User:Egm4313.s12.team19.R1

R1.1 A Spring-Dashpot-Mass Parallel System
Contributed by Gabriel Arab.

Problem Statement:
Derive the equation of motion of a spring-dashpot system in parallel with a mass and applied force f(t).

Solution:
Choosing a coordinate system with (+) y-direction being to the right, we develop the following equations from the analysis of forces occurring on the mass.

For the spring force where k is the spring constant,


 * {| style = "width: 73%"

! $$ f_k = k \ y $$ !! (Eq. 1)
 * }

For the dashpot damping force where c is the viscious friction,


 * {| style = "width: 73%"

! $$ f_c = c \ y' $$ !! (Eq. 2)
 * }

For the resultant force on the mass,


 * {| style = "width: 73%"

! $$ F = m \ y'' $$ !! (Eq. 3)
 * }

Since from Calculus, acceleration is the second derivative of position.

Newton’s Second Law can then be applied to the mass, summing all forces at assumable equal distances,


 * {| style = "width: 73%"

! $$ \Sigma F = m \ y'' = -f_k - f_c + f (t) $$ !! (Eq. 4)
 * }

Substituting (1) and (2) into (4) and rearranging into standard ODE form yields the equation of motion for a spring-dashpot-mass system:


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ y'' + {c \over m} y' + {k \over m} y = {1 \over m} f (t) $$ (Eq. 6)
 * }

R1.2 A Spring-Dashpot-Mass Series System
Contributed by Adam Burton.

Given:

We are given a mass-spring system, with a mass m on an elastic spring. This is shown in Figure 53 in Kreyszig 2011 page 85 (shown below). For this problem, $$ y(t) $$ is a function of time t of the displacement of the mass m from rest.

Problem Statement:
We are asked to derive the equation of motion of the mass spring dashpot system shown in the figure, with an external force $$ r(t) $$ applied on the ball.

Solution:
From Newton’s second law of motion:


 * {| style = "width: 73%"

! $$ \Sigma F = m a = m {d^2 y \over d t^2} $$ !! (Eq. 1)
 * }

In this system, there are three internal forces and one external force. The internal forces include the inertial force $$ m \ddot {y} $$, the damping force $$ c \ddot {y} $$, and the spring force $$ k y $$.

The external force is $$ r(t) $$.

We also know that the force of a spring is given as:


 * {| style = "width: 73%"

! $$ F_s = -k \ y $$ !! (Eq. 2)
 * }

And that the force caused by damping is given as:
 * {| style = "width: 73%"

! $$ F_D = c {d y \over d t} $$ !! (Eq. 3)
 * }

For conventional purposes, $$ \dot y = {d y \over d t} $$, and $$ \ddot y = {d^2 y \over d t^2} $$.

Using a Free Body Diagram:


 * {| style = "width: 73%"

! $$ \Sigma F = m {d^2 y \over d t^2} $$ !! (Eq. 4)
 * }


 * {| style = "width: 73%"

! $$ m \ddot y = {-c \dot y - k y + r (t)} $$ !! (Eq. 5)
 * }


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ \ m \ddot y + c \dot y + k y = r (t) $$ (Eq. 6)
 * }

R1.3 A spring–dashpot–mass system
Contributed by Joshua Campbell.

Given: A spring–dashpot(damper)–mass system

Problem Statement:
Use the FBD to derive the equation of motion (2) from lecture slide 1–4.

Solution:
Assuming no gravitational acceleration in the downward direction, the FBD of the mass is:



where f(t) is a force exerted on the mass as a function of time, and fx is the equal and opposite force asserted by Newton’s 3rd law.

An expression of Newton’s 2nd law states:


 * {| style = "width: 73%"

! $$ \Sigma F = {dv \over dt} = 0 $$ !! (Eq. 1)
 * }

Since
 * {| style = "width: 73%"

! $$ \ a = v' = y'' $$ !! (Eq. 2)
 * }

And
 * {| style = "width: 73%"

! $$ \ \Sigma F = m a $$ !! (Eq. 3)
 * }

Thus,
 * {| style = "width: 73%"

! $$ \ m y'' = \Sigma F $$ !! (Eq. 4)
 * }

or


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ \ m y'' + f_y = f (t) $$ (Eq. 5)
 * }
 * }

R1.4 RLC Voltage
Contributed by Charles Chiamchittrong.

