Wave motionMediumBloom L3
Question
A wave travels on a light string. The equation of the wave is Y=Asin(kx−wt+30°) .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
Options
A.Y=0.8Asin(kx−wt+30°+180°)
B.Y=0.8Asin(kx+wt+30°+180°)
C.Y=0.8Asin(kx+wt−30°)
D.Y=0.8Asin(kx+wt+30°)
Solution
Sol. Y = Asin(kx – ωτ + 30°)
Reflected wave will have opposite phase because it is reflected back from rigid end.
It will travel towards –ve x axis.
Ereff= 64% of initial energy
E ∝ A2
Y = A sin (kx – ωτ + 30°)
At x = 0
Y = A sin(–ωt + 30°)
Y = Asin(ωt + 150°)
Yref = (0.8A) sin(kx + ωτ + 150° + 180°)
= 0.8 A sin(kx + ωt + 330°)
= 0.8 A sin(kx + ωτ – 30°)
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