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

Report 1

## R1.1[edit | edit source]

### Question[edit | edit source]

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[edit | edit source]

Kinematic relationship:

Kinetic relationship:

Relations:

Spring:

Dash-pot:

The next relationship can be found as:

If the following are equivalent:

Then the final equation can be found:

### Author[edit | edit source]

This problem was solved and uploaded by Derik Bell

## R1.2[edit | edit source]

### Question[edit | edit source]

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[edit | edit source]

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

Where is the force due to the spring, is the force due to the dashpot, and is the applied force.

Therefore: and

Both the force due to the spring and the force due to the dashpot oppose motion, and is assumed to act in the negative y direction.

From

### Author[edit | edit source]

This problem was solved and uploaded by [Joshua House]

## R1.3[edit | edit source]

### Question[edit | edit source]

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

### Solution[edit | edit source]

**Figure 3**

*(spring)*

From the Free Body Diagrams we can see:

and ,

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:

- ;

- ;

Solving for f(t):

- ;

or

### Author[edit | edit source]

This problem was solved and uploaded by Radina Dikova

## R1.4[edit | edit source]

### Question[edit | edit source]

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

Derive:

Equation (3):

and

Equation (4):

### Solution[edit | edit source]

#### Derivation of Equation 4[edit | edit source]

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

#### Derivation of Equation 3[edit | edit source]

Taking a derivative of equation 2 yields:

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[edit | edit source]

This problem was solved and uploaded by [David Herrick]

## R1.5[edit | edit source]

### Question[edit | edit source]

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

### Solution[edit | edit source]

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

#### Problem 4[edit | edit source]

We start by using the characteristic equation of this ODE in order to find the roots. We use:

where a = 4 and b = (π^2 + 4). Taking the discriminant, we find that

A negative discriminant implies that there exists complex conjugate roots to this equation. The general solution for this type of equation is

Where

This yields

To verify that this is the correct solution, we first find the first and second derivatives of the solution:

We then plug these values into the original problem to get this somewhat lengthy equation:

Upon further inspection, all the terms on the left side of the equation cancel out to 0.

#### Problem 5[edit | edit source]

As in problem 4, we start by using the characteristic equation for this ODE to determine the nature of the determinant:

Using the equation of the determinant, we find that

A discriminant that equals zero suggests a real double root, and the solution is of the form

In order to check that this is the correct solution, we take the first and second derivative:

Plugging these values back into the original equation, we get:

Upon inspection, the terms on the left side of the equation cancel each other out and equal zero.

### Author[edit | edit source]

This problem was solved and uploaded by [William Knapper]

## R1.6[edit | edit source]

### Question[edit | edit source]

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[edit | edit source]

(1.6.1)

(1.6.2)

(1.6.3)

(1.6.4)

(1.6.5)

(1.6.6)

(1.6.7)

(1.6.8)

### Solution[edit | edit source]

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 w_{o} 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[edit | edit source]

This problem was solved and uploaded by Michael Wallace