Given:

Equation (2) p.2-2 which describes the voltage in an RLC circuit in series:


 * {| style = "width: 73%"

! $$ V = L C \ {d^2 v _C \over d t^2} + R C \ {d v _C \over d t} + v_C $$ !! (Eq. 1)
 * }

Problem Statement:
Derive the following alternate forms of the given equation:


 * {| style = "width: 73%"

! $$ L I'' + R I' + {1 \over C} I = V' $$ !! (Eq. 2) p. 2-2
 * }


 * {| style = "width: 73%"

! $$ L Q'' + R Q' + {1 \over C} Q = V $$ !! (Eq. 3) p. 2-2
 * }

Part 1
In order to derive equation (2) we start by taking the derivative of equation (1):


 * {| style = "width: 73%"

! $$ V' = L C {d^3 v_C \over d t^3} + R C {d^2 v_C \over d t^2} + {d v_C \over d t} $$ !! (Eq. 4)
 * }

The variables L (inductance), C (capacitance), and R (resistance) represent constant values in the circuit that do not change over time and therefore remain unaffected.

Next we use part of equation (1) p.2-2 which relates current and capacitance:


 * {| style = "width: 73%"

! $$ I = C {d v_C \over d t} $$ !! (Eq. 5)
 * }

By taking the derivative of this equation twice, we obtain the following:


 * {| style = "width: 73%"

! $$ I' = C {d^2 v_C \over d t^2} $$ !! (Eq. 6)
 * }

and


 * {| style = "width: 73%"

! $$ I''' = C {d^3 v_C \over d t^3} $$ !! (Eq. 7)
 * }

Also, we can manipulate equation (5) by dividing by C:


 * {| style = "width: 73%"

! $$ {I \over C} = {d v_C \over d t} $$ !! (Eq. 8)
 * }

Now, we can substitute equations (6), (7), and (8) back into equation (4), yielding the following:


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ V' = L I'' + R I' + {I \over C} $$ (Eq. 9)
 * }

Rewriting the third term on the right-hand side and rearranging yields the given equation (2) that we were trying to derive.

Part 2
In order to derive equation (3) we use another part of equation (1) p.2-2 which relates charge to capacitance:


 * {| style = "width: 73%"

! $$ \ Q = C v_C $$ !! (Eq. 10)
 * }

We manipulate this equation in a manner similar to that of equation (5). First we take the derivative twice, resulting in the following:


 * {| style = "width: 73%"

! $$ Q' = C {d v_C \over d t} $$ !! (Eq. 11)
 * }

and


 * {| style = "width: 73%"

! $$ Q'' = C {d^2 v_C \over d t^2} $$ !! (Eq. 12)
 * }

Next we divide equation (10) by C:


 * {| style = "width: 73%"

! $$ {Q \over C} = v_C $$ !! (Eq. 13)
 * }

Finally, we can substitute equations (11), (12), and (13) back into the original given equation (1):


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ V = L Q'' + R Q' + {Q \over C} $$ (Eq. 14)
 * }

Once again, rewriting the third term on the right-hand side and rearranging yields the equation (3) that we are looking for.

R1.5 Selected Problems from Kreyszig 2011
Contributed by Santiago Marin.

Problem Statement:
Solve the following questions from the book.

K 2011 p.59 pb. 4
Given: $$ \ y'' + 4 y' + (\pi^2 + 4) y = 0 $$

Find: A general solution

Solution:
The equation is a second-order homogeneous ODE taking the standard form:


 * {| style = "width: 73%"

! $$ \ y'' + a y' + b y = 0 $$ !! (Eq. 1)
 * }

Using the discriminant: $$ \ a^2 - 4 b $$
 * {| style = "width: 73%"

! $$ \ 4^2 - 4 (\pi^2 + 4) = -4 \pi^2 $$ !! (Eq. 2)
 * }

Since $$ \ -4 \pi^2 < 0 $$ it gives a general solution:

$$ \ y = e^{\left({-a x \over 2} \right)} (A \cos (\omega x) + B \sin (\omega x)) $$ where $$ \ \omega^2 = b - {1 \over 4} a^2 $$


 * {| style = "width: 73%"

! $$ \ \omega^2 = \pi^2 + 4 - {1 \over 4} (4^2) $$ !! (Eq. 3)
 * }


 * {| style = "width: 73%"

! $$ \ \omega = \pi $$ !! (Eq. 4)
 * }

Substituting back into the general solution gives:


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ \ y = e^{(-2x)} (A \cos (\pi x) + B \sin (\pi x)) $$ (Eq. 5)
 * }

K 2011 p.59 Problem 5
Given: $$ \ y'' + 2 \pi y' + \pi^2 y = 0 $$

Find: A general solution

Solution:
The equation is a second-order homogeneous ODE taking the standard form:


 * {| style = "width: 73%"

! $$ \ y'' + a y' + b y = 0 $$ !! (Eq. 6)
 * }

Using the discriminant: $$ \ a^2 - 4 b $$


 * {| style = "width: 73%"

