First Order Differential Equation for a Harmonic Oscillator A box with mass $m$ is attached to a spring with spring coefficient $k$. This system is then placed into a glass case filled with a liquid with drag coefficient $\alpha$. Now I have the following equation set-up: $m\ddot{x}+\alpha\dot{x}+kx=0$ Now the solution for this is with usage of $e^\left(\lambda t\right)$, but what I would like to know is when to use exponent and when to use $Acos(\omega t +\phi)$? Also, the frequency is $\omega=\sqrt\frac{k}{m} $ is that just by dividing the coefficient of $"\ddot{x}"$ term and the coefficient of the "x" term? Thanks in advance!

I would like to extend upon Aldon's comment on the first part.

It is important to distinguish between the two situations $\alpha >0$ and $\alpha = 0$. If there is friction, i.e. if $\alpha>0$, then the box will ultimately be at rest. This can only be captured by using the exponential form. Observe that the characteristic equation \begin{equation} m z^2 + \alpha z + k = 0 \end{equation} has the two roots $\lambda = \frac{-\alpha \pm \sqrt{\alpha^2 - 4 km}}{2m}$. The real parts are negative, stressing the exponential decay. On the other hand if $\alpha=0$, then the box will swing happily back and forth until the end of time and the harmonic form is sufficient. The phase shift $\phi$ is needed to accommodate specific initial conditions.

I have nothing to add to the second part beyond what Aldon has already written in the comments.