflected from a rarer medium, therefore there is no change in phase. Hence equation for the opposite direction can be written as
$y = 0.5 A sin (- k x - ω t + θ)$
$= - 0.5 A sin (k x + ω t - θ)$

Q. A travelling wave y=Asin(kx−ωt+θ) passes from a heavier string to a lighter string. The reflected wave has amplitude 0.5A. The junction of the strings is at x=0. The equation of the reflected wave is:

y′=0.5Asin(kx+ωt+θ)

y′=−0.5Asin(kx+ωt+θ)

y′=−0.5Asin(ωt−kx−θ)

y′=−0.5Asin(kx+ωt−θ)