# University of Florida/Egm4313/IEA-f13-team10/R1

## Report 1

[edit | edit source]### Problem 1.1

[edit | edit source]#### Problem Statement

[edit | edit source]Derive the equation of motion of a spring-dashpot system in parallel with a mass and applied force.

#### Solution

[edit | edit source]##### Kinematics

[edit | edit source]The spring and dashpot are in parallel, so the displacement of the spring is equal to the displacement of the dashpot.

(1)

##### Kinetics

[edit | edit source](2)

(3)

##### Constitutive Relations

[edit | edit source](4)

(5)

From (2), (3), (4) and (5), the equation of motion for the system is

From the first and second derivatives of (1),

So in terms of the displacement of the mass, the final equation of motion is,

#### Honor Pledge

[edit | edit source]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

[edit | edit source]#### Problem Statement

[edit | edit source]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

[edit | edit source]##### Kinematics

[edit | edit source]The displacement of the spring equals the negative displacement of the dashpot (constitutive relation)

##### Kinetics

[edit | edit source]Newton's second law

Substituting the relation above and kinematics section into the kinetics part, we get the equation of motion

Using the constitutive relations shown above we can write the equation of motion as

#### Honor Pledge

[edit | edit source]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

[edit | edit source]#### Problem Statement

[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]##### Step 1, Kinematics

[edit | edit source], therefore and

##### Step 2, FBD and Kinetics

[edit | edit source]

##### Step 3, Relations

[edit | edit source]

##### Step 4

[edit | edit source], therefore , and and

##### Step 5

[edit | edit source]Plugging the last equation from step 4 into the last equation from step 1,

##### Step 6

[edit | edit source]Plugging the equation from step 5 into the first kinetic equation as y",

#### Honor Pledge

[edit | edit source]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

[edit | edit source]#### Problem Statement

[edit | edit source]Derive (3) and (4) from (2).

(2)

(3)

(4)

#### Solution

[edit | edit source]##### Derivation of (3) from (2)

[edit | edit source](A)

(B)

Using equation A, we can show that and

Using equation B, we can show that

After these transformations, the new equation 2 is

Taking the derivative of both sides of equation 2, we get

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

[edit | edit source]Using equation B, we can show that and and

After these transformations, the new equation 2 is

#### Honor Pledge

[edit | edit source]### Problem 1.5

[edit | edit source]K 2011 p.59 pbs 3,12

#### Problem Statement

[edit | edit source]Find a general solution.

#### Solution

[edit | edit source]##### Problem 3

[edit | edit source]This is a linear, first order ODE with constant coefficients.

To find the general solution to this ODE set

so that and

Substituting in y to the ODE and factoring out we get:

Using the quadratic formula to solve for r we get

where a = 1, b = 6, and c = 8.96

Solving to get r = -3.2 and r = -2.8

We get the general solution

Now with the initial values y(0) = 1 and y'(0) = .5

,

##### Problem 12

[edit | edit source]

This is a linear, first order ODE with constant coefficients.

To find the general solution to this ODE set

so that and

Substituting in y to the ODE and factoring out we get:

Using the quadratic formula to solve for r we get

where a = 1, b = 9, and c = 20

Solving to get r = -5 and r = -4

We get the general solution

Now with the initial values y(0) = 1 and y'(0) = .5

,

#### Honor Pledge

[edit | edit source]### Problem 2.3

[edit | edit source]K 2011 p.3 Fig.2

#### Problem Statement

[edit | edit source]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

[edit | edit source]##### Falling stone

[edit | edit source](1)

Equation (1) is the ODE that describes the acceleration of the stone as it falls from some height 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 " (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 , which is comprised of the sum of the homogeneous solution

and the particular solution , should satisfy the original ODE. So if superposition applies,

(2) .

To determine whether superposition applies to (1), we consider

.

Now, using (2) and the linearity of the derivative,

.

satisfies (1), so **superposition can be applied**.

##### Parachutist

[edit | edit source](3)

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(w+y) does not equal f(w) + f(y) thus it is not linear.

Rearranging (3) we obtain
.

The homogeneous solution of the ODE is
.

The particular solution of the ODE is
.

Using (2) and the linearity of the derivative we obtain

but

Therefore, is not a solution to the ODE. So **superposition cannot be applied**.

##### Outflowing water

[edit | edit source](4)

(4) is a **first-order ODE** because of * h' *.

(4) is a **non-linear ODE** because of the square-rooted dependent variable *h*.

D(y+w) does not equal f(w)+f(y) so it is non-linear.

Rearranging (4) we obtain .

Summing the homogeneous and particular solutions we obtain .

But .

So is not a solution to (4), and **superposition cannot be applied**.

##### Vibrating mass on a spring

[edit | edit source](5)

(5) is a **second-order linear ODE**.

satisfies (5), so **superposition can be applied**.

##### Beats of a vibrating system

[edit | edit source](6)

(6) is a **second-order linear ODE**. Even with the and 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.

satisfies (6), so **superposition can be applied**.

##### Current I in an RLC circuit

[edit | edit source](7)

(7) is a **second-order linear ODE**.

satisfies (7), so **superposition can be applied**.

##### Deformation of a beam

[edit | edit source](8)

(8) is a **fourth-order linear ODE**.

satisfies (8), so **superposition can be applied**.

##### Pendulum

[edit | edit source](9)

(9) is a **second-order non-linear ODE**.

But

Therefore is not a solution to (9), and **superposition does not apply**.