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I was studying the underdamped harmonic motion and got curious about the fact that the decreasing exponentials $\pm Ae^{-\gamma t}$ are good approximations only for light damping $(\gamma<<\omega) $. So I've searched and found that $\pm Ae^{-\gamma t}$ are the envelope of the motion (i.e. tangent to points of the curve that represents it) and the points of tangency are not coincident to the maxima and minima of the curve, as shown in the image below (Source: leancrew.com): Maximum vs. Point of Tangency

The question that came immediately in my mind was "How to find the exponentials that contains all maximum/minimum points?", so I've searched more and found a footnote in Morin's Introduction to Classical Mechanics:

To be precise, the amplitude doesn't decrease exactly like $Ce^{-\gamma t}$, as Eq. (4.16) suggests, because $Ce^{-\gamma t}$ describes the envelope of the motion, and not the curve that passes through the extremes of the motion. You can show that the amplitude in fact decreases like $$Ce^{-\gamma t} \cdot cos\left(tan^{-1}(\gamma/ \tilde{\omega})\right) .$$ This is the expression for the curve that passes through the extremes.

Eq. (4.16): $$\space x(t)=e^{-\gamma t} \left(Ae^{i \tilde{\omega}t}+Be^{-i \tilde{\omega}t}\right) \equiv e^{-\gamma t}C \space cos\left(\tilde{\omega }t+\phi\right)$$ $$\tilde{\omega} \equiv \sqrt{\omega^2-\gamma^2}$$


So, now I know the equation of exponentials that contains all the points of maximum/minimum: $$\pm Ce^{-\gamma t} \cdot cos\left(tan^{-1}(\gamma/ \tilde{\omega})\right) $$

but, after trying a lot, I still couldn't figure out how to deduce them from Eq. (4.16). How can I do this?

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but, after trying a lot, I still couldn't figure out how to deduce them

Did you try:

$$\frac{dx}{dt} = -\gamma e^{-\gamma t}C\cos(\tilde\omega t + \phi) - \tilde\omega e^{-\gamma t}C\sin(\tilde\omega t + \phi)$$

which is zero when

$$\tan^{-1}\frac{\gamma}{\tilde\omega} = -(\tilde\omega t + \phi)$$

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    $\begingroup$ I'm sorry. I don't know why but it didn't occurred to me to derive the expression. Now, after your tip, I think I figured out how to deduce the equation of the exponentials. I'll post my deduction, could you please look to see if everything is ok? $\endgroup$ – Vinicius ACP Oct 17 '18 at 21:36
  • $\begingroup$ I posted the deduction. $\endgroup$ – Vinicius ACP Oct 18 '18 at 20:49
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    $\begingroup$ @ViniciusACP, and I've upvoted it. By the way, if use \tan rather than tan in your mathjax, you get $\tan$ rather than $tan$. Similarly for the other trig functions. Also, I believe you can accept your own answer. $\endgroup$ – Alfred Centauri Oct 18 '18 at 20:56
  • $\begingroup$ I fixed it now for all sin, cos and tan. Thank's for your help. $\endgroup$ – Vinicius ACP Oct 18 '18 at 21:05
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The image below (Source: Wikipedia) is a plot of the curve $\space x(t)=e^{-\gamma t}C \space \cos\left(\tilde{\omega }t+\phi\right)$:

                            

We need to make the cosine fixed in a way that the above expression picks all the maximum/minimum (extrema) values of the curve. To accomplish this, its necessary to find all the instants when turning points (extrema points) occur and substitute then inside of $\cos\left(\tilde{\omega }t+\phi\right)$.


Remembering Fermat's theorem from Calculus:

If $f(x)$ has a relative extrema at $x=c$ and $f(c)$ exists, then $x=c$ is a critical point of $f(x)$. In fact, it will be a critical point such that $f'(c)=0$

Using this theorem, we can find all the relative extrema (which is exactly what we're looking for):

$$\dot{x}(t)=0 \iff -\gamma e^{-\gamma t}C\cos(\tilde\omega t + \phi) - \tilde\omega e^{-\gamma t}C\sin(\tilde\omega t + \phi)=0$$ $$-\dfrac{\gamma}{\tilde\omega}=\frac{\sin(\tilde\omega t + \phi)}{\cos(\tilde\omega t + \phi)}=\tan(\tilde\omega t + \phi) \stackrel{(1)}{\iff} \tilde\omega t + \phi = \tan^{-1}\left(-\frac{\gamma}{\tilde\omega}\right)+n\pi,n\in\mathbb{Z} $$ $$\tilde\omega t + \phi = \tan^{-1}\left(-\frac{\gamma}{\tilde\omega}\right)+n\pi,n\in\mathbb{Z} \stackrel{(2)}{\iff} t=\frac{-\tan^{-1}\left(\gamma/\tilde\omega\right)-\phi+n\pi}{\tilde\omega},n\in \mathbb{N}$$

Note: Looking at the expression of $t$, we can have negative time. To avoid it, is enough to make: $$-\tan^{-1}\left(\gamma/\tilde\omega\right)-\phi+n\pi \geq 0 \stackrel{(3)}{\iff}n\geq\left \lceil{\frac{\tan^{-1}(-\gamma/\tilde\omega)+\phi}{\pi}}\right \rceil,n\in\mathbb{N} $$


Substituting all the instants when turn points occur inside the cosine of the initial expression, we have:

\begin{alignat}{1}e^{-\gamma t}C \space \cos(\tilde\omega t + \phi) &= e^{-\gamma t}C \space \cos\left[\tilde\omega\cdot \left(\frac{-\tan^{-1}(\gamma/\omega)-\phi+n\pi}{\tilde\omega}\right)+ \phi \right] \\&=e^{-\gamma t}C \space \cos \left[-\tan^{-1}(\gamma/\tilde\omega)+n\pi \right] \end{alignat}

Simplifying the cosine: \begin{alignat}{1} \cos \left[-\tan^{-1}(\gamma/\tilde\omega)+n\pi \right] &= \cos \left[-\tan^{-1}(\gamma / \tilde\omega)\right]\cdot \cos(n\pi)-\sin\left[-\tan^{-1}(\gamma / \tilde\omega)\right]\cdot \sin(n\pi) \\ &= \cos \left[\tan^{-1}(\gamma / \tilde\omega)\right]\cdot (\pm 1)-\sin\left[-\tan^{-1}(\gamma / \tilde\omega)\right]\cdot 0 \\ &= \pm \cos \left[\tan^{-1}(\gamma / \tilde\omega)\right] \end{alignat}


Therefore, the modulation of the amplitude is given by $M(t)$, where:

$$M(t)=\pm \space e^{-\gamma t}C \space \cos \left[\tan^{-1}(\gamma / \tilde\omega)\right]$$

If we have a light dumping $(\gamma << \omega)$, then the following simplification can be done:

\begin{alignat}{1}M(t) &\approx \pm \space e^{-\gamma t}C \space \cos \left[\tan^{-1}(\approx0)\right] \\&\approx \pm \space e^{-\gamma t}C \space \cos \left[\approx0\right] \\&\approx \pm \space e^{-\gamma t}C \end{alignat}


Notes:

(1)$\space \tan(y)=x \iff y=\tan^{-1}(x)+k\pi,k\in\mathbb{Z}$

(2)$\space \tan^{-1}(-x)=-\tan^{-1}(x)$ and $\cos(-x)=\cos(x)$, so is sufficient to use $n\in\mathbb{N}$ instead of $n\in\mathbb{Z}$

(3)$\space \left\lceil{Ceil}\right\rceil$ because $n\in\mathbb{N}$

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