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?

  • $\begingroup$ 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? $\endgroup$ Commented Jun 22, 2023 at 16:49
  • $\begingroup$ Then, you can write your own answer. $\endgroup$ Commented Jun 22, 2023 at 16:50

1 Answer 1


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

  • $\begingroup$ 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? $\endgroup$
    – Bunji
    Commented Jun 22, 2023 at 17:58
  • $\begingroup$ oops. yeah. hold on a second $\endgroup$
    – hft
    Commented Jun 22, 2023 at 17:59
  • $\begingroup$ @Bunji I updated my answer. Unfortunately, it is not as simple as before, thanks to that darn beta. $\endgroup$
    – hft
    Commented Jun 22, 2023 at 18:15
  • $\begingroup$ 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. $\endgroup$ Commented Jun 22, 2023 at 20:26
  • 1
    $\begingroup$ @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. $\endgroup$
    – Bunji
    Commented Jun 22, 2023 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.