Harmonic oscillation with friction

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Discussion Overview

The discussion centers on incorporating kinetic friction into the model of a harmonic oscillator, specifically a block attached to a horizontal spring on a table. Participants explore the forces involved and the implications for the differential equation governing the system's motion.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant introduces the need to account for kinetic friction in the harmonic oscillator's differential equation, noting that friction opposes velocity but is not proportional to it.
  • Another participant suggests a form for the frictional force as F_\mathrm{friction}(t) = \mu x' (t), indicating a potential misunderstanding of the friction model.
  • A further contribution clarifies that in the Coulomb model of friction, the force is constant and its direction depends on the velocity's sign, leading to different cases for solving the equation.
  • It is proposed that the solution involves switching between two separate cases based on the sign of the velocity, resulting in a graph of displacement over time that shows half-oscillations with linearly decreasing amplitudes.
  • One participant notes that the mass will eventually stop moving when the static friction force balances the spring's tension, indicating a finite number of oscillations.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the frictional force and its implications for solving the differential equation. There is no consensus on a single approach to model the friction in this context.

Contextual Notes

The discussion includes assumptions about the nature of friction and its mathematical representation, which remain unresolved. The dependence on initial conditions and the transition between different cases in the solution process are also noted but not fully explored.

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Hello,

I want to include kinetic friction into the harmonic oscillator.
A small blocks is attached to a horiontal spring on a table.
Because there is kinetic friction there are two forces on the blok that we need to describe the oscillation.

First, the force that the spring exerts and second, the kinetic friction.
The kinetic friction is always opposite to the velocity but it is not proportional to velocity.

The differential equation is:

[itex]x''(t)=-kx-F_{friction}[/itex]

How can I rewrite Fore of friction to solve this equation? force of friction is not just a constant since it depends on the direction of the velocity.
 
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You mean that [itex]F_\mathrm{friction}(t) = \mu x' (t)[/itex]?

How would you solve it without the friction term? I guess you would make an Ansatz for the form of the solution such as [itex]x(t) = e^{\lambda t}[/itex] ?
 
CompuChip said:
You mean that [itex]F_\mathrm{friction}(t) = \mu x' (t)[/itex]?

How would you solve it without the friction term? I guess you would make an Ansatz for the form of the solution such as [itex]x(t) = e^{\lambda t}[/itex] ?

Most engineers (at least in the UK and US) would call that "viscous damping", not "friction".

For the Coulomb model of friction, F in the OP's equation is constant, and its sign depends on the sign of the velocity.

You can easily solve the two separate cases where F is positive or negative. The solution is the same as if the mass and spring was vertical, and F was the weight of the mass.

For the complete solution, you start with one of the two solutions (depending on the initial conditiosn) until the velocity = 0, then you switch to the other solution, and so on. You can't easily get a single equation that gives the complete solution in one "formula" for x.

The graph of displacement against time will look like a sequence of half-oscillations of simple harmonic motion, with amplitudes that decrease in a linear progression (not exponentially). The mass will stop moving after a finite number of half-osciillations, at some position where the static friction force can balance the tension in the spring.
 
Thank you very much for replying:)
 

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