I was reading Halliday's section on Damped Simple Harmonic Motion, which stated that this equation:

$$-b\dot{x} - kx = m \ddot{x}$$

Is the differential equation that dictates the displacement of the object, and $b$ is the damping constant of the system.

The author claims that the solution of the equation is:

$$ x(t)=x_m \mathrm e^{-bt/2m}\cos(\omega't+\phi) $$

Suppose we want to set initial conditions $x(0) = x_0$ and $x'(0) = 0$. Giving the equations

$$ x_m \cos(\phi) = x_0 $$ $$ -\frac{x_m b}{2m} \cos(\phi) - x_m \omega' \sin(\phi) = 0 \implies \frac{b}{2m} \cos(\phi)+\omega' \sin(\phi) = 0 $$

Physically, I would love to say that the initial conditions imposed are, a priori, the maximum amplitude and the initial velocity. If that is correct, then the first equation would imply that $x_m = x_0$ and $\cos(\phi) = 1$ (and therefore $\sin(\phi) = 0$)

However, that does not makes sense, since if we substitute these values into the second equation, we obtain $$ \frac{b}{2m} = 0 $$ But both $m$(mass) and $b$ (damping constant) are non-zero. What is wrong with my intuition?

Also, if we begin with the second equation and solve for $\phi$, we obtain $\phi = \tan^{-1}\left(\frac{-b}{2m\omega'}\right)$. And then we would be bound to accept that $x_m = \frac{x_0}{\cos(\phi)}$. If correct, then what does $x_m$ represent, if not just a mathematical constant?


3 Answers 3


Here $x_m$ represents the displacement of the undamped oscillator; with right initial conditions you can equate that to maximum displacement, i.e, amplitude. Where as the product $x_m e^{-bt/2m}$ is the decaying amplitude of the damped oscillator.

Given initial condition, $x(t=0) = x_0$, you get

$$ x_m = \frac{x_0}{\cos(\phi)} $$

and the initial condition $x'(t=0) = 0$ means the initial velocity of the oscillator is zero, during which the there is no damping takes place, because damping factor $f \propto -\dot{x}$, because of this initial condition you are getting

$$ \frac{b}{2m} = 0 \implies b = 0 $$

the above tells that damping takes place once the oscillator starts moving.

  • $\begingroup$ Ok, I will get used to the fac that $x_m$ its not necesarily the amplitude. Thanks for your response. $\endgroup$
    – alexp9
    Sep 7, 2021 at 16:44

This is an algebraic trap...

You are assuming that xm always is the maximum amplitude of the movement. That is wrong.

Let's look at this equation:

$$a =- \omega x $$ $$x = Acos(\omega t) + Bsin(\omega t)$$

or to mimic your expression: $$x = x_mcos(\omega t + \phi)$$

If the initial amplitude, is x0:

$$A = x_mcos(\phi)$$

but with your logic, here we will have:

$$A = x_m$$

and as you can see that is not correct. bot phi and xm are variables that need to be calculated.

Most people end up forgetting the phase and intuitively assume the motion is just the amplitude times a sin/cosine, but that is not correct when using these mathematical expressions, you usually forget that

$$x_m = A$$ $$cos(\phi)=1$$

is a set of solutions, so are $$x_m = 3A$$ $$cos(\phi) = \frac{1}{3}$$

but you have other equations (other initial conditions) in which these values should hold correct as well …

xm, phi and all these are variables that you calculate.


Answers that came before me perfectly OK. I will just add one significant intuitive picture of what's going on. Picture in your mind the oscillator starting at $x_m$ but with initial velocity running away from equilibrium position. Why would you identify $x\left(0\right)$ with the amplitude? Interesting exercise: Fix initial velocity at zero and see what happens --as @147875 suggests.


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