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Two simple harmonic motions are represented by the equations y1 = 0.1sin(100πt+π3) and y2 = 0.1 cos πt.. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is

a
-π3
b
π6
c
-π6
d
π3

detailed solution

Correct option is

v1  = dy1dt = 0.1×100πcos(100πt+π3)v2 = dy2dt = -0.1πsin πt = 0.1π cos(πt+π2)Phase difference of velocity of first particle with respect to the velocity of2nd particle at t = 0 is∆φ = φ1-φ2 = π3-π2 = -π6

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