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When solving the damped oscillator, $\ddot x + 2\gamma\dot x + \omega^2 x = 0$ in the overdamped regime, we find that there are two decay scales, $\gamma\pm \sqrt{\gamma^2-\omega^2}$.

I am trying to understand if there is a physical origin to these two scales. What I mean by that is, is there a way to see that one expects two solutions beyond statements about solutions of differential equations.

Useful Analogy: When first doing projectile motion, we often use quadratic equation to find the times at which a ball passes a certain height. This gives us two roots. These roots have the physical interpretation that one of them is for when the ball is moving up. The other is for when the ball is moving downwards. I am trying to understand what the upwards and downwards corresponding roots are in this situation.

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    $\begingroup$ Consider the following. When first doing projectile motion, we often use quadratic equation to find the times at which a ball passes a certain height. This gives us two roots. These roots have the physical interpretation that one of them is for when the ball is moving up. The other is for when the ball is moving downwards. I am trying to understand what the upwards and downwards corresponding roots are in this situation. The solutions to ODEs answer is just math, one needs to interpret what these solutions mean physically. Hopefully this helps. $\endgroup$ Commented yesterday
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    $\begingroup$ I have to say I am a bit surprised that this question has attracted negative comments and two close votes. It seems very obvious what the OP is asking, though perhaps this is obvious only to people with some experience of fluid dynamics. $\endgroup$ Commented 20 hours ago
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    $\begingroup$ @Pelti I use the word timescale because that is exactly what I am looking for - its inverse if you want to be really specific. For the underdamped case, the real part of the roots gives us the decay timescale, and the imaginary part gives us the oscillation frequency. Since the complex roots come in conjugate pairs, we have this clear separation into decay and oscillation time scales. The same is not obvious for the overdamped situation. $\endgroup$ Commented 15 hours ago
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    $\begingroup$ @QuantumRingTheory I guess that depends what terminology one is used to. I would call them 'time constants'. In the context of circuit theory, the double solution is a consequence of the need for two different mechanisms of energy storage: in the electric field via capacitors, and in the magnetic field, via inductors. The two limiting time constants, for example in a series RLC circuit are RC and L/R and apply to the extreme cases of very low and very high frequencies (in a parallel RLC circuits the frequency dependence is reversed). The intermingling at a generic f leads to the actual formula $\endgroup$ Commented 12 hours ago
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    $\begingroup$ @Peltio I think you should write this as an answer: it will pair nicely with John Rennie's answer, which I suspect is the same thing mathematically, but in a different physical context. $\endgroup$ Commented 10 hours ago

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It's because in hydrodynamics we have two limits, viscosity dominated and inertia dominated, and these two processes have different timescales.

In the far overdamped region where $\gamma \gg \omega$ the equation of motion simplifies to:

$$ \ddot x + 2\gamma\dot x = 0 $$

because the term in $x$ is negligible, and we get:

$$ x(t) \propto e^{-2\gamma t} $$

which is the result you get from taking the plus sign in $\gamma\pm \sqrt{\gamma^2-\omega^2}$. This is the viscosity dominated regime where the drag is so great that it dominates the motion.

By contrast the inertia dominated regime is when the drag is negligible compared to the elastic forces, and in this regime we get the oscillatory behaviour with an angular frequency $\omega$ and this has a different timescale.

Now, you are considering only the overdamped region, in which by definition we cannot get even a single oscillation, but for $\gamma$ only slightly greater than $\omega$ we still get a significant contribution from the inertial forces, so our equation of motion will have two different timescales: one due to the viscous forces and one due to the inertial forces. That's why we get an equation of the form:

$$ x(t) = A e^{-a_+t} + B e^{-a_-t} $$

where $a_\pm = \gamma\pm \sqrt{\gamma^2-\omega^2} $. The coefficients $A$ and $B$ will be determined by the initial conditions.

I would guess what has puzzled you is that you are assuming the drag dominates in which case the motion is indeed dominated by $a_+$.

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