Dampened Harmonic Motion Derivation

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  • #1
Nabeshin
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Main Question or Discussion Point

I am attempting to derive the equation for dampened harmonic motion from the differential equation:
[tex]m\frac{d^2 x}{dt^2}+b\frac{dx}{dt}+kx=0[/tex]

It would be painful to type the whole derivation out, but there is one point relevant to the problem I am having. In order to get SHM, we assume that the dampening force is small compared to the restoring force, so I assumed:
[tex]b^2-4km<0[/tex]
Which allowed me to write:
[tex]\sqrt{b^2-4km}=i\sqrt{4km-b^2}[/tex]

Which aided in my analysis. Now, after a lot of algebra I get something that looks like this:

[tex]x(t)=Ae^{\frac{-bt}{2m}}\sqrt{\frac{4km}{4km-b^2}}Cos((\frac{1}{2m}*\sqrt{4km-b^2})t-ArcTan(\frac{b}{\sqrt{4km-b^2}}))[/tex]

The correct formula looks like this:
[tex]x(t)=Ae^{\frac{-bt}{2m}}Cos((\frac{1}{2m}*\sqrt{4km-b^2})t+\phi)[/tex]

Where phi is just the phase parameter and I'm not worried about that.

I can get close to this solution if I assume b is small compared to other quantities. For example, the square root factor in front of the cosine goes to 1, and the arctangent factor goes to zero. However, I cannot consistently apply this approximation without also changing the factor on t in the cosine.

My question is, is my derived formula wrong somewhere? I can post a longer derivation tomorrow, but for the time being we'll see if anyone notices something obvious here.
 

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  • #2
arildno
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Sure, this is a correct, albeit unstandard representation of the SHM solution.

Notice that you can rewrite your cosine with a sum as its argument as a sum of cos*cos+sin*sin, where one of the trigs will have a constant argument (i.e, the arctan part).


I would, however, be interested to see what you have actually done to derive this, since doing so is fairly trivial in the standard manner.
 
  • #3
Nabeshin
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Ok this is going to take a lot of latex but here's the full derivation:

An object moves with a restoring force proportional to its displacement and a drag force proportional to its velocity. (To get SHM at all we have to assume the restoring force is the dominant action, the drag force just dampens)

So the DE looks like:
[tex]m\frac{d^2 x}{dt^2}+b\frac{dx}{dt}+kx=0[/tex]
[tex]\frac{d^2 x}{dt^2}+\frac{b}{m}\frac{dx}{dt}+\frac{k}{m}x=0[/tex]

The characteristic polynomial of this 2nd order linear homogenous DE is:
[tex]r^2+\frac{b}{m}r+\frac{k}{m}=0[/tex]
And it has roots at:
[tex]r=\frac{-b}{2m}\pm\frac{\sqrt{b^2-4km}}{2m}[/tex]

Like I mentioned in my original post, if we assume [tex]b^2-4km < 0 [/tex] we can rewrite as:
[tex]r=\frac{-b}{2m}\pm i \frac{\sqrt{4km-b^2}}{2m}[/tex]

So our two exponential solutions to the DE are:
[tex]\tilde{x}_{1}(t)=e^{r_{1}t}[/tex]
[tex]\tilde{x}_{2}(t)=e^{r_{2}t}[/tex]

At this point to simplify a lot of typing, I'm going to let:
[tex]\mu=\frac{1}{2m}\sqrt{4km-b^2}[/tex]

Using Euler's formula:

[tex]\tilde{x}_{1}(t)=e^{\frac{-bt}{2m} - ti \frac{\sqrt{4km-b^2}}{2m}} = e^{\frac{-bt}{2m}}Cos(-\mu t})+ie^{\frac{-bt}{2m}}Sin(-\mu t)=e^{\frac{-bt}{2m}}Cos(\mu t})-ie^{\frac{-bt}{2m}}Sin(\mu t)[/tex]

And similarly,
[tex]\tilde{x}_{2}(t)=e^{\frac{-bt}{2m}}Cos(\mu t})+ie^{\frac{-bt}{2m}}Sin(\mu t)[/tex]

To get rid of the complex numbers, I'll define two new solutions:
[tex]x_{1}(t)=\frac{\tilde{x}_{1}(t)+\tilde{x}_{2}(t)}{2}[/tex]
[tex]x_{2}(t)=\frac{\tilde{x}_{2}(t)-\tilde{x}_{1}(t)}{2i}[/tex]

Which through some convenient process eliminates the imaginary parts! After simplification,

[tex]x_{1}(t)=e^{\frac{-bt}{2m}}Cos(\mu t)[/tex]
[tex]x_{2}(t)=e^{\frac{-bt}{2m}}Sin(\mu t})[/tex]

So our general solution to the DE is given by some linear combination of these two:
[tex]x(t)=Bx_{1}(t)+Cx_{2}(t)=Be^{\frac{-bt}{2m}}Cos(\mu t)+Ce^{\frac{-bt}{2m}}Sin(\mu t})[/tex]
 