! $$ \ 4 \pi^2 - 4 \pi^2 = 0 $$ !! (Eq. 7)
 * }

Since $$ \ 4 \pi^2 - 4 \pi^2 = 0 $$ it gives a general solution:


 * {| style = "width: 73%"

! $$ \ y = (c_1 + c_2 x) e^{\left({-a x \over 2}\right)} $$ !! (Eq. 8) Substituting back into the general solution gives:
 * }
 * {| class = wikitable style="width: 73%; text-align:left"

! $$ \ y = (c_1 + c_2 x) e^{(-\pi x)} $$ (Eq. 9)
 * }
 * }

R1.6 Selected Problems from Kreyszig 2011
Contributed by Grant (Kyle) Uppercue.

Problem statement:
"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 there of), and show whether the principle of superposition can be applied." Sec 2 (d)

Definitions:

Order = Highest order of derivative

"A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable." 

Principle of Superposition

If y1(t) and y2(t) are two solutions to a linear, homogeneous differential equation then so is


 * {| style = "width: 73%"

! $$ \ y (t) = c_1 y_1 (t) + c_2 y_2 (t) $$ !! (Eq. 1)
 * }

Solutions:

 * 1) Falling Stone: $$ \ y'' = g = constant $$
 * 2) Order = 2
 * 3) Linearity (or lack of): Linear.
 * 4) Show whether the principle of superposition can be applied:


 * {| style = "width: 73%"

! &#9; $$ \ y'' = g = constant $$ !! (Eq. 2)
 * $$ \ y_h'' = 0 $$
 * (Eq. 3)
 * $$ \ y_p'' = g $$
 * (Eq. 4)
 * $$ \ y_h + y_p = g $$
 * (Eq. 5)
 * }
 * $$ \ y_h + y_p = g $$
 * (Eq. 5)
 * }


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ \bar y (x) = (y_h + y_p)'' $$ (Eq. 6)
 * }

Superposition can be applied


 * 1) Parachutist: $$ \ m v' = m g - b v^2 $$
 * 2) Order = 1
 * 3) Linearity (or lack of): Non-linear
 * 4) Show whether the principle of superposition can be applied:


 * {| style = "width: 73%"

! $$ \ m v' = m g - b v^2 $$ !! (Eq. 7)
 * $$ \ m v' + b v^2 = m g $$
 * (Eq. 8)
 * $$ \ m v_h'+ b v_h^2 = 0 $$
 * (Eq. 9)
 * $$ \ m v_p'+ b v_p^2 = m g $$
 * (Eq. 10)
 * $$ \ m v^' + b v^2 = m g $$
 * (Eq. 11)
 * $$ \ m v_p' + b v_p^2 + m v_h' + b v_h^2 = m g $$
 * (Eq. 12)
 * $$ \ {m (v_p + v_h )}' + b (v_p^2 + v_h^2) = m g $$
 * (Eq. 13)
 * }
 * $$ \ m v_p' + b v_p^2 + m v_h' + b v_h^2 = m g $$
 * (Eq. 12)
 * $$ \ {m (v_p + v_h )}' + b (v_p^2 + v_h^2) = m g $$
 * (Eq. 13)
 * }
 * }


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ {m (v_p + v_h)}' + b (v_p + v_h)^2 \neq m g $$ (Eq. 14)
 * }

Superposition can NOT be applied


 * 1) Outflowing Water: $$ \ h' = -k \sqrt h $$
 * 2) Order = 1
 * 3) Linearity (or lack of): Non-linear.
 * 4) Show whether the principle of superposition can be applied:


 * {| style = "width: 73%"

! $$ \ h' = -k \sqrt h $$ !! (Eq. 15)
 * }


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ \ h' + k \sqrt h = 0 $$ (Eq. 16) Since h involves a square root, the solution would be a complex solution (involving imaginary numbers), and Superposition can Not be applied.
 * }


 * 1) Vibrating Mass on a Spring: $$ \ m y'' + k y = 0 $$
 * 2) Order = 2
 * 3) Linearity (or lack of): Linear
 * 4) Show whether the principle of superposition can be applied:


 * {| style = "width: 73%"

! $$ \ m y'' + k y = 0 $$ !! (Eq. 17)
 * $$ \ m y_h'' + k y_h = 0 $$
 * (Eq. 18)
 * $$ \ m y_p''+ k y_p = 0 $$
 * (Eq. 19)
 * $$ \ m (y_h + y_p)'' + k (y_h + y_p) $$
 * (Eq. 20)
 * }
 * $$ \ m (y_h + y_p)'' + k (y_h + y_p) $$
 * (Eq. 20)
 * }


