Two simple harmonic motions are represented by the equations y1=0.1sin(100πt+π3) and y2=0.1cosπt. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is
A
−π3
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B
π6
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C
−π6
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D
π3
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Solution
The correct option is C−π6 v1=dy1dt=0.1×100πcos(100πt+π3) v2=dy2dt=−0.1πsinπt=0.1πcos(πt+π2)
The phase difference of velocity of the first particle with respect to the velocity of 2nd particle at t=0 is Δφ=φ1−φ2=π3−π2=−π6