# University of Florida/Egm4313/s12.team5.R1

Report 1

## R1.1

### Question

Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force f(t). (See figure 1) Figure 1

### Solution

Kinematic relationship:

$\displaystyle y=y_{c}=y_{k}$ Kinetic relationship:

$\displaystyle f(t)=m{\ddot {y}}+f_{k}+f_{c}$ Relations:

Spring:

$\displaystyle f_{k}=ky_{k}$ Dash-pot:

$\displaystyle f_{c}=C{\dot {y}}_{c}$ The next relationship can be found as:

$\displaystyle {\ddot {y}}={\ddot {y}}_{k}={\ddot {y}}_{c}$ If the following are equivalent:

$\displaystyle f_{k}=ky_{k}=ky_{c}$ $\displaystyle f_{c}=C{\dot {y}}_{c}$ Then the final equation can be found:

$\displaystyle f(t)=m{\ddot {y}}+f_{k}+f_{c}=m{\ddot {y}}_{c}+C{\dot {y}}_{c}+ky_{c}$ ### Author

This problem was solved and uploaded by Derik Bell

## R1.2

### 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.

Figure 2

### Solution

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: $\displaystyle f_{k}=ky_{k}$ and $\displaystyle f_{c}=cy_{c}^{'}$ 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 $\displaystyle \sum F_{y}$ $\displaystyle my^{''}+f_{c}+f_{k}=r(t)$ $\displaystyle my^{''}+cy_{c}^{'}+ky_{k}=r(t)$ ### Author

This problem was solved and uploaded by [Joshua House]

## R1.3

### Question

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

### Solution

Figure 3
(spring)

Figure 4 (dashpot)

Figure 5 (mass)

From the Free Body Diagrams we can see:

$\displaystyle f_{k}=f_{c}$ and $\displaystyle f_{c}=f_{i}$ ,
thus by summing the forces of the mass, we will be able to derive the equation of motion.

Summing the forces in the x-direction:

$\displaystyle \sum F_{x}=ma$ ;

$\displaystyle f(t)-f_{i}=ma$ ;

Solving for f(t):

$\displaystyle f(t)=ma+f_{i}$ ;

or

$\displaystyle f(t)=my^{''}+f_{i}$ ## R1.4

### Question

Using equation (2) from note section pg 2-2,

Derive:

Equation (3):

and

Equation (4):

$f_{2}(x,y)={\frac {1}{\sqrt {2}}}{\begin{pmatrix}\cos 135^{\circ }&-\sin 135^{\circ }\\\sin 135^{\circ }&\cos 135^{\circ }\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}+{\begin{pmatrix}1\\0\end{pmatrix}}$ ### Solution

#### Derivation of Equation 4

In order to derive equation 4, we need only to plug in the relations related to charge (Q) listed above. This will yield:

$f_{2}(x,y)={\frac {1}{\sqrt {2}}}{\begin{pmatrix}\cos 135^{\circ }&-\sin 135^{\circ }\\\sin 135^{\circ }&\cos 135^{\circ }\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}+{\begin{pmatrix}1\\0\end{pmatrix}}$ #### Derivation of Equation 3

Taking a derivative of equation 2 yields:

$ay^{2}-bx^{2}=C$ In order to derive equation 3 from the derivative of equation 2, we need only to plug in the relations related to current (I) listed above. This will yield:

### Author

This problem was solved and uploaded by [David Herrick]

## R1.5

### Question

Complete problems 4 and 5 from p.59 of K 2011.

### Solution

The two problems have the same instructions: "Find a general solution. Check your answer by substitution."

#### Problem 4

$\displaystyle 4.\;{y}''+4{y}'+(\pi ^{2}+4)y=0$ We start by using the characteristic equation of this ODE in order to find the roots. We use:

$\displaystyle \lambda ^{2}+a\lambda +b=0$ where a = 4 and b = (π^2 + 4). Taking the discriminant, we find that

$\displaystyle 4^{2}-4(\pi ^{2}+4)=-4\pi ^{2}<0$ A negative discriminant implies that there exists complex conjugate roots to this equation. The general solution for this type of equation is

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

$\displaystyle \omega ={\sqrt {b-{\frac {1}{4}}a^{2}}}={\sqrt {(\pi ^{2}+4)-({\frac {1}{4}})4^{2}}}=\pi$ This yields

$\displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))$ To verify that this is the correct solution, we first find the first and second derivatives of the solution:

$\displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))$ $\displaystyle {y}'=e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))$ $\displaystyle {y}''=e^{-2x}[-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x)+2A\pi sin(\pi x)-2B\pi sin(\pi x)]-2e^{-2x}[-A\pi sin(\pi x)+B\pi cos(\pi x)-2Acos(\pi x)-2Bsin(\pi x)]$ We then plug these values into the original problem to get this somewhat lengthy equation:

$\displaystyle e^{-2x}A\pi ^{2}cos(\pi x)-e^{-2x}B\pi ^{2}sin(\pi x)+2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi sin(\pi x)+2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi cos(\pi x)+$ $\displaystyle 4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x)-4e^{-2x}A\pi sin(\pi x)+4e^{-2x}B\pi cos(\pi x)-$ $\displaystyle 8e^{-2x}Acos(\pi x)-8e^{-2x}Bsin(\pi x)-e^{-2x}A\pi ^{2}cos(\pi x)+e^{-2x}B\pi ^{2}sin(\pi x)+4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x)=0$ Upon further inspection, all the terms on the left side of the equation cancel out to 0.

