2nd order ordinary differential equation for damped harmonic motion

[halfoflessthan5]

Homework Statement

the equation of motion for a damped harmonic oscillator is

d^2x/dt^2 + 2(gamma)dx/dt +[(omega0)^2]x =0

...

show that

x(t) = Ae^(mt) + Be^(pt)

where

m= -(gamma) + [(gamma)^2 - (omega0)^2 ]^1/2
p =-(gamma) - [(gamma)^2 - (omega0)^2 ]^1/2

If x=x0 and dx/dt =v0 at t=0. show that

A= v0 - px0

m - p​

B =mx0 - v0

m - p​

In the case of very strong damping (i.e gamma >> omega0) show that

p (approx)= -2(gamma)

*************************************

m (approx)= -(omega0)^2

2(gamma)​

Hence show that if v0 = 0, the displacement of the oscillator is given approximately by

x(t) = x0 e^(q)

where q = (t(omega0)^2) / 2(gamma)

Homework Equations

The Attempt at a Solution

Im okay up until the asterixes. Dont understand how you get the approximation for p in the limit gamma>>omega0. It just tends towards 0 as far as i can see. I tried l'hopitals, isolating the dominant term etc, but couldn't get anywhere

Dont quite get the last bit either, but that might be because i don't get the step before.

(PS I am going to learn latex soon sorry for all the mess)

EDIT: Yes, youre right Aleph. thankyou

 

[AlephZero]

I think there is typo in your post (or the original question).

m and p should not be the same. One of them should be

-(gamma) - [(gamma)^2 - (omega0)^2 ]^1/2

 

[halfoflessthan5]

bump

anyone?

 

[AlephZero]

Sorry, I thought the typo was the reason you couldn't do the question.

sqrt(gamma^2 + omega0^2) = gamma sqrt(1 + omega0^2/gamma^2)

gamma >> omega0 so omega0/ gamma is small.

Use the biominal theorem to approximate the square root.

 

[halfoflessthan5]

yeh okay. that was obvious

thankyou