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ m \bar y (x)'' + k \bar y (x) = 0 $$ (Eq. 21)
 * }

Super Position can be applied


 * 1) Beats of a Vibrating System: $$ y'' + \omega_0^2 y = cos (\omega t), \ \omega_0 \asymp \omega $$
 * 2) Order = 2
 * 3) Linearity (or lack of): Linear
 * 4) Show whether the principle of superposition can be applied:


 * {| style = "width: 73%"

! $$ \ y'' + \omega_0^2 y = cos (\omega t), \ \omega_0 \asymp \omega $$ !! (Eq. 22)
 * $$ \ (y_h)'' + \omega_0^2 y_h = 0 $$
 * (Eq. 23)
 * $$ \ (y_p)'' + \omega_0^2 y_p = \cos (\omega t) $$
 * (Eq. 24)
 * $$ \ (y_h + y_p)'' + \omega_0^2 (y_h + y_p) = \cos (\omega t) $$
 * (Eq. 25)
 * }
 * $$ \ (y_h + y_p)'' + \omega_0^2 (y_h + y_p) = \cos (\omega t) $$
 * (Eq. 25)
 * }


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ \bar y (x)'' + \omega_0^2 \bar y (x) = \cos (\omega t) $$ (Eq. 26)
 * }

Super Position can be applied


 * 1) Current I in an RLC Circuit: $$ \ L I'' + R I' + {1 \over C} I = E' $$
 * 2) Order = 2
 * 3) Linearity (or lack of): Linear
 * 4) Show whether the principle of superposition can be applied:


 * {| style = "width: 73%"

! $$ \ L I'' + R I' + {1 \over C} I = E' $$ !! (Eq. 27)
 * $$ \ L (I_p)'' + R (I_p)' + {1 \over C} I_p = E' $$
 * (Eq. 28)
 * $$ \ L (I_h)'' + R (I_h)' + {1 \over C} I_h = 0 $$
 * (Eq. 29)
 * $$ \ L (I_p + I_h)'' + R (I_p + I_h)' + {1 \over C} (I_p + I_h) = E' $$
 * (Eq. 30)
 * }
 * $$ \ L (I_p + I_h)'' + R (I_p + I_h)' + {1 \over C} (I_p + I_h) = E' $$
 * (Eq. 30)
 * }


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ L \bar I'' + R \bar I' + {1 \over C} \bar I = E' $$ (Eq. 31)
 * }

Super Position can be applied


 * 1) Deformation of a Beam: $$ \ E I y^{iv} = f (x) $$
 * 2) Order = 4
 * 3) Linearity (or lack of): Linear
 * 4) Show whether the principle of superposition can be applied:


 * {| style = "width: 73%"

! $$ \ E I y^{iv} = f (x) $$ !! (Eq. 32)
 * $$ \ E I (y_h)^{iv} = 0 $$
 * (Eq. 33)
 * $$ \ E I (y_p)^{iv} = f (x) $$
 * (Eq. 34)
 * }
 * (Eq. 34)
 * }


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ E I (y_p + y_h)^{iv} = E I \bar y^{iv} $$ (Eq. 35)
 * }

Super Position can be applied


 * 1) Pendulum: $$ \ L \theta'' + g sin (\theta) = 0 $$
 * 2) Order = 2
 * 3) Linearity (or lack of): Non-linear
 * 4) Show whether the principle of superposition can be applied:


 * {| style = "width: 73%"

! $$ \ L \theta'' + g \sin (\theta) = 0 $$ !! (Eq. 36)
 * $$ \ L (\theta_h)'' + g \sin (\theta_h) = 0 $$
 * (Eq. 37)
 * $$ \ L (\theta_p)'' + g \sin (\theta_p) = 0 $$
 * (Eq. 38)
 * $$ \ L (\theta_h + \theta_p)'' + g \sin (\theta_h + \theta_p) = 0 $$
 * (Eq. 39)
 * }
 * $$ \ L (\theta_h + \theta_p)'' + g \sin (\theta_h + \theta_p) = 0 $$
 * (Eq. 39)
 * }


 * {| class = wikitable style="width: 73%; text-align:center"

! $$ L \bar \theta'' + g \sin (\bar \theta) \neq 0 $$ (Eq. 40)
 * }

Super Position can Not be applied

Microsoft Word source
Original Microsoft Word text compiled into one file.

Converted by Vladimir Horwitz.

Page information
Page designed by Vladimir Horwitz on 00:56, 28 January 2012 (UTC).

Last edited by Egm4313.s12.team19.Horwitz on 20:39, 1 February 2012 (UTC).