Background: What I know about simple harmonic oscillators
For a simple (undamped) harmonic oscillator, one expression for the position as a function of time is
$$
x(t) = x_0 \cos(\omega_0 t) + \frac{v_0}{\omega_0} \sin(\omega_0 t)
$$
where $\omega_0 = \sqrt{k/m}$ is the natural frequency and $x_0$ and $v_0$ are the initial position and velocity, respectively. We can rewrite this a more conceptually pleasing way by letting
$$
A \equiv \sqrt{x_0^2 + \left(\frac{v_0}{\omega_0}\right)^2} \qquad \text{and} \qquad \delta \equiv \arctan \left(\frac{v_0}{\omega_0 x_0}\right).
$$
Then the equation for the SHO is
$$
x(t)=A \cos(\omega_0 t – \delta).
$$
Question: Damped harmonic oscillators
If the damping coefficient $\beta$ has units of 1/s, and a new frequency $\omega_1 = \sqrt{\omega_0^2 – \beta^2}$ is defined, textbooks such as Taylor's Classical Mechanics write the equation for the damped harmonic oscillator as
$$
x(t)=A e^{-\beta t} \cos(\omega_1 t – \delta).
$$
My question is, how do we define $A$ and $\delta$ in this case? Are they the exact same definitions as in the SHO case, or do we have to replace both instances of $\omega_0$ with $\omega_1$?
Put another way, how do we write the equivalent of $x(t) = x_0 \cos(\omega_0 t) + \frac{v_0}{\omega_0} \sin(\omega_0 t)$ for the DHO?
Best Answer
The short answer is that you can not just replace $\omega_0$ with $\omega_1$.
You did not write the specific damped equation you are interested in, but I will assume it is: $$ \ddot x + 2\beta \dot x + \omega_0^2 x = 0\;,\tag{1} $$ where I have assumed you want a positive $2$ multiplying the $\beta$ coefficient.
You can check for yourself that one way to write the solution of the damped Eq. (1) is: $$ x(t) = \tilde A e^{-\beta t}\left(\tilde B \cos(\omega_1 t) + \tilde C\sin(\omega_1 t)\right) $$
If I want to combine the cosine and sine into a single phase-shifted cosine, I can define: $$ \cos(\delta)\equiv \frac{\tilde B}{\sqrt{\tilde B^2 + \tilde C^2}} $$ and $$ \sin(\delta)\equiv \frac{\tilde C}{\sqrt{\tilde B^2 + \tilde C^2}}\;. $$
With these definitions, my solution looks like: $$ x(t) = A e^{-\beta t}\cos(\omega_1 t - \delta)\;, $$ where $$ A = \tilde A \sqrt{\tilde B^2 + \tilde C^2} $$ and $$ \tan(\delta) = \frac{\tilde C}{\tilde B}\;. $$
By evaluating the solution for $x(t)$ and its derivative $\dot x(t)$ at $t=0$ you can find two equations, one for $x_0$: $$ x_0 = A\cos(\delta)\;,\tag{2} $$ and one for $v_0$: $$ v_0 = A(\omega_1\sin(\delta)-\beta\cos(\delta))\;. $$
Dividing one equation by the other cancels out the A term and yields an expression for $\delta$: $$ \tan(\delta) = \left(\frac{v_0}{x_0}+\beta \right)\frac{1}{\omega_1}\;. $$
Using the above equation to rewrite the expression for $v_0$ yields: $$ (v_0+\beta x_0)/\omega_1 = A\sin(\delta)\tag{3} $$
Squaring Eq (2) and Eq (3) and then adding them yields an expression for $A$: $$ A = \sqrt{x_0^2 + \frac{\left(v_0+\beta x_0\right)^2}{\omega_1^2}} $$