User:Egm4313.s12.team6.hickey

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:


 * {| style="width:100%" border="0"

$$ r(t)=my''+cy'+ky $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * 
 * }
 * }

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:
 * {| style="width:100%" border="0"

$$ y(x)=e^{-2x}[cos{3x}+\frac{2}{3}sin{3x}] $$  The plot for this equation is given as shown in plot 1:
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * 
 * }
 * }

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.