User:Egm4313.s12.team12/R1

Statement
Given: A spring-dashpot mass system in parallel.

Find: A second-order ordinary differential equation to describe the system's motion.

Diagram of the Entire System:



Free Body Diagram of the forces acting on the Spring:



Free Body Diagram of the forces acting on the Dashpot:



Free Body Diagram of the forces acting on the Mass:



Assumptions
1) Let the position direction (y) be positive as the mass system moves to the right.

2) The mass of the spring and the dashpot are negligible in terms of the force they apply on the mass.

3) Let Fc be the force caused by the dashpot on the mass.

4) Let Fs be the force caused by the spring on the mass.

5) f(t) will denote the variable pulling force that acts on the mass over time.

6) During motion, the spring and dashpot will not be stretched beyond their elastic limits, and will remain connected to the mass.

Solution
In order to develop an equation of motion, Newton's Second law will be used. The sum of all the forces in the direction of motion, y, will be balanced by the body's mass and acceleration. As shown in the above FBD, the forces acting on the body are the variable force, the force of the spring, and the force of the dashpot. The weight of the body also acts, but not in this given direction of motion, so it is not considered. A reference zero for the initial displacement, y0, will be considered from the location of the mass when the entire system is in equilibrium, i.e., when the body is not moving.

The force of the spring Fs is equation to the magnitude of the product of its stretched displacement, y, and a given spring constant, k. Therefore, we can write:


 * $$F_s = ky_k$$

The force of the dashpot is equal to its equivalent displacement, y, and a constant c. Therefore:


 * $$F_c = cy'_{c}$$

If the sum of all the forces acting on the body is equal to the body's mass and acceleration, which is equal to the second derivative of the body's displacement y, the following equation can be written:


 * $$\sum F_y = my'' = f(t) - F_c - F_k$$

If their equivalent values are substituted into this equation, the following is formed:


 * $$\sum F_y = my'' = f(t) - cy'_{c} - ky_k$$

However, in order to properly write a solvable second-order ODE (ordinary differential equation), the equation given must be in terms of one variable. In this instance, it can be shown that the equation can be written in terms of y. If the body is pulled a certain distance y due to the variable force f(t), then the dashpot and spring will be forced to extend by the same displacement amount y in order to remain in contact with the mass. Otherwise, the spring and the dashpot will be disconnected. Therefore, we can write:

Subsequently, the first and second derivatives of the variable displacement y can be taken to yield the following:


 * $$y' = y'_{k} = y'_{c}$$


 * $$y = y_{k} = y''_{c}$$

Therefore, it is shown that the displacement of both the dashpot and the spring is equal to the displacement of the mass. Therefore, we can write the following second-order ODE in terms of one variable y :



Then, if the equation is manipulated, the equation can be written to solve for the variable force:



Finally, the equation can be written in standard from by dividing by the mass m to yield the function:



Author & Resources
Author: Daniel Stewart 128.227.38.74 19:09, 1 February 2012 (UTC)

Resource: Kreyszig, Erwin, Herbert Kreyszig, and E. J. Norminton. Advanced Engineering Mathematics. Hoboken, NJ: Wiley, 2011. Print.

Statement
Derive the equation of motion of the spring-mass-dashpot in Fig. 53, in Kreyszig 2011 p.85, with an applied force $$r(t)\!$$ on the ball.



Source: Kreyszig, Erwin, Herbert Kreyszig, and E. J. Norminton. Advanced Engineering Mathematics. Hoboken, NJ: Wiley, 2011.

Solution
Kinematic Equations:

$$ \displaystyle y=y_k=y_c $$

Kinetics:

$$ \displaystyle my''+{{f}_{k}}+{{f}_{c}}= r(t)$$

Constitutive relations:

$$ \displaystyle {{f}_{k}}=k{{y}_{k}}$$

$$ \displaystyle {{f}_{c}}=c{{y}_{c}}'$$

Rearranging the equations:

$$ \displaystyle my''+k{{y}_{k}}+c{{y}_{c}}' = r(t)$$

Equation of motion:


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$$ \displaystyle my''+ky+cy' = r(t)$$
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Author & Resources
Author: Garrett Sutcliffe

Statement
For the spring dashpoint mass system on p1-4, draw the FBDs and derive the equation of motion (2). p1-4

Solution


Start with Equilibrium equation

$$ \sum F = ma $$

$$ a = \frac{d^2y}{dx^2} =y'' $$

$$ ma = m'' $$

also the forces of the spring internal force and dashpoint are equal

$$ f_I = f_k = f_c $$

so plugging into equilibrium equation

$$ f(t) - f_I = my'' $$

Which gives you equation 2 by solving for f(t)

$$ my'' + f_I = f(t) $$

Author & Resources
Author: Patrick Greivell

Resource(s): Kreyszig, Erwin, Herbert Kreyszig, and E. J. Norminton. Advanced Engineering Mathematics. Hoboken, NJ: Wiley, 2011.

