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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?

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  • $\begingroup$ 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

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

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  • $\begingroup$ 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
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    $\begingroup$ @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

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