- #1

Nabeshin

Science Advisor

- 2,205

- 16

## 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.

[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.