User:Egm4313.s12.team8/R1

=R1.1 - Equation of Motion for a Parallel Spring-Dashpot System=

Problem Statement


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

Solution
Solved on my own

Kinematics

 * $$\displaystyle\begin{align}

y = y_k = y_c \end{align}$$

Kinetics

 * $$\displaystyle m y'' + f_k + f_c = f(t)$$

Constitutive Relations

 * $$\displaystyle\begin{align}

f_k &= k y_k \\ f_c &= c y_c' \end{align}$$

Therefore
$$\displaystyle m y''(t) + c y'(t) + k y(t) = f(t)$$

The general solution of the above ODE is:


 * $$\displaystyle\begin{align}

y_g (t) &= C_1 \; e^{\frac{-c - \sqrt{c^2 - 4 m k}}{2 m} t} \\ &+ C_2 \; e^{\frac{-c + \sqrt{c^2 - 4 m k}}{2 m} t} \\ &- \frac{\int e^{\frac{\left( c + \sqrt{c^2 + 4 m k} \right) t}{2 m}} f(t) \,dt}{\sqrt{c^2 - 4 m k}} \; e^{\frac{-c - \sqrt{c^2 - 4 m k}}{2 m} t} \\ &+ \frac{\int e^{\frac{\left( c - \sqrt{c^2 + 4 m k} \right) t}{2 m}} f(t) \,dt}{\sqrt{c^2 - 4 m k}} \; e^{\frac{-c + \sqrt{c^2 - 4 m k}}{2 m} t} \end{align}$$

=R1.2 - Equation of Motion for a Series Spring-Dashpot System=

Problem Statement


Derive the equation of the spring-mass-dashpot in Fig.53 (shown above) in K 2011 p.85, with an applied force $$\displaystyle r(t) $$ on the ball.

Kinematics
$$\displaystyle y=y_{c}=y_{k} $$ (2-0)

Kinetics
$$\displaystyle r(t)=m y''+f_{c}+f_{k} $$ (2-1)

Where
$$\displaystyle f_{k}=k y_{k}$$ (2-2) and $$\displaystyle f_{c}=c y_{c}$$ (2-3) Plugging into (2-1) gives us: $$\displaystyle r(t)=my''+cy_{c}+ky_{k}$$ (2-4) Using (2-0) and substituting into (2-4), we find that the final equation of motion for the spring-mash-dashpot is: $$\displaystyle r(t)=my''+cy'+ky $$ (2-5)

=R1.3 - Draw FBD of System on p.1-4=

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

Solution:
Break the system into its three components: spring, dashpot, and mass.

Derivation of Equation of Motion
Looking at the FBD of the mass: $$f_I$$ represents the internal forces in the system from both the spring and dashpot. $$my''$$ represents the inertial force of the mass. $$f(t)$$ represents the excitation on the spring-dashpot-mass system. Sum the forces on this FBD and set equal to 0: $$ \sum f_y=-my''-f_{I}+f(t)=0$$ Rearrange to get the desired equation of motion: $$my''+f_{I}=f(t)$$

=R1.4 - Derive RLC Circuit Equations from p.2-2=

Problem Statement
Derive (i) Equation 3 ($$\displaystyle LI+RI'+\frac{1}{C}I=V'$$) and (ii) Equation 4 ($$\displaystyle LQ+RQ'+\frac{1}{C}Q=V$$) using Equation 2 ($$\displaystyle V=LC\frac{\mathrm{d^2v_{c}} }{\mathrm{d} t^2}+RC\frac{\mathrm{d} v_{c}}{\mathrm{d} t}+v_c$$) on page 2-2 in the lecture notes.

Part i
Equation 2:

Equation 1:

Equation 1 can be rewritten as Eqn 1-b:

Differentiating Eqn 1-b yields:

Differentiating Eqn 1-c gives:

Then we differentiate Equation 2 to get:

Then we subsitute Eqn. 1-b, Eqn. 1-c, and Eqn. 1-d into Equation 2-b to get:

By cancelling like terms we derive Equation 3:

Which can be rewritten as:

Part ii
To derive Equation 4, we again start with Equation 2:

Then we will use another version of Equation 1 on p. 2-2 in the lecture notes.

Then rearranging the variables, we get:

Differentiating this yields:

And differentiating again yields:

Substituting these derivatives into Equation 2 gives:

Cancelling like terms gives Equation 4:

=R1.5.1 - Find the General Solution to Kreyszig Pg. 59 Problem 4=

Problem Statement
'''Kreyszig 2011 Pg. 59 Problem 4''' Find a general solution for the following ODE.

