# User:Egm4313.s12.team5.house

### Report 1

My Report 1 Problem (1.2)

From Newton's Second Law and summing forces in the y-direction:

$\displaystyle \sum F_{y}=ma=my^{''}=-f_{k}-f_{c}+r(t)$ Where $\displaystyle f_{k}$ is the force due to the spring,$\displaystyle f_{c}$ is the force due to the dashpot, and $\displaystyle r(t)$ is the applied force.

Therefore: $\displaystyle f_{k}=ky$ and $\displaystyle f_{c}=cy^{'}$ Both the force due to the spring and the force due to the dashpot oppose motion, and $\displaystyle r(t)$ is assumed to act in the negative y direction.

From $\displaystyle \sum F_{y}$ $\displaystyle my^{''}+f_{k}+f_{c}=r(t)$ ### Report 2

My Report 2 problem (2.8)

15. $\displaystyle y^{''}+0.54y{'}+(0.0729+\pi )y=0$ Writing the characteristic equation:

$\displaystyle \lambda ^{2}+0.54\lambda +(0.0729+\pi )=0$ Which is now in the form:

$\displaystyle \lambda ^{2}+a\lambda +b=0$ Solving for the two roots:

$\displaystyle \lambda _{1}={\frac {1}{2}}(-a+{\sqrt {a^{2}-4b}}),\lambda _{2}={\frac {1}{2}}(-a-{\sqrt {a^{2}-4b}})$ We will find the discriminant to be less than 0, leading to complex conjugate roots:

$\displaystyle a^{2}-4b=0.54^{2}-4(0.0729+\pi )=-4\pi$ Which leads to a solution of the form:

$\displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))$ Where $\displaystyle \omega ^{2}=b-{\frac {1}{4}}a^{2}$ , therefore:

$\displaystyle \omega ^{2}=(0.0729-\pi )-{\frac {1}{4}}(0.54)^{2}=\pi \rightarrow \omega ={\sqrt {\pi }}$ Giving us a solution of:

$\displaystyle y=e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))$ ### Report 4

$\displaystyle {\begin{bmatrix}b&a&2&&&\\&b&2a&&&\\&&b&&&\\&&&b&a(n-1)&n(n-1)\\&&&&b&an\\&&&&&b\end{bmatrix}}$ 