What happens to the c2sin(t) part of the worked solution?

[StillAnotherDave]

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?

 

[PeroK]

Sine and cosine are essentially the same function, but $\pi/2$ out of phase. If you look at a table of trig identities, you'll find the relevant identity. See, for example:
https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Sine_and_cosine

 

[StillAnotherDave]

Aha, so:

I didn't know this!

 

[Delta2]

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 $$\cos\phi=\frac{a}{\sqrt{a^2+b^2}},\sin\phi=\frac{b}{\sqrt{a^2+b^2}}$$