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
4.1 Answer:

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