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Waves question

2004 · Q158
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Waves question

2004 · Q158

NEETPhysicsWavesMCQ+4 / −1
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.
  1. A
    1.07 radians
  2. B
    2.07 radians
  3. C
    0.5 radians
  4. D
    1.5 radians
View written solutionFree

Correct answer: A

y1 = 10–6sin[100t + (x/50) + 0.5]

y2 = 10–6cos[100t + (x/50)]
[using cosx = sin(x + π\pi π/2)]

= 10–6sin[100t + (x/50) + π\pi π/2]

= 10–6sin[100t + (x/50) + 1.57]

The phase difference = 1.57 – 0.5 = 1.07

[or using sinx = cos(π\pi π/2 – x). We get the same result.]

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