Given: $$\displaystyle y''+4y'+(\pi^2+4)y=0 $$

Solution
The characteristic equation for the given ODE is  $$\displaystyle \lambda^2+\lambda+(\pi^2+4)=0 $$

The roots of the characteristic equation are complex roots in the form $$\displaystyle \lambda=-\frac{1}{2}a\pm i \omega $$

The roots of the characteristic equation are

$$\displaystyle \lambda=\frac{1}{2}(-4\pm 2 \pi i) $$

Therefore this ODE falls under Case III which has a general solution in the form

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

The general solution to the ODE is

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

=R1.5.2 - Find the General Solution to Kreyszig Pg. 59 Problem 5=

Problem Statement
Problem found on page 59, problem number 5

$$ \displaystyle 5. \; {y}''+2\pi {y}'+\pi ^{2}y=0 $$

Solution
The characteristic equation that corresponds to this kind of problem is:

$$ \displaystyle \; {y}''+a {y}'+b y=0 $$

So for this problem:

$$ \displaystyle \;  a=2\pi  $$  and  $$ \displaystyle  \;   b=\pi ^{2} $$

Substiting into the equation we have a term $$ \displaystyle a^{2}-4b $$ that tells us the case number we are dealing with

$$ \displaystyle \;  a^{2}-4b= (2\pi)^{2}-4(\pi ^{2})= 4\pi ^{2}- 4\pi ^{2} =0 $$

Since this term equals 0, we have Case II and therefore a real double root.

Therefore:

$$ \displaystyle \;  \lambda_1= -a/2= - 2\pi/2 = -\pi $$

The first solution is:

$$ \displaystyle \;  y_1= e ^{\lambda_1x}= e ^{-\pi x} $$

You can obtain the second solution by using the method of reduction of order

$$ \displaystyle \; y_2= uy_1  $$

$$ \displaystyle \; {y_2}'= {u}'y_1 + u{y_1}'  $$

$$ \displaystyle \; {y_2}= {u}y_1 +2{u}'{y_1}'+ u{y_1}''  $$

Substituting these derivatives into the original characteristic equation we get:

$$ \displaystyle \; ({u}y_1 +2{u}'{y_1}'+ u{y_1}) + a({u}'y_1 + u{y_1}') + buy_1 =0  $$

Collecting $$ \displaystyle u, {u}', {u}''  $$ terms together you obtain:

$$ \displaystyle \; {u}y_1 + {u}'(2{y_1}'+ay_1) + u({y_1}+a{y_1}'+by_1)=0  $$

The second and third terms will disappear to leave you with:

$$ \displaystyle \; {u}y_1=0 $$ and therefore $$ \displaystyle  \; {u}=0  $$

Integrate twice to get the expressions:

$$ \displaystyle \; {u}'=c_1  $$

$$ \displaystyle \; u=c_1x+c_2  $$

We can now simply choose for $$ \displaystyle \; c_1=1 $$ and $$ \displaystyle  \; c_2=0 $$

So now:

$$ \displaystyle \; u=x  $$ and therefore $$ \displaystyle  \;y_2=xy_1   $$

Therefore:

$$ \displaystyle \; y_2= xe^ {-\pi x}  $$

And the General Solution becomes:

$$ \displaystyle \; y=(c_1+c_2x)e^ {-\pi x}   $$

=R1.6 - Determine Properties of ODEs=

Problem Statement
For each ODE in Fig.2 in K2011 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.

Falling Stone (Sec. 1.1)
$$ \displaystyle y''=g=const $$

Parachutist (Sec. 1.2)
$$ \displaystyle mv'=mg-bv^2 $$

Out-flowing Water (Sec.1.3)
$$ \displaystyle h'=-k\sqrt{h} $$

Vibrating Mass on a Spring (Sec. 2.4,2.8)
$$ \displaystyle my''+ky=0 $$

Beats of a vibrating system (Sec. 2.8)
$$ \displaystyle y''+w_0^2y= \cos wt$$

Current I in RLC circuit (Sec. 2.9)
$$ \displaystyle LI''+RI'+ \frac 1 c I=E' $$

Deformation of a Beam (Sec. 3.3)
$$ \displaystyle EIy^{iv} = f(x) $$

Pendulum (Sec. 4.5)
$$ \displaystyle L \theta''+g \sin \theta = 0 $$

Falling Stone (Sec. 1.1)
Order: 2nd Linearity: Linear Superposition: Yes The Homogeneous Solution, $$ \displaystyle y_h''=0 $$ plus the Particular Solution, $$ \displaystyle y_p''=g $$ yields $$ \displaystyle y_h+y_p=g $$ which is the same as $$ \displaystyle (y_h+y_p)= \bar y=g $$. Therefore, the Superposition Principle can be applied.

