Verifying a solution to Damped SHM

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Homework Help Overview

The discussion revolves around verifying that the function Ae^(-βt)cos(ωt) is a solution to the damped simple harmonic motion equation d^2(x)/dt^2 + ϒdx/dt + (ω_0)^2*x = 0. Participants are exploring the relationships between the parameters β and ω in terms of ϒ and ω_0.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating derivatives of the proposed solution and substituting them into the differential equation. There is an exploration of how to simplify the resulting expression and questions about the cancellation of terms involving ϒ, ω, and (ω_0)^2.

Discussion Status

Some participants suggest specific values for β and ω to facilitate finding a solution. There is recognition of a quadratic equation that arises from the simplification process, and participants are examining the implications of setting certain parameters to zero. Multiple interpretations of the relationships between the variables are being explored.

Contextual Notes

Participants are working under the assumption that they can choose values for β and ω, and there is a focus on ensuring that the coefficients of sin(ωt) and cos(ωt) equal zero for the equation to hold 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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