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NEET 2004 (AIPMT) · Progressive Wave Equation

Progressive Wave Equation question from NEET 2004 (AIPMT)

Q99NEET 2004 · AIPMTEasy

Progressive Wave Equation

The phase difference between two waves. represented by

y1 = 10−-6 sin[100t + (x/50) + 0.5] m

y2 = 10−-6 cos[100t + (x/50)] m,

where x is expressed in metres and t is exressed in secondss, is approximately.

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Answer: (a)

  1. y2=10−6cos⁡(100t+x50)=10−6sin⁡(100t+x50+1.57)y_2 = 10^{-6}\cos\left(100t + \frac{x}{50}\right) = 10^{-6}\sin\left(100t + \frac{x}{50} + 1.57\right)
  2. y1y_1 has phase constant 0.50.5, y2y_2 has 1.571.57.
  3. Δϕ=1.57−0.5=1.07\Delta\phi = 1.57 - 0.5 = 1.07 rad
  4. So the answer is (a) 1.07 radians.
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