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A wave travels on a light string. The eq...

A wave travels on a light string. The equation of the wave is `Y = A sin (kx - omegat + theta))`. It is reflected from a heavy string tied to an end of the light string at `x = 0`. If `64%` of the incident energy is reflected the equation of the reflected wave is

A

`y=0.5Asin(kx+omegat+theta)`

B

`y'=-0.5Asin(kx+omegat+theta)`

C

`y'=-0.5Asin(omegat-kx-theta)`

D

`y'=-0.5Asin(kx+omegat-theta)`

Text Solution

Verified by Experts

The correct Answer is:
D
As wave has benn reflected from a rares medium therefore there is no change in phase. Hence equation for the oppsite direction can be written as
`y=0.1Asin(-kz-omegat+phi)`
`=-0.5Asin(kx+omegat-theta)`
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