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Two particles P and Q are executing SHM across same straight line whose equations are given as `y_(P)=Asin (omegat+phi_(1)) and y_(Q)=Acos (omegat+phi_(2))` An observer, at `t=0` observer the particle P at a distance `A//sqrt(2)` moving to the right from mean position O while particle Q at `(sqrt(3))/(2)` A moving to the left from mean position O as shown in figure. Then `phi_(2)-phi_(1)(i n rad)` ie equal to A. `5pi//6`B. `3pi//4`C. `5pi//2`D. `7pi//12` |
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Answer» Correct Answer - D `y_(p)=a sin (omegat+phi_(1)),` For particle P, `t=0, y_(p)=(A)/(sqrt(2))` `:. (A)/(sqrt(2))=A sin (0+phi_(1))=Asin phi_(1)` or `sin phi_(1)=(1)/(sqrt(2)) or phi_(1)=(pi)/(4)rad` `y_(Q)=A cos (omegat +phi_(2)),` For particle `Q,t=0, y_(Q)=(-sqrt(3))/(2)A` `:. -(sqrt(3))/(2)A=A cos (0 +phi_(2))=A cosphi_(2)` or `cosphi_(2)=-(sqrt(3))/(2)=cos((5)/(6))pi or phi_(2)=(5pi)/(6)` `:. phi_(2)-phi_(1)=(5pi)/(6)-(pi)/(4)=(7pi)/(12)rad` |
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