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Obtain the formula of acceleration of a particle from the formula of displacement of SHM. |
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Answer» Solution :ACCELERATION of SHM particle is TWO times differentiation of displacement w.r.t. to time of particle OR Acceleration of SHMf particle is the differentiationl of velocity of particle w.r.t. time. The displacement of SHM particle at time t, `x(t)= A cos (omega t+ phi)` Differentiating with time t, `v(t)= -A omega sin (omega t+phi)` Again differentiating with time t, `a(t) = -A omega^(2) cos (omega t+ phi)` `therefore a(t)= -omega^(2) x(t) """..........."(1)` `[therefore -A cos (omega t+phi)= x(t)]` generally `a= -omega^(2) x`. Special cases : (1) At mean position x(t)=0, `a= -omega^(2) xx 0` `therefore a= 0` Acceleration at mean position is zero and velocity is maximum. (2) At extreme points, `x(t)= |A|` `a= -omega^(2) |A|` `therefore a= omega^(2)A" or "a= -omega^(2)A` Hence, at extreme points acceleration is maximum is called `a_("MAX")` `therefore a_("max")= A omega^(2)` Acceleration in +x-direction is negative and in -x-direction it is POSITIVE. |
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