Verifying a solution to Damped SHM

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The discussion focuses on verifying that the function Ae^(-βt)cos(ωt) is a solution to the damped simple harmonic motion equation. The user successfully calculates the first and second derivatives of the proposed solution and simplifies the resulting expression. They identify that to satisfy the equation, specific values for β and ω must be chosen, leading to the conclusion that β = ϒ/2 and ω = 0. This results in the relationship (ω_0)^2 = ϒ^2/4, confirming the solution. The key takeaway is that the coefficients of sin(ωt) and cos(ωt) must equal zero for the equation to hold true for all t.
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Homework Statement


Verify that Ae^(-βt)cos(ωt) is a possible solution to the equation:

d^2(x)/dt^2+ϒdx/dt+(ω_0)^2*x = 0

and find β and ω in terms of ϒ and ω_0.

Homework Equations


N/a, trig identities I suppose.

The Attempt at a Solution


I think this is simply a 'plug and chug' type equation, but I'm having alll sorts of difficulty canceling things.

I first calculated the first and second derivative of the given possible solution.

First derivative = Ae^(-βt)*(-ωsin(ωt))+(-Aβe^(-βt)cos(ωt))
Second derivative = -Ae^(-βt)ω^2cos(ωt)+Aβe^(-βt)ωsin(ωt)+Aβ^2e^(-βt)cos(ωt)+Aβe^(-βt)ωsin(ωt)

I then plugged them into their respective spots into the equation and tried to simplify.
I factored and divided both sides by Ae^(-βt).

I am at:
2βωsin(ωt)+β^2cos(ωt)-ϒ(ωsin(ωt)+βcos(ωt))+(ω_0)^2cos(ωt)-ω^2cos(ωt) = 0

The problem I am having is visualizing how the terms can cancel.
How does a term with ϒ cancel with terms with ω and terms with (ω_0)^2

I have to be missing something.
 
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You a free to choose values for β and ω so that you have a solution.
 
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Ah, that is the trick I was overlooking. I do recall that from my notes. We let ω = 0 such that we are left with

β^2- ϒβ+(ω_0)^2 = 0

Now we let β = ϒ/2 and we KNOW that (ω_0)^2 = ω^2+ϒ^2/4 but since ω^2 = 0 we are left with the following from the above

ϒ^2/4-ϒ^2/2+ϒ^2/4 which indeed equals 0.

So, in short, choose:
ω = 0
β = ϒ/2
and know
(ω_0)^2 = ϒ^2/4 + ω^2

and the first part is solved!
 
The second part comes directly from the first
β = ϒ/2
(ω_0) = sqrt(ϒ^2/4 + ω^2)
 
RJLiberator said:
We let ω = 0 such that we are left with

β^2- ϒβ+(ω_0)^2 = 0

Now we let β = ϒ/2 and we KNOW that (ω_0)^2 = ω^2+ϒ^2/4

You had shown the relation that must be satisfied:
2βωsin(ωt)+β^2cos(ωt)-ϒ(ωsin(ωt)+βcos(ωt))+(ω_0)^2cos(ωt)-ω^2cos(ωt) = 0

This must be satisfied for all values of t. The only way this can happen is if the overall coefficient of sin(ωt) is zero and the overall coefficient of cos(ωt) is zero.

This will tell you the values required for both β and ω.
 
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