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Obtain the expression of displacement from force law for simple harmonic motion. |
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Answer» Solution :Force law for SHM `F= -kx (t)` `ma = -m omega^(2) X(t) ""[therefore F= ma(t), K= m omega^(2)]` `therefore a(t) = -omega^(2)x (t)` Integrating w.r.t time t on both sides, `int a(t) DT = -omega^(2) int x (t) dt` `= -omega^(2) int A COS (omega t+ phi) dt` `a(t) XXT = -A omega^(2) int cos (omega t+ phi) dt` `v(t) = -A omega^(2)xx (sin (omega t+ phi))/(omega)` `therefore v(t) = -A omega sin (omega t+ phi)` integrating again w.r.t. time t, `int v(t) dt= -A omega int sin (omega t+phi)dt` `therefore v(t)xxt = -A omega [-(cos (omega t+phi))/(omega)]` `therefore x(t) = -A cos (omega t+phi)` is the expression of displacement. |
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