A travelling wave Y=A sin (kx−ωt+θ) passes from a heavier string to lighter string. The reflected has amplitude 0.5A. The junction of the string is x=0. The equation of the reflected wave is:
A
y′=0.5Asin(kx+ωt+θ)
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B
y′=−0.5Asin(kx+ωt+θ)
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C
y′=−0.5Asin(ωt−kx−θ)
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D
y′=−0.5Asin(kx+ωt−θ)
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Solution
The correct option is Dy′=−0.5Asin(kx+ωt+θ) Y=Asin(Kx−ωt+θ)y′=0.5Asin(Kx+ωt+θ+π)=−0.5Asin(Kx+ωt+θ)Hence,optionBiscorrectanswer.