## Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u” + 1 4 u’ + 2u = 2 cos ωt

Question

Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u” + 1 4 u’ + 2u = 2 cos ωt, u(0) = 0, u'(0) = 2 (a) Determine the steady state part of the solution of this problem.

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1 year 2021-08-21T07:30:00+00:00 1 Answers 37 views 0

Therefore the required solution is

$$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$$

Explanation:

Given vibrating system is

$$u”+\frac{1}{4}u’+2u= 2cos \omega t$$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= – A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U”(t) =  – A ω² cosωt -B ω² sin ωt

Putting this in given equation

$$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$$

$$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$$

Equating the coefficient of sinωt and cos ωt

$$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$$

$$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$$………(1)

and

$$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$$

$$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$$……..(2)

Solving equation (1) and (2) by cross multiplication method

$$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$$

$$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$$

$$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$$   and        $$B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$$

Therefore the required solution is

$$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$$