# Amplitude of a Damped Harmonic Oscillator

## 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?

• The coefficients of the sine and cosine (and thus the amplitude and phase) of the undamped solution are found from setting $t=0$ in $x(t)$ and $v(t)$ and plugging in the initial conditions. Why not try the same thing for the damped solution? Commented Jun 22, 2023 at 16:49
• Then, you can write your own answer. Commented Jun 22, 2023 at 16:50

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 $$\omega_1$$?

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 $$A$$: $$A = \sqrt{x_0^2 + \frac{\left(v_0+\beta x_0\right)^2}{\omega_1^2}}$$

• Thanks! But I'm guessing there should be a $e^{-\beta t}$ in your final $x(t)$ equation or embedded in the amplitude $A$, right? Commented Jun 22, 2023 at 17:58
• oops. yeah. hold on a second
– hft
Commented Jun 22, 2023 at 17:59
• @Bunji I updated my answer. Unfortunately, it is not as simple as before, thanks to that darn beta.
– hft
Commented Jun 22, 2023 at 18:15
• The equations aren't squared correctly. We should get $A=x_0\sqrt{1+r^2}$, where $r=\frac{v_0}{\omega x_0}$ in the undamped case and $r=\frac{v_0+\beta x_0}{\omega_1x_0}$ in the damped case. Commented Jun 22, 2023 at 20:26
• @Chemomechanics thank you--I've been checking this too, and I now get $\tilde{C} = (v_0 + x_0 \beta)/\omega_1$ which agrees with the final $A$ given here. I appreciate the multiple updates which set me down the right path, hft. Commented Jun 22, 2023 at 20:41