Verified Solution • 2025–26 Session moderate

QAIPMT-2004-Ph-2633: NEET 2004 Physics - Class 11 AIPMT-2004-Ph-2633 Solution: The phase difference between two waves represented by…

The phase difference between two waves represented by y1=10−6sin⁡[100t+(x/50)+0.5]y_1=10^{-6}\sin[100t+(x/50)+0.5] m, y2=10−6cos⁡[100t+(x/50)]y_2=10^{-6}\cos[100t+(x/50)] m, where x is expressed in metres and t is expressed in seconds, is approximately.

Multiple Choice Options

  • A 1.07 radians
  • B 2.07 radians
  • C 0.5 radians
  • D 1.5 radians

Step-by-Step Verified Solution

y 1 = 10 �6 sin[100t + (x/50) + 0.5] y 2 = 10 �6 cos[100t + (x/50)] [using cosx = sin(x + � /2)] = 10 �6 sin[100t + (x/50) + � /2] = 10 �6 sin[100t + (x/50) + 1.57] The phase difference = 1.57 � 0.5 = 1.07 [or using sinx = cos( � /2 � x). We get the same result.]

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