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Chris L T521 · #9 August 22nd, 2008, 11:36 PM

It's been a while since I've last posted. I finally found the time right now to do this, since it has been on my mind for a while. I put this together a little quickly, but I will come back later and make changes [if necessary].

Applications of Homogeneous and Non Homogeneous Differential Equations (Spring Mass Systems)

A useful application to the real world would be spring mass systems. We prefer to use homogeneous or non-homogeneous differential equations to model the systems.

There are two types of systems. One models damped motion [due to a dashpot of some sort or due to friction], and the other models undamped motion [sometimes known as free motion].

Undamped Spring Mass Systems

Consider a spring of length $l$ hanging from the ceiling. The spring has a mass $m$ attached to its end. The system comes to rest at its equilibrium position.

When the system is place in motion, the spring varies in length about the equilibrium position. We will denote this distance from the equilibrium position as $x$ .

When a free body analysis is done, we see two different forces:

$F=ma$ [Newton's 2nd Law] and $F=-kx$ [Hooke's Law].

Thus, we see that $\sum F_y=0$ when $-kx-ma=0\implies ma+kx=0$

Since $a=\frac{d^2x}{dt^2}$ , our equation becomes the differential equation $m\frac{d^2x}{dt^2}+kx=0\implies x''+\frac{k}{m}x=0$ .

Letting $\frac{k}{m}=\omega^2$ , the DE becomes $x''+\omega^2x=0$ .

However, we are dealing with systems that are being acted upon my external forces, so we make a modification to our DE. If an external force is applied, then the DE takes on the form $x''+\omega^2x=F(t)$ , which is non-homogeneous. When no external force is applied, $F(t)=0$ , which then makes the equation homogeneous.

Let's take a look at an example:

Example 20:

A mass weighing 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 $\tfrac{4}{3}~ft/s$ . Determine the equation of free motion.

This can be seen as an initial value problem.

Since the weight of the mass is 2 pounds, we can determine the mass, because $w=mg\implies m=\frac{w}{g}\implies m=\frac{2~lbs}{32~ft/s^2}\implies m=\frac{1}{16}~slug$

Since the string is displaced by 6 inches = 1/2 a foot, we can determine k:

$F=kx\implies 2~lbs=k\left(\tfrac{1}{2}~ft\right)\implies k=4~lbs/ft$

At $t=0$ , the mass is released 8 inches below the equilibrium position. This implies that the condition is $x(0)=\tfrac{2}{3}~ft$ .

At $t=0$ , the mass is released with an upward velocity of $\tfrac{4}{3}~ft/s$ . This implies that the condition is $x'(0)=-\tfrac{4}{3}~ft/s$ .

Keep in mind the signs of these conditions. If the mass starts off above the equilibrium position, the sign applied to the condition is negative, and if the mass starts off below the equilibrium poistion, the sign applied to the condition is positive. The same idea is applied to upward and downward velocities.

Let's start to set up our DE:

Since $\frac{k}{m}=\omega^2$ , we see that $\omega^2=64$

So our IVP is:

$x''+64x=0;~~x(0)=\tfrac{2}{3},~~x'(0)=-\tfrac{4}{3}$

We see that the characteristic equation has the form

$r^2+64=0$

This implies that we have complex conjugate roots. Thus, we see that $r=\pm8i\implies x(t)=A\cos(8t)+B\sin(8t)$

Applying the condition $x(0)=\tfrac{2}{3}$ , we see that $A=\tfrac{2}{3}$ .

Before we apply the second condition, let's find $x'(t)$

$x'(t)=-8A\sin(8t)+8B\cos(8t)$

Applying the condition $x'(0)=-\tfrac{4}{3}$ , we see that $-\tfrac{4}{3}=8B\implies B=-\tfrac{1}{6}$ .

Thus, the equation of free motion is $\color{red}\boxed{x(t)=\tfrac{2}{3}\cos(8t)-\tfrac{1}{6}\sin(8t)}$

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Damped Motion

Let's go back to the equation we came up with for undamped motion:

$mx''+kx=F(t)$

For damped motion, we need to introduce a damping term. The term is $\beta x'$ .

When introduced into the system, we see from a free body diagram that $ma=-kx-\beta x'$

Thus, we see that $ma+\beta x'+kx=0$ .

Since $a=x''$ the DE becomes $mx''+\beta x'+kx=0\implies x''+\frac{\beta}{m}x'+\frac{k}{m}x=0$ .

Letting $\frac{k}{m}=\omega^2$ and $\frac{\beta}{m}=2\lambda$ , the DE is then transformed into

$x''+2\lambda x'+\omega^2 x=0$

Given a particular external force, F(T), we can say the DE takes on the form

$x''+2\lambda x'+\omega^2 x=F(t)$

In the case that there is no external force, $F(t)=0$ .

