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I have read that the during overdamping the damped forces or the resistance to movement of an object in S.H.M is so high that if the object is displaced from ita mean position then it returns to it very slowly without any oscillations. I can see why it moves slowly but ,why does is exactly stop at the mean position and why no oscillations are ovserved about the mean position.

My thoughts on this was-

If we Imagine a box at moved from its equilibrium position(mean positon) towards any direction,assuming that there are resistive forces through out the space which acts opposite to the movement of the box.So if restoring force acts on the box and makes it move towarda it mean point then we can say that as the box moves closer to the point the restoring force decreases eventually but due to the movement the resistive forces are still acting on the box and it remains the same(as they have nothing to do with the box being closer to the mean point) so a point will come before reaching the mean point where the two forces will be equal and thus we can observe a dynamic equilibrium(where the box still continues to move towards the mean point). Again as the box is moving towards the mean point the restoring force starts decreasing further and then eventually becoming zero at mean point. Now if there were no force the box would have moved further due to inertia but in our cases the resistive forces will act on the box until and unless it's velocity is zeroed thus even after the mean point as the box tries to move further the resistive forces still acts on it and thus stops it(also the restoring force is also in the direction of resistive force thus there are 2 forces opposing the motion). Here, what I am in doubt is whether the box stops exactly at the mean point or it just moves a bit ahead of the point and then comes back and stop (cause this is oscillation),or what.I need an explanation in any of the case.

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    $\begingroup$ That no oscillations occur is the definition of the overdamped case, as opposed to damped oscillations, which decrease in amplitude with time. It all comes out from solving the damped oscillator equation: $\ddot{x}+\gamma\dot{x}+\omega_0^2x=0$ $\endgroup$
    – Roger V.
    Commented Sep 6, 2021 at 8:54
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    $\begingroup$ @RogerVadim already gave you a full answer, for details see this: en.wikipedia.org/wiki/Damping $\endgroup$
    – Quillo
    Commented Sep 6, 2021 at 9:02
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    $\begingroup$ You are right, the motion of an overdamped system does not always stop at the mean position without overshooting. If you apply a large initial velocity, it might overshoot once and then return to the mean and stop. But it will never overshoot more than once, so we don't say that it "oscillates". $\endgroup$
    – alephzero
    Commented Sep 6, 2021 at 11:49
  • $\begingroup$ Yes thats what even I thought , as the damping is too high in overdamping case thus it will move a bit ahead from the mean point due to inertia and then the restoring and the heavy damped forces will pull it towards the mean postion very quickly.But there's a very littile oscillation, which I think is ignored and thus considered as non oscillatory as compaired to oscillations in the underdamped and crictically damped cases of damped harmonic oscillator.... $\endgroup$
    – PATRICK
    Commented Sep 6, 2021 at 13:47

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As it nears the equilibrium position $x=0$ the position of an overdamped system is approximately an exponential decay

$$x = e^{-\lambda t}$$

so an ideal overdamped system never actually reaches its equilibrium position in finite time. In real systems, of course, after a long enough time $x$ and $\frac {dx}{dt}$ are both effectively zero for all practical purposes.

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  • $\begingroup$ Sir ,can you provide and example for making it crystal clear,I am not able to imagine a sitution for overdamping.Thank you ... $\endgroup$
    – PATRICK
    Commented Sep 6, 2021 at 10:15

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