User:Eml4507.s13.team3.chen/R2

=R2.4 (sec.53a, p.53-2b, Pb-53.4)= On my honor, I have neither given nor received unauthorized aid in doing this assignment.

Problem Statement

 * Consider an L2-ODE-CC with the following roots: $$\lambda_1,_2 = -0.5+i2$$. Find the system natural frequency $$\omega$$.


 * Let the system be excited by a periodic force of $$f(t) = f_0cos(\bar \omega t)$$. Find the expression for the particular solution $$y_p(t)$$.


 * Consider:
 * $$f_0/m = r_0 = 1$$


 * $$\bar \omega = 0.9\omega$$


 * Find the particular solution $$y_p(t)$$ for this case and the amplification factor $$\mathbb A$$.


 * Find the complete solution $$y(t)$$ satisfying $$y(0) = 1, y'(0) = 0$$.


 * Plot in separate figures: $$y_h(t), y_p(t), y(t)$$
 * Plot in the same figure: $$y_h(t), y_p(t), y(t)$$

Find the system natural frequency $$\omega$$
From R1.5 (Case 3) :

$$ (\lambda-\lambda_1)(\lambda-\lambda_2)$$ $$ (\lambda+.5+i2)(\lambda+.5-i2)$$ $$ \lambda^2 + .5\lambda - i2\lambda + .5\lambda + .25 + .5i2 +i2\lambda - .5i2 + 4$$ $$ \lambda^2 + \lambda + 4.25 = 0$$

The standard L2-ODE-CC is then $$y'' + y' + 4.25y = 0 $$

Relate L2-ODE-CC to standard form of equation of motion:

$$y + y' + 4.25y = y + ay' + by$$ $$y'' + 2\zeta \omega y' + \omega^2 y$$ $$a = 2\zeta \omega$$ $$b = \omega^2$$ $$\omega = \sqrt{4.25}$$ $$\zeta = \frac1{2\sqrt{4.25}}$$

The system natural frequency: $$\omega = \sqrt{4.25}$$

Find expression for particular solution $$y_p(t)$$
First find the homogenous solution $$y_h(t)$$. $$y_h(t) = C_1 e^{(\alpha t)} sin(\beta t)+C_2 e^{(\alpha t)} cos(\beta t)$$, $$\lambda_1 = \bar \lambda_2 = \alpha + i\beta$$, where $$\alpha = -0.5, \beta = 2$$ $$y_h(t)=C_1 e^{(-0.5 t)} sin(2 t)+C_2 e^{(-0.5 t)} cos(2 t)$$

By setting $$y''(t) + y'(t) + 4.25y(t) = f_0cos(\bar \omega t)$$ and solving for the particular solution:

$$y_p(t) = \frac{(f_0 cos(t \bar \omega)(-\bar \omega^2 sin^2(2 t)-\bar \omega^2 cos^2(2 t)+4.25)+f_0 \bar \omega sin^2(2 t) sin(t \bar \omega)+f_0 \bar \omega cos^2(2 t) sin(t \bar \omega))}{(\bar \omega^4-7.5 \bar \omega^2+18.0625)}$$

Find the particular solution $$y_p(t)$$ for the case and amplification factor $$\mathbb A$$
Consider the case:
 * $$f_0/m = r_0 = 1$$


 * $$\bar \omega = 0.9\omega$$

$$y_p(t) = \mathbb A \frac{( cos(t 0.9\omega)(- 0.81\omega^2 sin^2(2 t)- 0.81\omega^2 cos^2(2 t)+4.25)+ 0.9\omega sin^2(2 t) sin(t 0.9\omega)+ 0.9\omega cos^2(2 t) sin(t 0.9\omega))}{( 0.6561\omega^4-6.075\omega^2+18.0625)}$$ $$\mathbb A := \frac1{\left[(1 - \rho^2)^2+(2 \zeta \rho)^2 \right]^{1/2}}$$, where $$\rho = \frac {\bar \omega}{\omega} = 0.9$$ $$\mathbb A = 2.10032$$

Find the complete solution $$y(t)$$ satisfying $$y(0) = 1, y'(0) = 0$$
By superposition: $$y(t)=y_h(t)+y_p(t)$$ $$y_h(t)=C_1 e^{(-0.5 t)} sin(2 t)+C_2 e^{(-0.5 t)} cos(2 t)$$ $$y_p(t) = \mathbb A \frac{( cos(t 0.9\omega)(- 0.81\omega^2 sin^2(2 t)- 0.81\omega^2 cos^2(2 t)+4.25)+ 0.9\omega sin^2(2 t) sin(t 0.9\omega)+ 0.9\omega cos^2(2 t) sin(t 0.9\omega))}{( 0.6561\omega^4-6.075\omega^2+18.0625)}$$ $$y(t)=C_1 e^{(-0.5 t)} sin(2 t)+C_2 e^{(-0.5 t)} cos(2 t)$$
 * $$+\mathbb A \frac{( cos(t 0.9\omega)(- 0.81\omega^2 sin^2(2 t)- 0.81\omega^2 cos^2(2 t)+4.25)+ 0.9\omega sin^2(2 t) sin(t 0.9\omega)+ 0.9\omega cos^2(2 t) sin(t 0.9\omega))}{( 0.6561\omega^4-6.075\omega^2+18.0625)}$$

Simplify with: $$\omega = \sqrt{4.25}$$ $$1 = C_2 + \frac{2.10032(-0.81 \omega^2+4.25)}{4.09456}$$

$$C_2 = 0.58579$$ $$0 = C_1(2)+0.58579(-0.5)+ \mathbb A \frac{3.4425}{4.09456}$$ $$C_1 = -0.73648$$ The complete solution is: $$y(t)=-0.73648 e^{(-0.5 t)} sin(2 t)+0.58579 e^{(-0.5 t)} cos(2 t)$$
 * $$+\mathbb A \frac{( cos(t 0.9\omega)(- 0.81\omega^2 sin^2(2 t)- 0.81\omega^2 cos^2(2 t)+4.25)+ 0.9\omega sin^2(2 t) sin(t 0.9\omega)+ 0.9\omega cos^2(2 t) sin(t 0.9\omega))}{( 0.6561\omega^4-6.075\omega^2+18.0625)}$$

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