# Damped oscillator equation

Homework Statement:
Why does the final solution of the equation of motion for damped oscillation not have a sine function?
Relevant Equations:
$$x(t)=Ae^{-bt/2m} cos⁡(ωt+φ)$$
Hello folks,

So the solution of the equation of motion for damped oscillation is as stated above. If we were to take an specific example such as:

$$\frac{d^2x}{dt^2}+4\frac{dx}{dt}+5x=0$$

then the worked solution to the second order homogeneous is:

$$x\left(t\right)=e^{-2t}\left(c_1cos(t)+c_2sin(t)\right)$$

What happens to the $$c_2sin(t)$$ part of the worked soution? Why is it not part of the actual solution of the equation of motion?

Or does the sine function give the phase constant phi?

Last edited:

Aha, so:

I didn't know this!

Delta2
Delta2
Homework Helper
Gold Member
Aha, so:

I didn't know this!
Your question was mostly trigonometry than physics. In the identity you give above it is $$c={\sqrt{a^2+b^2}}$$ and ##\phi## an angle such that $$\cos\phi=\frac{a}{\sqrt{a^2+b^2}},\sin\phi=\frac{b}{\sqrt{a^2+b^2}}$$