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Consider an undamped mass and spring system with friction. The friction is not proportional to the velocity of the mass. If we solve the corresponding differential equation,the spring will oscillate indefinitely as there is no friction. However, for a damped system with friction, the energy in the system goes down and the mass will eventually stop. Is this correct? Here's my solution: enter image description here

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  • $\begingroup$ can you show your solution? you likely forgot to reverse the direction of the friction when v changes sign $\endgroup$ – Wolphram jonny Jan 15 at 19:27
  • $\begingroup$ Why do you think that the diff eq is that same as that w/o friction? You must have done something wrong. You really should show how you set up the diff eq. Otherwise we cannot really help you. $\endgroup$ – ggcg Jan 15 at 19:46
  • $\begingroup$ Is there a motor or magnet or something putting energy back in? $\endgroup$ – PhysicsDave Jan 15 at 19:50
  • $\begingroup$ @Wolphramjonny I considered half of ocillation from -A to A. If only this part be the same this goes on, the object wil again go back to -A and so on. $\endgroup$ – Pouya Esmaili Jan 15 at 19:58
  • $\begingroup$ The general solution is now x=C_1sin(wt+psi)+C_2 $\endgroup$ – Wolphram jonny Jan 15 at 22:01
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The general solution now is $x=c_1 \cos(\omega t+\phi)+c_2$.

If you use initial conditions $x(0)=-A$, $v(0)=0$ and $a(0)=\omega^2A-f_k/m$, then you get: $-A=c_1 \cos(\phi)+c_2$

$0=-c_1\omega\sin(\phi)$

$\omega^2A-f_k/m=-A\omega^2\cos(\phi)$

From the second equation we get $\phi=0$

Then, after some algebra, we get: $c_1=f_k/(m\omega^2)-A$ and $c_2=-f_k/(m\omega^2)$

The motion starts at -A but does not reach A. See for instance the plot here

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  • $\begingroup$ I confused c1 to c2. Is that F friction over m? Can you show your solution for c1 and c2? $\endgroup$ – Pouya Esmaili Jan 16 at 3:54
  • $\begingroup$ it is very straightforward, but I will do it tonight. yes, F is the friction divided by m, a bad choice of nomenclature $\endgroup$ – Wolphram jonny Jan 16 at 11:36
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Your solution to the DE is incorrect. Here's how it works, from the Equation of Motion:

$$m\ddot{x}=-kx\pm f_k$$ $$\frac{m}{k}\ddot{x}+x\pm \frac{f_k}{k}=0$$ Use the following substitution: $$u=x\pm\frac{f_k}{k}$$ $$\mathrm{d}u=\mathrm{d}x$$ $$\mathrm{d^2}u=\mathrm{d^2}x$$

$$\Rightarrow \frac{m}{k}\ddot{u}+u=0$$ Now set: $$\frac{m}{k}=\frac{1}{\omega^2}$$ $$\Rightarrow\ddot{u}+\omega^2 u=0$$ This classic DE has the solution, with $c_1$ and $c_2$ integration constants: $$u=c_1\cos \omega t+c_2 \sin \omega t$$ The values of $c_1$ and $c_2$ are found from the initial conditions: $t=0 \to (u_0,0)$ and $t=0 \to (\dot{u}_0,0)$

Back substitution with $u=x\pm\frac{f_k}{k}$ then delivers $x(t)$, which due to the oscillating nature of $f_k$ needs to be done half-amplitude by half-amplitude. So let's do it for the first half-amplitude.

$$u=x+\frac{f_k}{k}$$ At $t=0$:

$$-A+\frac{f_k}{k}=c_1$$

At $t=0$, first derive to $t$:

$$\dot{x}=-c_1\omega \sin\omega t+c_2\omega \cos\omega t$$

So with $\dot{x}=0$ for $t=0$, we get:

$$0=0+c_2\omega \Rightarrow c_2=0$$

$$\Rightarrow x(t)+\frac{f_k}{k}=\Big(-A+\frac{f_k}{k}\Big) \cos \omega t$$ $$\boxed{x(t)=\Big(\frac{f_k}{k}-A\Big) \cos \omega t-\frac{f_k}{k}}$$

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