Statement
Derive

$$ V = LQ'' + RQ' + \frac{Q}{C} $$           (1)

$$ V' = LI'' + RI' + \frac{I}{C} $$          (2)

From

$$ V= LC\frac{dV^2_c}{dt^2} + RC\frac{dV_c}{dt} + V_c $$

Solution
Solving for (1)

If $$ Q = CV_c $$  (3)

We can differentiate twice to find:

$$ Q' = C\frac{dV_c}{dt} $$ (4)

$$ Q'' = C\frac{d^2V_c}{dt^2} $$ (5)

and we can also rearrange (3) to find

$$ V_c = \frac{Q}{C} $$ (6)

Equations (4), (5), and (6) can then be substituted into the initial equation to find:

$$ V = LQ'' + RQ' + \frac{Q}{C} $$

Solving for (2)

We first differentiate the initial equation to find:

$$ V' = LC\frac{d^3V_c}{dt^3} + RC\frac{d^2V_c}{dt^2} + \frac{dV_c}{dt} $$

We then differentiate

$$ I = C\frac{dV_c}{dt} $$ (7)

to get

$$ I' = C\frac{d^2V_c}{dt^2} $$ (8)

$$ I'' = C\frac{d^3V_c}{dt^3} $$ (9)

Equations (7), (8), and (9) can then be substituted into the initial equation to find:

$$ V' = LI'' + RI' + \frac{I}{C} $$

Author & Resources
Cpettigrew 01:40, 1 February 2012 (UTC)

Statement
K 2011 p.59 pbs. 4-5 Find a general solution. Check your answer by substitution.

Solution
Problem 4: Consider an ODE of the form:  $$ \displaystyle {y}'' + 4y' + (\pi ^{2}+4)y = 0$$  First we notice that the ODE is homogenous, linear, and has constant coefficients.  Therefore, we find the auxiliary equation, which is of the form:  $$ \displaystyle \lambda^2 + a\lambda + b = 0 $$  Looking at the coefficients we obtain:  $$ \displaystyle \lambda^{2} + 4\lambda + (\pi^{2} + 4) = 0 $$  Next we obtain the roots of this equation with the quadratic formula: $$ \displaystyle \ \frac{-4 \pm \sqrt{(-4)^2 - 4(1)(\pi^2 + 4)}}{2(1)} $$  This can be simplified to:  $$ \displaystyle -2\pm\frac{\sqrt{16 - 4\pi^2 - 16}}{2} $$  $$ \displaystyle -2\pm\frac{\sqrt{-4\pi^2}}{2} $$<br\> We notice that by looking inside the radical, we will be obtaining complex roots. Continuing with simplification we get: <br\><br\> $$ \displaystyle -2\pm{i\pi} $$<br\><br\> The general solution to the case of complex roots is: <br\><br\> $$ \displaystyle y = e^{\frac{-ax}{2}}(Acos\omega{x} + Bsin\omega{x}) $$ <br\> <br\> Where complex roots are of the form: <br\><br\> $$ \displaystyle -\frac{1}{2}a\pm{i\omega} $$<br\><br\> Knowing $$ \displaystyle a = 4 $$ and $$ \displaystyle \omega = \pi $$, we get the final general solution: <br\><br\> $$ \displaystyle y = e^{-2x}(Acos\pi{x} + Bsin\pi{x}) $$<br\><br\>

Problem 5: Consider an ODE of the form: <br\><br\> $$ \displaystyle {y}'' + 2\pi y' + \pi ^{2}y = 0$$ <br\> This ODE is linear, and has constant coefficients. Therefore, its general solution is of the form: <br\><br\> $$ \displaystyle {y}=e^{\lambda x} $$ <br\> Therefore: <br\><br\> $$ \displaystyle {y}'=\lambda e^{\lambda x} $$ <br\> And: <br\><br\> $$ \displaystyle {y}''={\lambda}^2 e^{\lambda x} $$ <br\> The ODE can be rearranged so that: <br\><br\> $$ \displaystyle e^{\lambda x}(\lambda^2 + 2\pi \lambda + \pi^2)=0 $$ <br\> And since: <br\><br\> $$ \displaystyle e^{\lambda x} \neq 0 $$ <br\> Then: <br\><br\> $$ \displaystyle \lambda^2 + 2\pi \lambda + \pi^2=0 $$ <br\> $$ \displaystyle (\lambda + \pi)=0 $$ <br\> $$ \displaystyle \lambda =- \pi $$ <br\> Since: <br\><br\> $$ \displaystyle \lambda_{1} = \lambda_{2} $$ <br\> The general solution then becomes: <br\><br\> $$ \displaystyle y = c_{1}e^{\lambda x} + c_{2}xe^{\lambda x} $$ <br\> $$ \displaystyle y = e^{\lambda x} (c_{1} + c_{2}x) $$ <br\> Therefore: <br\><br\> $$ \displaystyle y = e^{-\pi x} (c_{1} + c_{2}x) $$ <br\>

