# User:Egm4313.s12.team6.hickey

## Report 1

### Question R1.2

#### Question

Derive the equation of motion of the spring - mass - dashpot in Fig. 53, in K 2011 p. 85, with an applied force r(t) on the ball.

#### Solution

There are 2 possible cases that can occur in this problem, depending on the direction of the applied force.
In both cases,

$F_{applied}=r(t)$ $F_{inertia}=my''$ $F_{damping}=cy'$ $F_{spring}=ky$ ##### Case 1

The applied force is in the positive direction, and therefore the displacement is in the positive direction.

From the Free Body Diagram, we get the equation

$F_{applied}+F_{inertia}-F_{damping}-F_{spring}=0$ Rearranging the equation, we get

$F_{applied}=-F_{inertia}+F_{spring}+F_{damping}$ Replacing Force variables, we get

$r(t)=my''+cy'+ky$ ##### Case 2

The applied force is in the negative direction, and therefore the displacement is in the negative direction.

From the Free Body Diagram, we get the equation

$-F_{applied}-F_{inertia}+F_{damping}+F_{spring}=0$ Rearranging the equation, we get

$F_{applied}=-F_{inertia}+F_{spring}+F_{damping}$ Replacing Force variables, we get

$r(t)=my''+cy'+ky$ #### Conclusion

Since both cases return the same solution, the equation of motion is derived as:

 $r(t)=my''+cy'+ky$ ## Report 2

### Question R2.9

#### Question

Find and plot the solution for the Linear Second order ODE with constant coefficients corresponding to equation (1).

Equation (1):

$\lambda ^{2}+4\lambda +13=0$ Initial Conditions:

$y(0)=1,y'(0)=0$ No excitation: $r(x)=0$ In another Figure, superpose 3 Figures: (a) This figure, (b)The figure in R2.6 p. 5-6, (c) The Figure in R2.1 p. 3-7.

#### Solution

The question implied by the problem statement is that we are solving the following ODE:

$y''+4y'+13=r(x)$ Where we know that $r(x)=0$ and the corresponding characteristic equation is $\lambda ^{2}+4\lambda +13=0$ To find the roots of this equation, we must use the Quadratic Equation, giving us $\lambda _{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$ Inserting our corresponding values for these variables, we have $\lambda _{1,2}={\frac {-4\pm {\sqrt {4^{2}-4(1)(13)}}}{2(1)}}={\frac {-4\pm {\sqrt {16-52}}}{2}}=-2\pm {\frac {\sqrt {-36}}{2}}=-2\pm 3i$ Knowing that we have complex roots, we can now insert the roots of our characteristic equation into the proper form for the homogeneous solution:

$y(x)=e^{-2x}[C_{1}cos{3x}+C_{2}sin{3x}]$ And taking the first derivative of this equation, we get

$y'(x)=-2e^{-2x}[C_{1}cos{3x}+C_{2}sin{3x}]+3e^{-2x}[-C_{1}sin{3x}+C_{2}cos{3x}]$ Now, we can use our initial conditions to find the values of our C coefficients:
$y(0)=1=e^{0}[C_{1}cos{0}+C_{2}sin{0}]=1[C_{1}+0]=C_{1}$ $C_{1}=1$ $y'(0)=0=-2e^{0}[C_{1}cos{0}+C_{2}sin{0}]+3e^{0}[-C_{1}sin{0}+C_{2}cos{0}]=-2[C_{1}+0]+3[0+C_{2}]$ $-2C_{1}+3C_{2}=0$ $3C_{2}=2$ $C_{2}={\frac {2}{3}}$ Therefore, we now know that our final solution is:

 $y(x)=e^{-2x}[cos{3x}+{\frac {2}{3}}sin{3x}]$ The plot for this equation is given as shown in plot 1:

Comparing the graphs between problems R2.1, R2.6, and R2.9, we get the following plot:

Where the solution for R2.1 is given in red, the solution for R2.6 is given in green, and the solution for R2.9 is given in blue.