# Critical Damped motion

1. Oct 10, 2008

### Benzoate

1. The problem statement, all variables and given/known data
Find the general solution of the damped SHM equation(5.9) for the special case of critical damping , that is , when K = $$\Omega$$. Show that , if the particle is initially relaeased from rest at x= a , then the subsequent motion is given by

x=a*(e^-($$\Omega$$*t))*(1+$$\Omega$$*t)

2. Relevant equations

x''+2Kx'+($$\Omega$$)2*x=0
x=A*cos($$\Omega$$*t) + B*sin($$\Omega$$*t)

3. The attempt at a solution

x=e$$\lambda$$*t

x'= $$\lambda$$*e$$\lambda$$*t

x''=$$\lambda$$2*e$$\lambda$$*t

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

x''= -A*$$\Omega$$2*cos($$\Omega$$*t) - B*$$\Omega$$2*sin($$\Omega$$*t)

Not sure how to proceed...

2. Oct 11, 2008

### gabbagabbahey

Take a look at the characteristic polynomial of the DE:

$$x''(t)+2 \Omega x'(t) + \Omega ^2 x(t)=0 \Rightarrow \lambda ^2 +2 \Omega \lambda + \Omega ^2 =0$$

where $$\lambda ^n x(t) \equiv \frac{d^n x(t)}{dt^n}$$.

What are the solutions for $\lambda$? Are they linearly independent? What does that tell you about the general solution for $x(t)$?

Last edited: Oct 11, 2008
3. Oct 11, 2008

### Benzoate

That $$\lambda$$=-$$\Omega$$; its a double root

I remember from my ODE class that if a polynomial had more than two root, then x(t)=c1*e-r1*t+c2-r2*t; Otherwise , if the polynomial has only one solution, then x(t)=c1*e-r1*t+c2*t*e-r1*t

How would I go about showing that x(t)=c1*e-r1*t+c2*t*e-r1*t
rather than loosely remembering what I learned about the conditions for ODE equations certain solutions

Last edited: Oct 11, 2008
4. Oct 11, 2008

### gabbagabbahey

I assume you mean $x(t)=c_1 e^{\lambda_1 t}+c_2te^{\lambda_1 t}=c_1 e^{-\Omega t}+c_2te^{-\Omega t}$?

If so, you can show this by substituting this general solution into the ODE and showing that it satisfies it.

Then, just use the fact that x(0)=a to show that the specific solution is the one given in the problem.

5. Oct 11, 2008

### Benzoate

Should I start off by assuming that the original ODE for a double root is $x(t)=c_1 e^{\lambda_1 t}+c_2te^{\lambda_1 t}=c_1 e^{-\Omega t}+c_2te^{-\Omega t}$? or do I need to go about proving that I arrived at this solution ?

6. Oct 11, 2008

### gabbagabbahey

I don't think you'll need to prove that it's the general solution, the solutions to constant coefficient ODE's are well known and can be found in almost any text on DE's; but you may want to check with your prof to make sure.

7. Oct 11, 2008

### Benzoate

x(0)=a

I think the initial conditions for v is zero as well , since the damped motion is undriven.

x' = -$$\Omega$$*a*e-$$\Omega$$*t + c2*(e-$$\Omega$$*t-x(0)=a

I think the initial conditions for v is zero as well , since the damped motion is undriven.

x' = -$$\Omega$$*a*e-$$\Omega$$*t + c2*(e-$$\Omega$$*t-$$\Omega$$*t*ex(0)=a

I think the initial conditions for v is zero as well , since the damped motion is undriven.

x' = -$$\Omega$$*a*e-$$\Omega$$*t + c2*(e-$$\Omega$$*t-$$\Omega$$*t*e*t*$$\Omega$$

How would I go about finding c2?

8. Oct 11, 2008

### gabbagabbahey

Your last post didn't make much sense.

You can determine $c_1$ by using x(0)=a, and $c_2$ by using x'(0)=0 (since the particle is released from rest at t=0)

9. Oct 11, 2008

### Benzoate

sorry for the confusion , but I was essentially saying what you said in your last post: Should I assume for an undriven damped oscillator , that the initial conditions for x' and x are zero at t=0.

10. Oct 11, 2008

### gabbagabbahey

There's no need to assume anything here; you are given the initial conditions:

that means x(t=0)=a and x'(t=0)=0.