Heavily Damped Oscillator Equation

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The discussion focuses on solving the Heavily Damped Oscillator Equation, specifically transforming the expression for α into the form α = C(1+y)^n. The user identifies that α^2 = (ϒ^2/4) - w_0^2 and simplifies it to α = ϒ/2[1 - (4*w_0^2/ϒ^2)]^(1/2). The conversation highlights the importance of recognizing leading terms and first-order corrections in the context of overdamped systems, ultimately leading to a successful solution.

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Homework Statement


Screen Shot 2016-02-03 at 6.19.37 PM.png


Homework Equations

The Attempt at a Solution


I'm not really sure how to go about this.

They tell me the hint, and to use the simplification. I assume when they say (1+y)^n they take y to be in general, anything. It was confusing at first to see a y in the format when no y was present in any of the previous equations.

So, we know ϒ^2/4 >> w_0^2
So, could we say α^2 = ϒ^2/4 -w_0^2 can turn into α^2 = ϒ^2/4 => α = ϒ/2.
But this is not of the form α = C(1+y)^n.

So, we know ϒ = b/m, but that doesn't seem to help too much as α = b/2m is still not in that form.

Any hints on how I can start this off on the right way? I have a lot of equations in front of me and I feel with a good start up hint ill be off to the races.
 
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Read the given hint. It is there for a reason. You have only kept the leading term and not the first order correction.

I also disagree with the problem, I find it extremely intuitive that an overdamped system decays slower.
 
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You have only kept the leading term and not the first order correction.

α^2=(ϒ^2/4)-w_0^2 and we want it in the form α = C(1+y)^n

I'm not quite sure what you mean with this quote.

The only way I can think of it is if we immediately take a square root of both sides.

α = sqrt(ϒ^2/4-w_0^2)

This is somewhat like what the hint desires, but I feel like I won't get anywhere with this (As I've tried some calculations with this).

I also disagree with the problem, I find it extremely intuitive that an overdamped system decays slower.

I also did as well.
 
RJLiberator said:
but I feel like I won't get anywhere with this (As I've tried some calculations with this).
Rewrite the square root as an exponent ...
 
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Ah, so you are suggesting there is more here.

α = sqrt(ϒ^2/4-w_0^2)
α = [ϒ^2/4-w_0^2]^(1/2)
In this case, C = 1
But, we need it to be in the form (1+y)^n
So, perhaps let's try to factor out a ϒ^2/4 from both terms
So we get
α = ϒ/2[1-4*w_0^2/ϒ^2]^(1/2)

Now it is appearing to be in correct form.
So,
α = ϒ/2[1-(1/2)*4*w_0^2/ϒ^2]

Simplifying we get
α = [ϒ/2-w_0^2/ϒ]Oh my god.
I just solved it.

I love math. I also love you.
Why oh why did this take me hours.
 
RJLiberator said:
Why oh why did this take me hours.

Maybe this particular type of series expansion is not fundamental in your research subject - it is in mine. :rolleyes:
 
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