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Question

Two simple harmonic motions are represented by the equations y1= 0.1 sin(100π t+π3) and y2 = 0.1 cosπ tThe phase difference of the velocity of particle 1 with respect to the velocity of particle 2 at t= 0 is

A
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B
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C
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D
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Solution

The correct option is C

v1=dy1dt=0.1× 100 π cos(100π t+π3)
v2=dy2dt=0.1 π sin π t=0.1π cos(π t+π2)
Phase difference of velocity of first particle with respect to the velocity of 2nd particle at
t = 0 is
ϕ=ϕ1ϕ2=π3π2=π6.


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