## Constant damping force on springsystem

If we consider a block of mass m attached to a spring, where the system oscillates on a table with friction f, the friction force f on the block would depend on the direction of the velocity, as

$$m\ddot{x} = \begin{cases} -kx+f & \text{if } \dot{x}<0\\ -kx-f & \text{if } \dot{x} > 0 \end{cases}$$

If I just look at one equation at a time and solve them both separatly first, I get equations where the amplitude doesn't drop with time. But that should be the case (energy in that closed system isn't conserved). So how can I solve this system?
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 Quote by P3X-018 If we consider a block of mass m attached to a spring, where the system oscillates on a table with friction f, the friction force f on the block would depend on the direction of the velocity, as $$m\ddot{x} = \begin{cases} -kx+f & \text{if } \dot{x}<0\\ -kx-f & \text{if } \dot{x} > 0 \end{cases}$$ If I just look at one equation at a time and solve them both separatly first, I get equations where the amplitude doesn't drop with time. But that should be the case (energy in that closed system isn't conserved). So how can I solve this system?
The direction of $\text{f}$ changes abruptly every time the velocity canges direction. You solve this by solving x(t) for every time interval that has constant $\text{f}$ separately. Solving for each half period separately may appear to give no change in amplitude, as if the mass were hanging on a spring in the presence of gravity, but in fact you are being fooled into thinking that is the case. The change in direction of $\text{f}$ every time the velocity chages direction is going to reduce the amplitude. You can find out the magnitude of this effect from energy considerations, or you can think about what a constant applied force does to the equilibrium position of the oscillator.