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Obtain instantaneous acceleration of a SHM particle with the help of reference circle. |
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Answer» SOLUTION :Acceleration of a particlel executing uniform circular motion with ANGULAR speed `omega` and amplitude A is `(v^2)/(A)" or "A omega^(2)` and direction of acceleration is toward centre so it is negative. The acceleration of projected particle in uniform circular motionl at time t `a(t)= -omega^(2) x(t)` This acceleration is towards the centre :  The CENTRIPETAL force at point P is `omega^(2)A`, its projection on X-axis is `A omega^(2) cos (omega t+phi)` and its direction is opposite to +X-axis hence INSTANTANEOUS acceleration of projected particle at time t is `a(t)= -A omega^(2) cos (omega t+ phi)` `therefore a(t) = -omega^(2) x(t)""[therefore A cos (omega t+ phi)= x(t)]` is the acceleration of SHM particle. |
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