SHMHard
Question
Two SHM′s are represented by y = a sin (ωt − kx) and y = b cos (ωt − kx). The phase difference between the two is :
Options
A.π/2
B.π/4
C.π/6
D.3π/4
Solution
y = a sin (ωt − kx)
y = b cos (ωt − kx) ⇒ y = b sin (ωt − kx + π/2)
So phase difference is π/2
y = b cos (ωt − kx) ⇒ y = b sin (ωt − kx + π/2)
So phase difference is π/2
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