# 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.

Figure 53, page 85 K 2011

#### Solution

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

${\displaystyle F_{applied}=r(t)}$

${\displaystyle F_{inertia}=my''}$

${\displaystyle F_{damping}=cy'}$

${\displaystyle F_{spring}=ky}$

##### Case 1
Case 1: Positive Force, Positive Displacement

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

${\displaystyle F_{applied}+F_{inertia}-F_{damping}-F_{spring}=0}$

Rearranging the equation, we get

${\displaystyle F_{applied}=-F_{inertia}+F_{spring}+F_{damping}}$

Replacing Force variables, we get

${\displaystyle r(t)=my''+cy'+ky}$

##### Case 2
Case 2: Negative Force, Negative Displacement

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

${\displaystyle -F_{applied}-F_{inertia}+F_{damping}+F_{spring}=0}$

Rearranging the equation, we get

${\displaystyle F_{applied}=-F_{inertia}+F_{spring}+F_{damping}}$

Replacing Force variables, we get

${\displaystyle r(t)=my''+cy'+ky}$

#### Conclusion

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

 ${\displaystyle r(t)=my''+cy'+ky}$

## 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):

${\displaystyle \lambda ^{2}+4\lambda +13=0}$

Initial Conditions:

${\displaystyle y(0)=1,y'(0)=0}$

No excitation: ${\displaystyle r(x)=0}$

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:

${\displaystyle y''+4y'+13=r(x)}$

Where we know that ${\displaystyle r(x)=0}$

and the corresponding characteristic equation is ${\displaystyle \lambda ^{2}+4\lambda +13=0}$

To find the roots of this equation, we must use the Quadratic Equation, giving us ${\displaystyle \lambda _{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$

Inserting our corresponding values for these variables, we have ${\displaystyle \lambda _{1,2}={\frac {-4\pm {\sqrt {4^{2}-4(1)(13)}}}{2(1)}}={\frac {-4\pm {\sqrt {16-52}}}{2}}=-2\pm {\frac {\sqrt {-36}}{2}}=-2\pm 3i}$

Knowing that we have complex roots, we can now insert the roots of our characteristic equation into the proper form for the homogeneous solution:

${\displaystyle y(x)=e^{-2x}[C_{1}cos{3x}+C_{2}sin{3x}]}$

And taking the first derivative of this equation, we get

${\displaystyle y'(x)=-2e^{-2x}[C_{1}cos{3x}+C_{2}sin{3x}]+3e^{-2x}[-C_{1}sin{3x}+C_{2}cos{3x}]}$

Now, we can use our initial conditions to find the values of our C coefficients:
${\displaystyle y(0)=1=e^{0}[C_{1}cos{0}+C_{2}sin{0}]=1[C_{1}+0]=C_{1}}$

${\displaystyle C_{1}=1}$

${\displaystyle y'(0)=0=-2e^{0}[C_{1}cos{0}+C_{2}sin{0}]+3e^{0}[-C_{1}sin{0}+C_{2}cos{0}]=-2[C_{1}+0]+3[0+C_{2}]}$

${\displaystyle -2C_{1}+3C_{2}=0}$

${\displaystyle 3C_{2}=2}$

${\displaystyle C_{2}={\frac {2}{3}}}$

Therefore, we now know that our final solution is:

 ${\displaystyle y(x)=e^{-2x}[cos{3x}+{\frac {2}{3}}sin{3x}]}$

The plot for this equation is given as shown in plot 1:

Plot 1

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

Plot 2

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.