Applications to Springs and Circuits An object of mass 1 kg is attached to a spr

Question

Applications to Springs and Circuits

An object of mass 1 kg is attached to a spring with spring constant k = 4 N/m. There is a resistive force due to friction with a damping constant of b = 4 kg/sec. There Is also a time dependent force acting on the object of F(t) = 3te^-2t. Let x(t) denote the position of the object relative to the equilibrium position after t seconds. Assume that at t = 0, the object begins at its equilibrium position with a velocity of 0. Set up and solve the differential equation that determines x(t).

Explanation / Answer

For the given system , the equation of motion is given by

mx'' + bx' + kx = 3te-2t

x'' + 4x' + 4x = 3te-2t

Solving for the differential equation x'' + 4x' + 4x = 0

Let x = ert

=> r2 + 4r + 4 = 0 => r = -2

=> xh = c1e-2t + c2te-2t

For particular solution ,

Let xh = ( a + bt + ct2 + dt3 )e-2t

x'h  = (-2)( a + bt + ct2 + dt3 )e-2t + ( b + 2ct + 3dt2 )e-2t

x''h  = 4( a + bt + ct2 + dt3 )e-2t + (-2)( b + 2ct + 3dt2 )e-2t + (-2)( b + 2ct + 3dt2 )e-2t + ( 2c + 6dt )e-2t

x''h + 4x'h + 4xh = e-2t( 4a + 4bt + 4ct2 + 4dt3 - 8a - 8bt - 8ct2 - 8dt3 + 4b + 8ct + 12dt2 + 4a + 4bt + 4ct2 + 4dt3 - 2b - 4ct -6dt2 - 2b - 4ct - 6dt2 + 2c + 6dt )

By comparing with original equation

a = 0 , b = 0 , c = 0 , d = 1/2

=> xp = (1/2)dt3

=> x = xh + xp = (1/2)dt3 + c1e-2t + c2te-2t

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