Time period of a back performing `SHM` is `2` second and it can travel to and fro from equilibrium position upto maximum `5cm`. At start the pendulum is at maximum displacement on right side of equilibrium position. Find displacement and time relation.
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Displacement expression for `S.H.M. , x = A sin (omegat + phi)` Time period of `SHM's T = (2pi)/(omega) = 2s :. omega = rad//s` Amplitude of `SHM A = 5 cm , :. x = 5 sin(pit + phi)` Now at `t = 0`, displacement `x = 5 cm :. 5 = 5 sin(pi xx 0 + phi) rArr sin phi = 1 rArr phi = (pi)/(2)` Therefore, `x = 5 "sin" pi(t + 1/2) rArr x = 5 sin(pit + (pi)/(2)) rArr x = 5 cos pit`
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