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Damped oscillators

Page history last edited by Ben Dreyfus 12 years ago

Class content > Oscillations and waves > Harmonic oscillation > Mass on a spring

 

Prerequisites:

 

In our discussion of a simple harmonic oscillator (Mass on a spring, Hanging mass on a spring), as is usual in physics, we began our analysis by assuming idealizations: an ideal spring satisfying Hooke’s law, no resistive forces (friction, viscosity, drag), etc., etc. We found that despite the fact that the force was continually changing (dependent on position), that there was a fairly simple solution, both mathematically and physically. Mathematically, the solution came out to be proportional to a sine or cosine function of the time, with various dimensioned constants put in to make the units come out right. Physically, the ideas were that an oscillation arises from

 

  • a stable point -- a point where all forces are balanced,
  • a restoring force -- when there is a displacement from the stable point in any direction the force changes so that it pushes it back towards the stable point, and
  • overshoot -- there is no force to stop it at when it gets back to the stable point so it keeps on going past.

 

There are many interesting examples where a system set in oscillation “rings” – oscillates very many times before stopping – such as any object that when struck puts out a tone. Since audible sounds are on the order of thousands of hertz (cycles per second) the oscillation of these objects die away very slowly compared to the time a single oscillation takes.

 

But there are many other examples of interest where resistive forces cannot be neglected. Let’s extend our analysis to a case where the oscillation dies out, as we know it often does.

 

The damped oscillator

The motion of an oscillator arises mathematically from Newton’s 2nd law in the case where the acceleration (second derivative of position) is proportional to the position.

Once the units and constants are sorted out, the solution to this differential equation of motion is straightforward: x(t) = A cos(ω0 t), where A gives the amplitude of the oscillation, ω0 is the angular frequency (= 2πf = 2π/T where f is the frequency and T is the period), and φ is a phase that tells where on the oscillation the object starts at t = 0.

 

This oscillation goes on forever, a very unreasonable result. A simple model of a damped oscillator is a hanging mass on a spring attached to something that is moving through a resistive medium. This could be the mass itself, or, as shown in the figure at the right, something attached to the mass that is moving in the resistive medium. (This is typical of how we do models in physics: we take the simplest possible model we can make that shows the phenomenon we are exploring, with the different aspects of the phenomenon separated in different idealized parts of the model system. In reality all of the pieces -- here, the springiness, the inertia, and the damping, might be mixed together in ways that are conceptually complex, but mathematically identical to the system considered here).

 

We have chosen a model system that has a simple form of damping, as long as our oscillations are small and not so fast as to induce turbulence in the fluid. Let’s suppose that we can get by with adding a simple viscous damping term (see Viscosity) that is proportional to the velocity and which we will assume to be small.

 

 

This small change – the extra v term – makes the math a lot messier. We won’t go through it here. (You can check out the details at Damped oscillators – the math.) But the idea is straightforward: we assume that it will still oscillate, so we still have a cosine term, but now we let the amplitude become a function of t. Since we are not sure what’s going to happen, we let the frequency possibly be different, but we assume the frequency is not a function of time. If that didn’t work out, we might have had to change that assumption, but in this model it does work. Turning all the handles, we get



 

We note that if gamma is too big (if γ/2 > ω0) then ω12 will become negative. Taking the square root of a negative number requires complex arithmetic. What this means is that the physical character of the solution changes and there are no more oscillations. As long as γ/2 < ω0, the system will oscillate. This situation is called underdamped.

 

Interpreting the result

The amplitude: We see that our guess that the amplitude would be a function of time worked out well. The math (the differential equation) told us that the solution would be a decaying exponential with the decay constant proportional to the damping factor, gamma. When there is no viscous damping, gamma becomes 0 and A just becomes A0, a constant (since e0 = 1).

 

The damping constant: It’s useful to make sense of the damping constant by considering its units. Since b has dimensionality of [force/velocity], gamma has dimensionality [force/(mass x velocity)] = [ma/mv]=MLT/MT2L = 1/T, the inverse of a time. Since it comes in with a factor of 2, let’s define the decay time (tau) as

τ = 2/γ.

Then the amplitude looks like A(t) = A0 e-t/τ. Every time the time grows by a factor of τ, the amplitude decreases by a factor of e (2.7…)

 

The frequency: We note that we did not need our frequency to become a function of time, but it did shift a bit becoming slightly smaller.

 

Overdamped and critically damped systems

When γ/2 ≥ ω0 we can't find a value of a frequency at which the system can oscillate. Redoing the differential equations, what we find is that the damping is cutting the amplitude down so fast that the mass slows as it approaches the stable point and doesn't overshoot. The damping wins the battle with the restoring force; the restoring force trying to speed up the mass as it approaches the stable point while the damping force is trying to slow it down. The mass never overshoots past the stable point but approaches it asymptotically.

 

The case that γ/2 = ω0 is called critically damped. Any weaker damping would allow the mass to pass the stable point at least once and show a bit of oscillation. This solution has no cosine part -- just a dying exponential (times a term A + Bt for technical reasons that will be left to a differential equations class). This case actually approaches the stable point maximally quickly and is the condition that car manufacturers try to produces in putting in damped springs (shock absorbers) to soften the ride.

 

The case in which γ/2 > ω0 is called overdamped. The solution turns out to be two exponential terms, one dying with a time constant longer than τ = 2/γ  and one dying with one shorter. The presence of the longer-time exponential is why the overdamped system does not die as fast as the critically damped one. The damping is so strong that the restoring force is not as effective in bringing it to the stable point as a lesser damping force.

 

Follow-ons:

 

Joe Redish 3/23/12

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