User:Egm4313.s12.team6.berthoumieux

Contents

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 of), and show wether the principle of superposition can be applied.

Given

${\displaystyle (1){y}''=g=constant}$

${\displaystyle (2)m{v}'=mg-bv^{2}}$

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

${\displaystyle (4)m{y}''+ky=0}$

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

${\displaystyle (6)L{I}''+R{I}'+{\frac {1}{c}}I={E}'}$

${\displaystyle (7)EIy^{iv}={\mathit {f(x)}}}$

${\displaystyle (8){\mathit {L{\theta }''+gsin{\theta }=0}}}$

Solution

${\displaystyle (1){y}''=g=constant}$
Order:2nd order. The highest derivative is a 2nd derivative on the y.
Linearity:Linear.
Superposition:Yes.

${\displaystyle (2)m{v}'=mg-bv^{2}}$
Order:1st order. The highest derivative is a 1st on the v.
Linearity:Non-linear.
Superposition:No.

${\displaystyle (3){h}'=-k{\sqrt {h}}}$
Order:1st order
Linearity:Non-linear.
Superposition:No.

${\displaystyle (4)m{y}''+ky=0}$
Order:2nd order.
Linearity:Linear.
Superposition:Yes.

${\displaystyle (5)y''+\omega _{0}^{2}=cos(\omega t)}$, ${\displaystyle \omega _{0}=\omega }$
Order:2nd order
Linearity:Linear.
Superposition:Yes.

${\displaystyle (6)L{I}''+R{I}'+{\frac {1}{c}}I={E}'}$
Order:2nd order
Linearity:Linear.
Superposition:Yes.

${\displaystyle (7)EIy^{iv}={\mathit {f(x)}}}$
Order:4th oder
Linearity:Linear.
Superposition:Yes.

${\displaystyle (8){\mathit {L{\theta }''+gsin{\theta }=0}}}$
Order:2nd order
Linearity:Non-linear.
Superposition:Yes.

Statement

Realize spring-mass-dashpot systems in series as shown in the figure.

With characteristic: ${\displaystyle {\lambda }^{2}-10{\lambda }+25=0}$

And with double real root at ${\displaystyle {\lambda }=-3}$

Find the values of k, c, m.

Solution

First, we write down the equations we can figure out from kinematics and from kinetics.
Kinematics: ${\displaystyle y=y_{k}+y_{c}}$
Kinetics:${\displaystyle m{y}''+f_{l}=f(t)}$
${\displaystyle f_{l}=f_{k}=f_{c}}$

Next, we can find any relations that exist:
${\displaystyle f_{k}=ky_{k}}$
${\displaystyle f_{c}=c{y_{c}}'}$

Since we already established that ${\displaystyle y=y_{k}+y_{c}}$, we can say that ${\displaystyle {y}''={y_{k}}''+{y_{c}}''}$

We also established that ${\displaystyle f_{k}=f_{c}}$ so we can say that ${\displaystyle ky_{y}=c{y_{c}}'n\to {y_{c}}'={\frac {k}{c}}y_{k}}$

Next, we plug in the value for ${\displaystyle {y_{c}}'}$:
${\displaystyle {y}''={y_{k}}''+{({y_{c}}')}'={y_{k}}''+{\frac {k}{c}}{y_{k}}'}$

Finally, plug ${\displaystyle {y}''}$ back into the kinetics equation:
${\displaystyle m({y_{k}}''+{\frac {k}{c}}{y_{k}}')+ky_{k}=f(t)}$
${\displaystyle m{y_{k}}''+m{\frac {k}{c}}{y_{k}}'+ky_{k}=f(t)}$

We are given that ${\displaystyle {\lambda }=-3}$ so the characteristic equation must be:
${\displaystyle (\lambda -(-3))^{2}=0}$
${\displaystyle \lambda ^{2}+6\lambda +9=0}$

This relates to the second order ODE:
${\displaystyle {y}''+6{y}'+9y=0}$

Comparing this to the following equation:
${\displaystyle m{y_{k}}''+m{\frac {k}{c}}{y_{k}}'+ky_{k}=f(t)}$

We find that

      ${\displaystyle m=1}$
${\displaystyle k=9}$
${\displaystyle c={\frac {9}{6}}={\frac {3}{2}}}$