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Q.

Two simple harmonic motions are represented by the equations y1=0â1sinâ¡100Ït+Ï3 and y2=0â1cosâ¡(100Ït) The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is

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a

Ï/6

b

Ï/3

c

âÏ/6

d

âÏ/3

answer is C.

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Detailed Solution

From the given eqs. dy1dt=v1=0â1Ã100Ïcosâ¡100Ït+Ï3Similarly v2 = dy2dt = -0.1Ã100Ï sin 100Ït = 0.1Ã100Ï cos (100Ït +Ï2)Therefore â³Ï = Ï3-Ï2 = -Ï6

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