# University of Florida/Egm4313/IEA-f13-team10/R1
## Report 1
### Problem 1.1
#### Problem Statement

Derive the equation of motion of a spring-dashpot system in parallel with a mass and applied force.

#### Solution
##### Kinematics

The spring and dashpot are in parallel, so the displacement of the spring is equal to the displacement of the dashpot.
(1) $y=y_{k}=y_{c}$

##### Kinetics

(2) $my''+f_{I}=f(t)$ (3) $f_{I}=f_{k}+f_{c}$

##### Constitutive Relations

(4) $f_{k}=ky_{k}$ (5) $f_{c}=cy_{c}'$
From (2), (3), (4) and (5), the equation of motion for the system is $my''+cy_{c}'+ky_{k}=f(t)$
From the first and second derivatives of (1), $y'=y_{k}'=y_{c}'$ $y''=y_{k}''=y_{c}''$
So in terms of the displacement of the mass, the final equation of motion is, $my''+cy'+ky=f(t)$

#### Honor Pledge

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

### Problem 1.2
#### Problem Statement

Derive the equation of motion of the springmass- dashpot in Fig.53, in K 2011 p.85, with an applied force r(t) on the ball.

#### Solution
##### Kinematics

The displacement of the spring equals the negative displacement of the dashpot (constitutive relation)
$y=y_{k}=-y_{c}$
$y''=y''_{k}=-y''_{c}$

##### Kinetics

Newton's second law ${\vec {F}}=m{\vec {a}}$
$r(t)=my''+r_{k}-r_{c}$
$r_{k}=ky_{k}$
$r_{c}=cy'_{c}$
Substituting the relation above and kinematics section into the kinetics part, we get the equation of motion $-my''_{c}-cy'_{c}+-ky_{c}=r(t)$
Using the constitutive relations shown above we can write the equation of motion as $my''+cy'+ky=r(t)$

#### Honor Pledge

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

### Problem 1.3
#### Problem Statement

For the spring-dashpot-mass system on p.1-4, draw the FBDs and derive the equation of motion (2) p.1-4.

#### Solution
##### Step 1, Kinematics

$y=y_{k}+y_{c}$, therefore $y'=y'_{k}+y'_{c}$ and $y''=y''_{k}+y''_{c}$

##### Step 2, FBD and Kinetics

$my''+f_{I}=f(t)$
$f_{I}=f_{k}=f_{c}$

##### Step 3, Relations

$f_{k}=ky_{k}$
$f_{c}=cy'_{c}$

##### Step 4

$f_{k}=f_{c}$, therefore $ky_{k}=cy'_{c}$, and $y'_{c}={\frac {k}{c}}$ and $y''_{c}={\frac {k}{c}}y'_{k}$

##### Step 5

Plugging the last equation from step 4 into the last equation from step 1, $y''=y''_{k}+{\frac {k}{c}}y'_{k}$

##### Step 6

Plugging the equation from step 5 into the first kinetic equation as y", $m(y''_{k}+{\frac {k}{c}}y'_{k})+ky_{k}=f(t)$

#### Honor Pledge

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

### Problem 2.1
#### Problem Statement

Derive (3) and (4) from (2).
(2) $V=LC{\frac {d^{2}v_{c}}{dt^{2}}}+RC{\frac {dv_{c}}{dt}}+v_{c}$
(3) $LI''+RI'+{\frac {1}{C}}I=V'$
(4) $LQ''+RQ'+{\frac {1}{C}}Q=V$

#### Solution
##### Derivation of (3) from (2)

(A) $I={\frac {dQ}{dt}}=Cv'_{c}$
(B) $Q=Cv_{c}$
Using equation A, we can show that $LCv''_{c}=LI'$ and $RCv'_{c}=RI$
Using equation B, we can show that $v_{c}=Q/C$
After these transformations, the new equation 2 is $LI'+RI+{\frac {Q}{C}}=V$
Taking the derivative of both sides of equation 2, we get $LI''+RI'+{\frac {1}{C}}I=V'$

##### Derivation of (4) from (2)

Using equation B, we can show that $LCv''_{c}=LQ''$ and $RCv'_{c}=RQ'$ and $v_{c}={\frac {Q}{C}}$
After these transformations, the new equation 2 is $LQ''+RQ'+{\frac {Q}{C}}=V$

#### Honor Pledge

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

### Problem 1.5

K 2011 p.59 pbs 3,12

#### Problem Statement

Find a general solution.