Three Cases for Damped Motion

Solving $x''+2\lambda x'+\omega^2 x=0$ , we see that we have the auxiliary equation $r^2+2\lambda r+\omega^2=0$

Using the quadratic formula, we see that $r=\frac{-2\lambda\pm\sqrt{4\lambda^2-4\omega^2}}{2}\implies r=\frac{-2\lambda\pm2\sqrt{\lambda^2-\omega^2}}{2}\implies r=-\lambda\pm\sqrt{\lambda^2-\omega^2}$

We have three cases, depending on the value of $r=-\lambda\pm\sqrt{\lambda^2-\omega^2}$

Case 1 : Overdamped

We have overdamped motion when r consists of real and distinct roots.

Other factors for determining the overdamped case are when $\beta>>k\implies \lambda^2>\omega^2$

Thus, the general solution for overdamped motion has the form $\color{red}\boxed{x(t)=c_1e^{(-\lambda+\sqrt{\lambda^2-\omega^2})t}+c_2e^{(-\lambda-\sqrt{\lambda^2-\omega^2})t}}$

Case 2 : Critically damped

We have critically damped motion when r consists of real, repeated roots.

One major factor that reveals this is when $\lambda^2=\omega^2$ .

Thus, the general solution for critically damped motion has the form $\color{red}\boxed{x(t)=c_1e^{-\lambda t}+c_2te^{-\lambda t}}$

Case 3 : Underdamped

We have underdamped motion when r consists of complex, conjugate roots.

Major factors that reveal this is when $k>>\beta$ and when $\lambda^2<\omega^2$ .

This then implies that the solution to the characteristic equation has the form $r=-\lambda\pm\sqrt{\omega^2-\lambda^2}i$ .

Thus, the general solution for underdamped motion has the form $\color{red}\boxed{x(t)=e^{-\lambda t}\left[c_1\cos\left(\sqrt{\omega^2-\lambda^2}t\right)+c_2\sin\left(\sqrt{\omega^2-\lambda^2}t\right)\right]}$

Let us take a look at another example.

Example 21:

A 4-foot spring measures 8 feet long after an 8-pound weight is attatched
to it. The medium through with the weight moves offers a resistance
numerically equal to $\sqrt{2}$ times the instantaneous velocity. Find the equation
of motion if the weight is released from the equilibrium postion with a downward
velocity of 5 ft/s. Find the time at which the weight attains its extreme
displacement from the equilibrium position.
Identify the system as overdamped, underdamped, or critically damped.

$w=mg\implies m=\frac{w}{g}$

Since $w=8~lbs$ and $g=32~ft/s^2$ , $m=\tfrac{1}{4}~slug$

$F=kx\implies k=\frac{F}{x}$

Since $F=w=8~lbs$ , and $x=4~ft$ , $k=2~lbs/ft$

Since the medium through with the weight moves offers a resistance numerically equal to $\sqrt{2}$ times the instantaneous velocity, this implies that $\beta=\sqrt{2}$ .

Now, let us determine the conditions:

Since the mass is being released from the equilibrium position, we see that $x(0)=0$

Since the mass will have a downward velocity of $5~ft/s$ , this implies that $x'(0)=5$

Since $\frac{k}{m}=\omega^2$ and $\frac{\beta}{m}=2\lambda$ , we see that $\omega^2=8$ and $2\lambda=4\sqrt{2}$

Now, we have the IVP:

$x''+4\sqrt{2}x'+8x=0;~~x(0)=0,~~x'(0)=5$

Solving, we see that the DE has the characteristic equation $r^2+4\sqrt{2}r+8=0$ . It turns out to be a perfect square:

$r^2+4\sqrt{2}r+8=0\implies\left(r+2\sqrt{2}\right)^2=0\implies r=-2\sqrt{2}$ with multiplicity 2.

Thus, $x(t)=c_1e^{-2\sqrt{2}t}+c_2te^{-2\sqrt{2}t}\implies x(t)=(c_1+c_2t)e^{-2\sqrt{2}t}$

When $x(0)=0$ , $c_1=0$ .

Now we need to find $x'(t)$

$x'(t)=-2\sqrt{2}(c_1+c_2t)e^{-2\sqrt{2}t}+c_2e^{-2\sqrt{2}t}$

When, $x'(0)=5$ , $5=-2\sqrt{2}c_1+c_2$

But $c_1=0$ , so $c_2=5$

Therefore, the equation of motion is $\color{red}\boxed{x(t)=5te^{-2\sqrt{2}t}}$ .

Now, we need to find the time when the greatest displacement is achieved.

This is the case when $x'(t)=0$

Since $x'(t)=5e^{-2\sqrt{2}t}-10\sqrt{2}te^{-2\sqrt{2}t}$ , the time when the greatest displacement is achieved when $0=5-10\sqrt{2}t\implies t=\frac{1}{2\sqrt{2}}\implies \color{red}\boxed{t=\frac{\sqrt{2}}{4}}$ .

Since we saw that the auxiliary equation had real repeated roots, we can tell that the equation of motion will be critically damped.

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My next post may be on another application of second order DE's [Electronic Circuits], or I may dive into systems of DEs and matrix methods.

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