Forces in Damped Forced Oscillations?

[applestrudle]

The example I'm thinking of is a mass spring system.

x = Ae^([itex]\gamma[/itex]/2)t cos(wt +a) + Ccos(wt)

If the steady state has been reached, the displacement due to the free oscillations will be negligible, so does that mean that the only force acting on the mass is the driving force, F0cos(wt)? Has the restoring force (-kx) disappeared or is it still acting (so that the resultant force = -kx + F0cos(wt))?

I'm having trouble understanding how the resultant force on the mass changes in the steady state.
It would make sense (to me) if -kx disappears andthe driving force is the only one present but then where does the restoring force go?




Thanks

 

[dauto]

No, the restoring force doesn't magically disappear.

 

[applestrudle]

No, the restoring force doesn't magically disappear.

But when the free oscillations have died out the restoring force must be very small, right? Because the displacement due to the free oscillation is small?

 

[mikeph]

The restoring force is -kx, where x is the total displacement, that's it. I think you're getting confused by splitting the oscillations into forced and free.

 

[sophiecentaur]

I'm not sure what the original expression is supposed to represent. If a natural oscillator is driven with a periodic force at (or near) its natural frequency, the result will be a resonance. The amplitude (at the driving frequency) will build up until the input power is the same as the power being dissipated. The 'original' energy in the freely oscillating system will gradually be dissipated, according to the Q of the system. The final amplitude will depend upon where the driving frequency sits on the response curve. The energy in the oscillator could be many times the energy input, per cycle, of the driving source.

 

[Pythagorean]

The example I'm thinking of is a mass spring system.

x = Ae^([itex]\gamma[/itex]/2)t cos(wt +a) + Ccos(wt)

If the steady state has been reached, the displacement due to the free oscillations will be negligible, so does that mean that the only force acting on the mass is the driving force, F0cos(wt)? Has the restoring force (-kx) disappeared or is it still acting (so that the resultant force = -kx + F0cos(wt))?

I'm having trouble understanding how the resultant force on the mass changes in the steady state.
It would make sense (to me) if -kx disappears andthe driving force is the only one present but then where does the restoring force go?

Thanks

Hyperphysics does your exact problem (assuming your gamma is negative so that it is actually damped). Here they give a break down of how each parameter of the original equation contributes to the final solution, therefore giving you an idea of which physics is dominating:

http://hyperphysics.phy-astr.gsu.edu/hbase/oscdr.html#c2

Notice that the final steady state solution depends on the natural frequency of the system, wo (determined by both k and m) as well as the driving force frequency, w, and even the damping term, gamma. So the final solution really depends on all physical parameters involved (and thus all physical phenomena involved).

 

[Redbelly98]

All three forces are still present in steady state: driving, damping, and restoring force.

As long as x≠0, the restoring force -kx is nonzero.

A similar argument holds for the damping force provided v≠0