  • #4
Nabeshin
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Note: At this point in typing up my derivation I realized that I really didn't need to do something. In the original derivation I used the method of amplitude and phase to rewrite the form of x(t) to the form:
[tex]x(t)=M Cos(\mu t -\delta \mu)[/tex]
[tex]M=\sqrt{(Be^{\frac{-bt}{2m}})^2+(Ce^{\frac{-bt}{2m}})^2}[/tex]
[tex]\delta=\frac{1}{\mu}ArcTan(\frac{Ce^{\frac{-bt}{2m}}}{Be^{\frac{-bt}{2m}}})[/tex]

But I guess this really only complicates matters. It leads directly to the form I wrote in the above post, so I'm not going to rewrite it. However, if I just leave the solution of the form:
[tex]x(t)=Be^{\frac{-bt}{2m}}Cos(\mu t)+Ce^{\frac{-bt}{2m}}Sin(\mu t})[/tex]

And couple this with the initial conditions:
[tex]x(0)=A[/tex] , [tex]x'(0)=0[/tex]
Where A is the amplitude of oscillation (At this point I'm ignoring the phase parameter because I couldn't figure out a good way to deal with it, so I just chose this initial condition).

I solve for B and C:
[tex]B=A[/tex]
[tex]C=\frac{Ab}{2m\mu}[/tex]

So finally, after some algebra:

[tex]x(t)=Ae^{\frac{-bt}{2m}}(Cos(\mu t)+ \frac{b}{\sqrt{4km-b^2}}Sin(\mu t))[/tex]

But I have a similar problem as before, this silly sin factor is deviating my answer from what it's supposed to be! Not sure how to get rid of it either.

Well that's the full derivation with boring algebra steps skipped. It drew heavily on what we've been learning in my ODE class lately, which is why the method of amplitude and phase for simplification came to mind.
 
  • #5
arildno
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No, C=0, since x'(0)=0.
 
  • #6
Nabeshin
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No, C=0, since x'(0)=0.
How do you figure?

[tex]
x(t)=Be^{\frac{-bt}{2m}}Cos(\mu t)+Ce^{\frac{-bt}{2m}}Sin(\mu t})
[/tex]
[tex]x(t)=e^{\frac{-bt}{2m}}(BCos(\mu t)+ CSin(\mu t))[/tex]
[tex]x'(t)=\frac{-b}{2m}e^{\frac{-bt}{2m}}(BCos(\mu t)+ CSin(\mu t))+e^{\frac{-bt}{2m}}(-\mu B Sin(\mu t)+ \mu C Cos (\mu t))[/tex]
[tex]x'(0)=0=\frac{-b}{2m}(B)+(\mu C)[/tex]
[tex]C=\frac{-bA}{2m\mu}[/tex]

Since B=A, unless I did some differentiation horribly wrong =/
 
  • #7
arildno
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Sorry about that, you are right!

But, it doesn't matter!

Remember that you can just rewrite this sum of sine cosine with into a single cosine with a phase adjustment.
 
  • #8
Nabeshin
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Sorry about that, you are right!

But, it doesn't matter!

Remember that you can just rewrite this sum of sine cosine with into a single cosine with a phase adjustment.
Your confidence is disheartening, because I don't see how this simplifies at all. If I rewrite it as one cosine, I get precisely what I posted in my first post. Your suggestion was to, at that point, split the cosine by the angle addition formula, but that doesn't help either. Perhaps you could be a little more explicit as to how things are supposed to clean up?
 
  • #9
arildno
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Okay, I've looked a bit closer.

You said you had the initial conditions x(0)=A and x'(0)=0.

But your so-called "correct solution doesn't manage this!

Noter that this says that [tex]x(0)=A\cos\phi[/tex],
but this is certainly not equal to A unless [itex]\phi=0[/itex]!!

And if phi equals 0, then it follows that the derivative is non-zero..

Thus, both of us have been blinded by the so-called "correct" solution to your problem.
 
Last edited:
  • #10
Nabeshin
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Okay, I've looked a bit closer.

You said you had the initial conditions x(0)=A and x'(0)=0.

But your so-called "correct solution doesn't manage this!

Noter that this says that [tex]x(0)=A\cos\phi[/tex],
but this is certainly not equal to A unless [itex]\phi=0[/itex]!!

And if phi equals 0, then it follows that the derivative is non-zero..

Thus, both of us have been blinded by the so-called "correct" solution to your problem.
Yeah, I realized the phi problem. I just ignored it. (My statement of the initial condition is equivalent to stating phi equals zero. Like I said I didn't really know how to work in some phase shift, it's not really a big deal IMO).

Also, you're right. I think the "correct" solution given in the book is driving me too much. I'm like 98% sure my solution is accurate, because I've made no assumptions whatsoever and there are no (obvious) problems with my algebra, so the solution must follow. Thanks for looking it over though, I really appreciate it :)
 
  • #11
arildno
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98% is not good enough; be 100% sure. :smile:
 

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