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Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. A point mass oscillates along the x-axis according to the law x=x_(0)cos omegat-(pi)/(4). If acceleration of the particle is written as a=A cos (omega t+delta), then |
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Answer» `A=x_(0)oemga^(2),delta=(pi)/(4)` `:.` ACCELERATION `a=(dv)/(dt)=-x_(0)omega^(2)cos(omegat-(pi)/(4))` `=x_(0)omega^(2)cos(pi+omega t-(pi)/(4))` Now compparing it with `a=A cos (omega t+delta)`, We have `A=x_(0)omega^(2)` and `delta=(3pi)/(4)`. So correct choice is ( c ). |
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