Modelling with higher c.c. LODE’s

class: center, middle, inverse, title-slide

.title[
# Modelling with higher c.c. LODE’s
]
.subtitle[
## Session 05
]
.author[
### Alejandro Ucan
]
.date[
### 2023-10-04
]

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# Goals

* We will use the Newton law to model a spring-mass system.
* Using the Auxiliary polynomial and Undeterminated Coefficients we will solve those models.

---
# Spring-Mass system

Suppose that you a spring that one side is attached to a platform and in the other there is a suspended object of mass `$m_w.$` We whant to know the function that describe the position of the object in the time `$t.$`

![Spring-mass model](vertspringmodel.jpg)

---
# Hypothesis:

We will model this system with the following hypothesis:

1. _Conservation_ the sum of all (interacting) forces is equal to cero.
1. _Newton law's_ the force is proportional to the mass(weight) and the acceleration, i.e. `$$F=m\frac{d^2x}{dt^2}.$$`
1. _Hooke's law_ the force of the spring's restablishment is opposite to the direction of the stretch and proportional to this quantity, i.e. `$$F=-ks.$$`
1. If the medium offers a resistance, this is directly proportional to the instant velocity of the spring, i.e. `$$F=-bv.$$`
1. If the platform is not fixed, so the force that induce this movement is given by a function that only depends on the displacement.

---
# The equation of motion

> The sum of all forces is equal to cero, i.e. `$$m\frac{d^2x}{dt^2}=-kx-b\frac{dx}{dt}+f(t),$$` where:
  * `$m$` is the mass (slugs) of the object.
  * `$k$` is the spring constant.
  * `$b$` is the damping constant.
  * `$f(t)$` is the external force.

---
 # Example 1:

> A mass weighing (force) 2 pounds stretches a spring 6 inches. At `$t=0$` the mass is released
from a point 8 inches below the equilibrium position with an upward velocity of `$3/4 ft/s$`. Determine the function that describe the mass motion.

--
<br/><br/>
The conversion of `$2lbs$` to slugs is `$m=\frac{2}{32}=1/16\, slugs.$` The conversion of `$6in$` to feet is `$s=\frac{6}{12}=1/2\, ft.$` The conversion of `$8in$` to feet is `$x_0=\frac{8}{12}=2/3\, ft.$` Now we are in the same units, we have that:
  * `$m=1/16\, slugs.$`
  * `$|F|=k|s|,$` then `$k=\frac{|F|}{|s|}=\frac{2}{1/2}=4\, lb/ft.$`
  * `$b=0$` because there is no damping.
  * `$f(t)=0$` because there is no external force.
  * `$x(0)=2/3\, ft.$`
  * `$x'(0)=3/4\, ft/s.$`

---
# Example 1:

So my ODE is `$$\frac{1}{16}x''+4x=0.$$` The auxiliary polynomial is `$$\frac{1}{16}m^2+4=0$$` and the roots are `$m_1=8i$` and `$m_2=-8i.$` Then the general solution is `$$x(t)=c_1\cos(8t)+c_2\sin(8t).$$`

* `$x(0)=2/3\, ft.$` then `$c_1=2/3.$`
* `$x'(0)=3/4\, ft/s.$` then `$c_2=-1/16.$`
---

# Example 2:

A mass weighing 20pounds, attached to the end of a spring, stretches it 4 in. Initially, the mass is released from rest from a point 3 inch above the equilibrium position. Find the function that describes the motion.
--
<br/><br/>
The conversion of `$20lbs$` to slugs is `$m=\frac{20}{32}=5/8\, slugs.$` The conversion of `$4in$` to feet is `$s=\frac{4}{12}=1/3\, ft.$` The conversion of `$3in$` to feet is `$x_0=\frac{3}{12}=1/4\, ft.$` Now we are in the same units, we have that:
  * `$m=5/8\, slugs.$`
  * `$|F|=k|s|,$` then `$k=\frac{|F|}{|s|}=\frac{20}{1/3}=60\, lb/ft.$`
  * `$b=0$` because there is no damping.
  * `$f(t)=0$` because there is no external force.
  * `$x(0)=1/4\, ft.$`
  * `$x'(0)=0\, ft/s.$`
---
# Example 2:

So the ODE is `$$\frac{5}{8}x''+60x=0.$$` The auxiliary polynomial is `$$\frac{5}{8}m^2+60=0$$` and the roots are `$m_1=4\sqrt{15}i$` and `$m_2=-4\sqrt{15}i.$` Then the general solution is `$$x(t)=c_1\cos(4\sqrt{15}t)+c_2\sin(4\sqrt{15}t).$$`

* `$x(0)=1/4\, ft.$` then `$c_1=1/4.$`
* `$x'(0)=0\, ft/s.$` then `$c_2=0.$`

Therefore the function that models the displacement is `$$x(t)=\frac{1}{4}\cos(4\sqrt{15}t).$$`

---
# Example 3:

A mass weighing `$8lbs$` stretches a spring `$2ft.$` Assuming that the damping force is equal to `$2$` times the instant velocity. Find the function that describes the motion if the mass is initially released from the equilibrium with a velocity (negative) of `$3ft/s.$`

--
<br/><br/>
The conversion of `$8lbs$` to slugs is `$m=\frac{8}{32}=1/4\, slugs.$` The conversion of `$2ft$` to feet is `$s=2\, ft.$` Now we are in the same units, we have that:
  * `$m=1/4\, slugs.$`
  * `$|F|=k|s|,$` then `$k=\frac{|F|}{|s|}=\frac{8}{2}=4\, lb/ft.$`
  * `$b=2$` because there is damping.
  * `$f(t)=0$` because there is no external force.
  * `$x(0)=0\, ft.$`
  * `$x'(0)=-3\, ft/s.$`

---
# Example 3:

So the differential equation is given by `$$\frac{1}{4}x''+2x'+4x=0.$$` The auxiliary polynomial is `$$\frac{1}{4}m^2+2m+4=0$$` and the roots is `$m=-4,$` and it is a unique root. Then the general solution is `$$x(t)=c_1e^{-4t}+c_2te^{-4t}.$$`

* `$x(0)=0\, ft.$` then `$c_1=0.$` 
* `$x'(0)=-3\, ft/s.$` then `$c_2=-3.$`

So the function, that models the displacement, is `$$x(t)=-3te^{-4t}.$$`
---
# Example 4:

A mass weighing 64 pounds stretches a spring 0.32 foot.
The mass is initially released from a point 8 inches
above the equilibrium position with a downward velocity
of 5 ft /s.
(a) Find the equation of motion.<br/>
(b) What are the amplitude and period of motion? <br/>
(c) How many complete cycles will the mass have completed
at the end of 3p seconds? <br/>
(d) At what time does the mass pass through the equilibrium
position heading downward for the second
time?