#### Problem 5

$\displaystyle 5.\;{y}''+2\pi {y}'+\pi ^{2}y=0$ As in problem 4, we start by using the characteristic equation for this ODE to determine the nature of the determinant:

$\displaystyle \lambda ^{2}+2\pi \lambda +\pi ^{2}$ Using the equation of the determinant, we find that

$\displaystyle {a}^{2}-4b={(2\pi )}^{2}-4({\pi }^{2})=0$ A discriminant that equals zero suggests a real double root, and the solution is of the form

$\displaystyle y=e^{\frac {-a}{2}}(c_{1}+c_{2}x)=e^{-\pi x}(c_{1}+c_{2}x)$ In order to check that this is the correct solution, we take the first and second derivative:

$\displaystyle y=e^{-\pi x}(c_{1}+c_{2}x)$ $\displaystyle {y}'=e^{-\pi x}(-c_{1}\pi -c_{2}\pi +c_{2})$ $\displaystyle {y}''=e^{-\pi x}(c_{1}\pi ^{2}+c_{2}\pi ^{2}-2c_{2}\pi )$ Plugging these values back into the original equation, we get:

$\displaystyle e^{-\pi x}(c_{1}\pi ^{2}+c_{2}\pi ^{2}-2c_{2}\pi -2c_{1}\pi ^{2}-2c_{2}\pi ^{2}+2c_{2}\pi +c_{1}\pi ^{2}+c_{2}\pi ^{2})=0$ Upon inspection, the terms on the left side of the equation cancel each other out and equal zero.

### Author

This problem was solved and uploaded by [William Knapper]

## R1.6

### Question

For each ODE in Fig.2 in K 2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show whether the principle of superposition can be applied.

### Given

(1.6.1)

$\displaystyle y''=g=const.$ (1.6.2)

$\displaystyle mv'=mg-bv^{2}$ (1.6.3)

$\displaystyle h'=-k{\sqrt {h}}$ (1.6.4)

$\displaystyle my''+ky=0$ (1.6.5)

$\displaystyle y''+w_{0}^{2}y=\cos wt,w_{0}/approx$ (1.6.6)

$\displaystyle LI''+RI'+{\frac {1}{C}}I=E'$ (1.6.7)

$\displaystyle EIy''''=f(x)$ (1.6.8)

$\displaystyle L\Theta ''+g\sin \Theta =0$ ### Solution

For equations (1.6.1-1.6.8) the order and linearity of each equation need to be determined. An ODE is said to be of nth order if the nth derivative of the function is the highest derivative in the equation. A first-order ODE is considered linear if it can be written into the form,

                                            y'+ p(x)y = r(x)


A second-order ODE is also considered linear if it can be written into the form,

                                           y"+ p(x)y' + q(x)y = r(x)


It also needs to be determined if the principle of superposition can be applied, or often called the linearity principle.

Fundamental Theorem for the Homogeneous Linear ODE(2) For a homogeneous linear ODE(2), any linear combination of two solutions on an open interval I is again a solution of (2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions.

(1.6.1) This is a second-order linear ODE based on the second derivative of y". The principle of superposition can also be applied because any constant or multiple applied to y" will still produce a constant.

                                                cy" = cg = c*constant


(1.6.2) This is a first-order nonlinear ODE because there is only one order of derivative on v. It is nonlinear because the powers of v are not all the same, and because the ODE is nonlinear then the principle of superposition can not be applied.

(1.6.3) This is a first-order nonlinear ODE. It is nonlinear because of the half-power affixed to the second h and thus the principle of superposition is inapplicable.

(1.6.4) This is a second-order linear ODE. All the powers of y are the same so linearity is confirmed and the principle of superposition is valid.

(1.6.5) This is a second-order linear ODE. Even though wo is affixed with a power of 2, the powers of y are still the same implying linearity. The principle of superposition can be applied for this ODE.

(1.6.6) This is a second-order linear ODE. The highest degree of derivation is present on the first I value and all of the powers of I are congruent implying linearity and the principle of superposition.

(1.6.7) This is a fourth-order linear ODE in which the principle of superposition can be applied.

(1.6.8) This is a second-order nonlinear ODE. Since it is nonlinear the principle of superposition can not be applied.

### Author

This problem was solved and uploaded by Michael Wallace