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Consider a block of mass $m$ moving with initial velocity $v_o$ attached to a spring with spring constant $k$, on a terrain which has a coefficient of kinetic friction $\eta$ and coefficient of static friction $\epsilon$. Find the time taken for oscillations to die off.

If we write the force equation of block when it's moving to right, we get:

$$ ma = -kx - \eta mg$$

Or,

$$ a = -\frac{k}{m} x - \eta g$$

For a shifted harmonic oscillator of form:

$$ x(t) = A \cos(\omega t + \phi) + x_0 \tag{1}$$

$$ \ddot{x} = -\omega^2 ( x(t) - x_0) $$

Comparing with previous equation,

$$ - \eta g = - \omega^2 x_0$$

Hence,

$$ \frac{ \eta g}{\omega^2} = x_0 \tag{2}$$

By the fundamental equation of springs,

$$ \omega^2 = \sqrt{\frac{k}{m}} \tag{3}$$

Combining 1,2,3:

$$ x = A \cos( \frac{k}{m} t + \phi) + \frac{m \eta g}{k}$$

Now the weird part:

This would suggest that the oscillation would go on forever! However is well known that friction is a dissipative force and removes energy from the system, so if energy is being removed from the system in every cycle, why does the equation not show it?

Possible resolutions

Deeply thinking about the problem, I realized that my differential equation breaks whenever the velocity of the block drops to zero because then all of a sudden the static friction replaces the kinetic friction. I think so this sudden shift shouldn't cause too many problems but I'm not sure. How do you deal with the differential equation of motion suddenly shifting? Or is it some other problem which caused this strange result which I got?

I'm mainly looking for an answer which discusses the break ups of the equation governing motion at when v drops to zero and sign of friction

Update: I found a paper discussing this, may write an answer later based on it (see here)

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  • $\begingroup$ Your final equation does not look like damped harmonic motion. For much more info, see hyperphysics.phy-astr.gsu.edu/hbase/oscda.html $\endgroup$ – David White Nov 21 '20 at 22:51
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    $\begingroup$ Your problem is that the force equation is $ma=-k(x-x_0)-\eta mg$. So the term $\eta g$ is new and you have nothing to compare it with. It's an extra element that will change your dynamics. $\endgroup$ – FGSUZ Nov 22 '20 at 0:16
  • $\begingroup$ Manipulating it a little bit makes it look the same, what's the problem with that? $\endgroup$ – Buraian Nov 22 '20 at 5:16
  • $\begingroup$ Even if it's not damped oscillator, doesn't friction disspate energy @DavidWhite $\endgroup$ – Buraian Nov 22 '20 at 5:23
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    $\begingroup$ I think I got the problem but I'm not sure how to fix it.. it is that the friction keeps flipping sign depending on sign of velocity but I'm not sure how to account for that in diff eqn $\endgroup$ – Buraian Nov 22 '20 at 5:44
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Or is it some other problem which caused this strange result which I got?

The problem is that your “friction” force, $\eta m g$, always points in the negative $x$ direction. It does not behave like friction which always points in the direction opposite $v$.

I'm mainly looking for an answer which discusses the break ups of the equation governing motion at when v drops to zero and sign of friction

That is not the problem here. However, it is certainly possible to include such effects. Usually, a force law like that will not have an analytical solution and you will have to rely on numerical methods

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