Author & Resources
Problems 4 and 5 were solved by [James Stadick] and [Wagner Schulz]

Statement
For each ODE determine the order, linearity (or lack of), and show whether the principle of superposition can be applied:

1) $$   \displaystyle  y''=g=constant $$

2) $$   \displaystyle  mv' = mg - bv^2 $$

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

4) $$   \displaystyle  my'' + ky = 0 $$

5) $$   \displaystyle y'' + \omega_{0}^2y = \cos \omega t, \omega_{0}\approx\omega $$

6) $$   \displaystyle  LI'' + RI' + \frac{C}I = E' $$

7) $$   \displaystyle  EIy = f(x) $$

8) $$   \displaystyle  L\theta'' + g\sin\theta = 0 $$

Solution
1) $$   \displaystyle  y'' = g = constant $$<br\><br\> a) Order of ODE<br\> - The order of an ODE is just the highest derivative seen in the equation<br\>
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This ODE is of the 2nd order
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b) Linearity of ODE<br\> - An ODE is said to be linear if the variable in the equation is always seen to the first power<br\>
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This ODE is linear
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c) Principle of Superposition<br\> - The Principle of Superposition states that for a linear homogeneous Ordinary Differential Equation where $$ \displaystyle y_1(t) $$ and $$ \displaystyle y_2(t) $$ are solutions then so is $$ \displaystyle y(t) = c_1y_1(t) + c_2y_2(t) $$. Therefore in order for this principle to be applied the equation has to be a linear homogeneous ODE.<br\>
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Yes the Principle of Superposition can be applied
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2) $$ \displaystyle mv' = mg - bv^2 $$<br\><br\> a) Order of ODE<br\>
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This ODE is of the 1st order
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b) Linearity of ODE<br\>
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This ODE is non-linear
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c) Superposition Principle<br]>
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No the Principle of Superposition cannot be applied
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3) $$ \displaystyle h' = -k\sqrt{h} $$<br\><br\> a) Order of ODE<br\>
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This ODE is of the 1st order
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b) Linearity of ODE<br\>
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This ODE is non-linear
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c) Principle of Superposition<br\>
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No the Principle of Superposition cannot be applied
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4) $$ \displaystyle my'' + ky = 0 $$<br\><br\> a) Order of ODE<br\>
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This ODE is of the 2nd order
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b) Linearity of ODE<br\>
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This ODE is linear
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c) Principle of Superposition<br\>
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Yes the Principle of Superposition can be applied
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5) $$ \displaystyle y'' + \omega_{0}^2y = \cos \omega t, \omega_{0} \approx \omega $$<br\><br\> a) Order of ODE<br\>
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This ODE is of the 2nd order
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b) Linearity of ODE<br\>
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This ODE is Linear
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c) Principle of Superposition<br\>
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Yes the Principle of Superposition can be applied
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6) $$ \displaystyle LI'' + RI' + \frac{1}{c}I = E' $$<br\><br\> a) Order of ODE<br\>
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This ODE is of the 2nd order
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b) Linearity of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is linear
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c) Principle of Superposition<br\>
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Yes the Principle of Superposition can be applied
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7) $$ \displaystyle EIy' = f(x) $$<br\><br\> a) Order of ODE'''<br\>
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This ODE is of the 4th order
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b) Linearity of ODE<br\>
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This ODE is linear
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c) Principle of Superposition<br\>
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Yes the Principle of Superposition can be applied
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8) $$ \displaystyle L\theta'' = g\sin\theta = 0 $$<br\><br\> a) Order of ODE<br\>
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This ODE is of the 2nd order
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b) Linearity of ODE<br\>
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This ODE is non-linear
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c) Principle of Superposition<br\>
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No the Principle of Superposition cannot be applied
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Author & Resources
Author: Lauren Anders

Resource: Kreyszig, Erwin, Herbert Kreyszig, and E. J. Norminton. Advanced Engineering Mathematics. Hoboken, NJ: Wiley, 2011. Print.