Question

For the given values of m, c, k, and f(t), assume the forced vibration is initially at equilibrium. For t 0, find the motion

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Answer #1

m ck f(t) 4x5 20 cos(3t)

4 5=0 D2 4D 5 (D 2)2 -12 D =-2 t i

So r(t)c1e2 cos(t)ce2 sin(t) and

A cos(3t B sin(3t) 4x 5 (-4B 12.A) sin(3t) + (t) (12B-4A) cost (3t)

Thus, we have 12B 4A 20, 4B12A 0 B=-3A so that 3 -40.A 20 A B = meaning

(t)cos(3 - sin(3t) 2 is the particular solution

The general solution is 3 -2t C cos(t)esn(t)cos(3t)sin(3t) -2t

Initial conditions imply r(0)(0)0 0,-2c1+c2- l so

C2 C1

Thus, x(t)=-\frac{1}{2}e^{-2t}\cos(t)+\frac{7}{2}e^{-2t}\sin(t)+\frac{1}{2}\cos(3t)-\frac{3}{2}\sin(3t)

Where transient solution is sin(t) -2t cos(t) --e Ttrans (t) e 2 and steady state solution is

x_{\text{steady}}(t)=\frac{1}{2}\cos(3t)-\frac{3}{2}\sin(3t)

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