#### Solution
##### Problem 3

$y''+6y'+8.96y=0$
This is a linear, first order ODE with constant coefficients.
To find the general solution to this ODE set $y=e^{rx}$
so that $y'=re^{rx}$ and $y''=r^{2}e^{rx}$
Substituting in y to the ODE and factoring out $e^{rx}$ we get:
$r^{2}+6r+8.96=0$
Using the quadratic formula to solve for r we get
$r={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$ where a = 1, b = 6, and c = 8.96
Solving to get r = -3.2 and r = -2.8
We get the general solution $y(x)=C_{1}e^{-2.8x}+C_{2}e^{-3.2x}$
Now with the initial values y(0) = 1 and y'(0) = .5
$y(0)=C_{1}+C_{2}=1$
$y'(0)=-2.8C_{1}+-3.2C_{2}=.5$
$C_{1}=9.25$ , $C_{2}=8.96$
$y(x)=9.25e^{-2.8x}+8.96e^{-3.2x}$

##### Problem 12

y′′+9 y′+20y=0
This is a linear, first order ODE with constant coefficients.
To find the general solution to this ODE set $y=e^{rx}$
so that $y'=re^{rx}$ and $y''=r^{2}e^{rx}$
Substituting in y to the ODE and factoring out $e^{rx}$ we get:
$r^{2}+9r+20=0$
Using the quadratic formula to solve for r we get
$r={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$ where a = 1, b = 9, and c = 20
Solving to get r = -5 and r = -4
We get the general solution $y(x)=C_{1}e^{-5x}+C_{2}e^{-4x}$
Now with the initial values y(0) = 1 and y'(0) = .5
$y(0)=C_{1}+C_{2}=1$
$y'(0)=-5C_{1}+-4C_{2}=.5$
$C_{1}=-4.5$ , $C_{2}=5.5$
$y(x)=9.25e^{-4.5x}+8.96e^{-5.5x}$

#### Honor Pledge

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

### Problem 2.3

K 2011 p.3 Fig.2

#### Problem Statement

For each ODE in Fig. 2 (except for the last one), determine the order; determine if it is linear; and show whether the principle of superposition can be applied.

#### Solution
##### Falling stone

(1) $y''=g=const.$
Equation (1) is the ODE that describes the acceleration of the stone as it falls from some height $y=0$ to the ground. The highest order derivative of the dependent variable in an ODE determines the order of the ODE. In (1) the highest order derivative of the dependent variable y is second order. So (1) is a second-order ODE.
"A second-order ODE is called linear if it can be written $y''+p(x)y'+q(x)y=r(x)$" (Kreyszig 2011, p.46). Since (1) is in this form, with p(x) and q(x) = 0 and r(x) = g, then (1) is a linear ODE.
If the principle of superposition applies, then a solution ${\bar {y}}(x)$, which is comprised of the sum of the homogeneous solution $y''_{h}+p(x)y'_{h}+q(x)y_{h}=0$
and the particular solution $y''_{p}+p(x)y'_{p}+q(x)y_{p}=r(x)$, should satisfy the original ODE. So if superposition applies, (2) $y(x)={\bar {y}}(x)=y_{h}(x)+y_{p}(x)$. To determine whether superposition applies to (1), we consider
$y''_{h}+y''_{p}=0+g$. Now, using (2) and the linearity of the derivative, $(y_{h}+y_{p})''=g$ ${\bar {y}}''=g$. ${\bar {y}}$ satisfies (1), so superposition can be applied.

