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Obtain the force law for SHM from the displacement of SHM particle. |
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Answer» Solution :Displacement of SHM particle at time t `X(t)= A cos (OMEGA t+ phi)"""……."(1)` where A= amplitude, `omega` = ANGULAR frequency and `phi`= initiall PHASE. Differentiating equation (1) w.r.t. time t `v(t)= (d[x(t)])/(dt)= (d)/(dt)[ A cos (omega t+phi)]` `v(t)= -A omega sin (omega t+phi)` Differentiating again w.r.t., time t `a(t) = (d[v(t)])/(dt)= (d)/(dt) [-A omega sin (omega t + phi)]` `therefore a(t) = -A omega^(2) cos (omega t + phi)` but `A cos (omega t+phi) = x(t)` `therefore a(t) = -omega^(2)x (t)` MULTIPLY by mass m on both sides `ma (t) = -m omega^(2) x(t)` `therefore F= -kx(t)" where "ma= F, m omega^(2) = k` `therefore F propto -x(t)`. |
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