A wave is incident on fixed target as y=5 2 (t- x 2 ) , the equation of reflected wave…
Physics · JEE Advanced · NTA Exams — Oscillations and Waves
A wave is incident on fixed target as \(y=5 \sin 2 \pi\left(t-\frac{x}{2}\right)\), the equation of reflected wave through rigid target
- y = –5 sin \(2 \pi\left(t+\frac{x}{2}\right)\)
- y = +5 sin \(2 \pi\left(t+\frac{x}{2}\right)\)
- y = +5 sin\(2 \pi\left(t-\frac{x}{2}\right)\)
- y = –5 sin \(2 \pi\left(t-\frac{x}{2}\right)\)
Answer
(A) y = –5 sin 2 (t+ x 2 )
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