##### Parachutist

(3) $mv'=mg-bv^{2}$
Since v' is the highest order derivative of the dependent variable v, (3) is a first-order ODE.
Because the dependent variable v is squared, (3) is a non-linear ODE.
$D(.)=m(.)'+b(.)^{2}$
$D(w+y)=m(w+y)'+b(w+y)^{2}$
D(w+y) does not equal f(w) + f(y) thus it is not linear.
Rearranging (3) we obtain $mv'+bv^{2}=mg$. The homogeneous solution of the ODE is $mv'_{h}+bv_{h}^{2}=0$. The particular solution of the ODE is $mv'_{p}+bv_{p}^{2}=mg$. Using (2) and the linearity of the derivative we obtain $m(v'_{h}+v'_{p})+b(v_{h}^{2}+v_{p}^{2})=0+mg$ $m(v_{h}+v_{p})'=m{\bar {v}}'$ but $b(v_{h}^{2}+v_{p}^{2})\neq b(v_{h}+v_{p})^{2}$ Therefore, ${\bar {v}}$ is not a solution to the ODE. So superposition cannot be applied.

##### Outflowing water

(4) $h'=-k{\sqrt {h}}$
(4) is a first-order ODE because of h' .
(4) is a non-linear ODE because of the square-rooted dependent variable h.
$D(.)=(.)'+k{\sqrt {.}}$ $D(w+y)=(w+y)'+k{\sqrt {w+y}}$ D(y+w) does not equal f(w)+f(y) so it is non-linear.
Rearranging (4) we obtain $h'+k{\sqrt {h}}=0$. Summing the homogeneous and particular solutions we obtain $(h'_{h}+h'_{p})+k({\sqrt {h_{h}}}+{\sqrt {h_{p}}})=0$. But ${\sqrt {h_{h}}}+{\sqrt {h_{p}}}\neq {\sqrt {h_{h}+h_{p}}}$. So ${\bar {h}}$ is not a solution to (4), and superposition cannot be applied.

##### Vibrating mass on a spring

(5) $my''+ky=0$ (5) is a second-order linear ODE.
$m(y''_{h}+y''_{p})+k(y_{h}+y_{p})=0$ $m{\bar {y}}''+k{\bar {y}}=0$ ${\bar {y}}$ satisfies (5), so superposition can be applied.

##### Beats of a vibrating system

(6) $y''+\omega _{0}^{2}y=cos\omega t,\omega _{0}=\omega$
(6) is a second-order linear ODE. Even with the $\omega _{0}^{2}$ and $cos\omega t$ terms, (6) is still linear because these are functions of the independent variable. We only look at the dependent variable when determining the linearity of an ODE.
$(y_{h}''+y_{p}'')+\omega _{0}^{2}(y_{h}+y_{p})=cos\omega t$ ${\bar {y}}''+\omega _{0}^{2}{\bar {y}}=cos\omega t$ ${\bar {y}}$ satisfies (6), so superposition can be applied.

##### Current I in an RLC circuit

(7)$LI''+RI'+{\frac {1}{C}}I=E'$
(7) is a second-order linear ODE.
$L(I_{h}''+I_{p}'')+R(I_{h}'+I_{p}')+{\frac {1}{C}}(I_{h}+I_{p})=E'$ $L{\bar {I}}''+R{\bar {I}}'+{\frac {1}{C}}{\bar {I}}=E'$ ${\bar {I}}$ satisfies (7), so superposition can be applied.

##### Deformation of a beam

(8) $EIy^{iv}=f(x)$
(8) is a fourth-order linear ODE.
$EI(y_{h}^{iv}+y_{p}^{iv})=f(x)$ $EI{\bar {y}}^{iv}=f(x)$ ${\bar {y}}$ satisfies (8), so superposition can be applied.

##### Pendulum

(9) $L\theta ''+gsin\theta =0$
(9) is a second-order non-linear ODE.
$L(\theta _{h}''+\theta _{p}'')+g(sin\theta _{h}+sin\theta _{p})=0$ But $sin\theta _{h}+sin\theta _{p}\neq sin(\theta _{h}+\theta _{p})$ Therefore ${\bar {\theta }}$ is not a solution to (9), and superposition does not apply.

#### Honor Pledge

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.