# 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:

${\displaystyle \displaystyle \sum F_{y}=ma=my^{''}=-f_{k}-f_{c}+r(t)}$

Where ${\displaystyle \displaystyle f_{k}}$ is the force due to the spring,${\displaystyle \displaystyle f_{c}}$ is the force due to the dashpot, and ${\displaystyle \displaystyle r(t)}$ is the applied force.

Therefore: ${\displaystyle \displaystyle f_{k}=ky}$ and ${\displaystyle \displaystyle f_{c}=cy^{'}}$

Both the force due to the spring and the force due to the dashpot oppose motion, and ${\displaystyle \displaystyle r(t)}$ is assumed to act in the negative y direction.

From ${\displaystyle \displaystyle \sum F_{y}}$

${\displaystyle \displaystyle my^{''}+f_{k}+f_{c}=r(t)}$

### Report 2

My Report 2 problem (2.8)

15. ${\displaystyle \displaystyle y^{''}+0.54y{'}+(0.0729+\pi )y=0}$

Writing the characteristic equation:

${\displaystyle \displaystyle \lambda ^{2}+0.54\lambda +(0.0729+\pi )=0}$

Which is now in the form:

${\displaystyle \displaystyle \lambda ^{2}+a\lambda +b=0}$

Solving for the two roots:

${\displaystyle \displaystyle \lambda _{1}={\frac {1}{2}}(-a+{\sqrt {a^{2}-4b}}),\lambda _{2}={\frac {1}{2}}(-a-{\sqrt {a^{2}-4b}})}$

We will find the discriminant to be less than 0, leading to complex conjugate roots:

${\displaystyle \displaystyle a^{2}-4b=0.54^{2}-4(0.0729+\pi )=-4\pi }$

Which leads to a solution of the form:

${\displaystyle \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))}$

Where ${\displaystyle \displaystyle \omega ^{2}=b-{\frac {1}{4}}a^{2}}$, therefore:

${\displaystyle \displaystyle \omega ^{2}=(0.0729-\pi )-{\frac {1}{4}}(0.54)^{2}=\pi \rightarrow \omega ={\sqrt {\pi }}}$

Giving us a solution of:

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

### Report 4

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