Parachutist (Sec. 1.2)
Order: 1st Linearity: Nonlinear Superposition: No

Rewriting the original equation with $$ \displaystyle \mathit{v}$$ on one side yields $$ \displaystyle mv'+bv^2=mg $$

The Homogeneous Solution is $$ \displaystyle mv_h'+bv_h^2=0 $$.

The Particular Solution is $$ \displaystyle mv_p'+bv_p^2=mg $$. Summing both solutions yields $$ \displaystyle mv_h'+bv_h^2 + mv_p'+bv_p^2 = m(v_h+v_p)' + b(v_h^2+v_p^2) = mg $$.

$$ \bar v = v_h+v_p $$, however, $$ v_h^2+v_p^2 \neq (v_h+v_p)^2 $$

Thus, the Superposition Principle CANNOT be applied.

Out-flowing Water (Sec.1.3)
Order: 1st Linearity: Nonlinear Superposition: No

The Homogeneous Solution $$ \displaystyle h_h'=-k\sqrt{h_h} $$

plus the Particular Solution $$ \displaystyle h_p'=-k\sqrt{h_p} $$

yields $$ \displaystyle h_h'+h_p'=-k\sqrt{h_h}-k\sqrt{h_p}$$

or $$ \displaystyle (h_h+h_p)'= \bar h' = -k(h_h^{1/2}+h_p^{1/2})$$.

Yet, $$ \displaystyle -k(h_h^{1/2}+h_p^{1/2}) \neq -k(h_h+h_p)^{1/2}$$.

Therefore, The Superposition Principle CANNOT be applied.

Vibrating Mass on a Spring (Sec. 2.4,2.8)
Order: 2nd Linearity: Linear Superposition: Yes

The Homogeneous Solution $$ \displaystyle my_h''+ky_h=0 $$

plus the Particular Solution $$ \displaystyle my_p''+ky_p=0 $$

yields $$ \displaystyle m(y_h+y_p)''+k(y_h+y_p)=0 $$.

or $$ \displaystyle m \bar y''+k \bar y=0 $$.

Therefore, the Superposition Principle applies.

Beats of a vibrating system (Sec. 2.8)
Order: 2nd Linearity: Linear Superposition: Yes

The Homogeneous Solution is $$ \displaystyle y_h''+w_0^2y_h=0 $$

The Particular Solution is $$ \displaystyle y_p''+w_0^2y_p= \cos wt $$ Summing both solutions yields $$ \displaystyle (y_h+y_p)'' + w_0^2(y_h+y_p) = \cos wt $$

We know $$ \displaystyle \bar y = y_h+y_p $$,

so $$ \displaystyle \bar y'' + w_0^2 \bar y = \cos wt $$

Therefore, the Superposition Principle can be applied.

Current I in RLC circuit (Sec. 2.9)
Order: 2nd Linearity: Linear Superposition: Yes

The Homogeneous Solution is $$ \displaystyle LI_h''+RI_h'+ \frac 1 c I_h=0 $$

The Particular Solution is $$ \displaystyle LI_p''+RI_p'+ \frac 1 c I_p=E' $$ Summing both solutions yields $$ \displaystyle L(I_h+I_p)''+R(I_h+I_p)'+ \frac 1 c (I_h+I_p)=E' $$

We know $$ \displaystyle \bar I = I_h+I_p $$,

so $$ \displaystyle L \bar I''+R \bar I'+ \frac 1 c \bar I=E' $$

Therefore, the Superposition Principle can be applied.

Deformation of a Beam (Sec. 3.3)
Order: 4th Linearity: Linear Superposition: Yes

The Homogeneous Solution $$ \displaystyle EIy_h^{iv} = 0 $$

plus the Particular Solution $$ \displaystyle EIy_p^{iv} = f(x) $$

yields $$ \displaystyle EI(y_h^{iv}+y_p^{iv}) = EI(y_h+y_p)^{iv} = f(x) $$.

Knowing $$\bar y = y_h+y_p$$, we have

$$ \displaystyle EI(\bar y)^{iv} = f(x) $$.

Thus, the Superposition Principle can be applied.

Pendulum (Sec. 4.5)
Order: 2nd Linearity: Nonlinear Superposition: No

The Homogeneous Solution $$ \displaystyle L \theta_h''+g \sin \theta_h = 0 $$

plus the Particular Solution $$ \displaystyle L \theta_p''+g \sin \theta_p = 0 $$

yields $$ \displaystyle L ( \theta_h+ \theta_p)+g{( \sin \theta_h+ \sin \theta_p)} = L ( \theta_h+ \theta_p)''+g{( \sin \theta_h+ \sin \theta_p)} = 0 $$

But, $$ g{( \sin \theta_h+ \sin \theta_p)} \neq g \sin ( \theta_h + \theta_p) $$

Therefore, the Superposition Principle CANNOT be applied.

=Team Contributions=