# Damped oscillators

#### Prerequisites

Perhaps the most important application of simple oscillators in biology is hearing. External sound waves (traveling vibrations) creating waves in the fluid in the inner ear. These vibrations cause small oscillators (hair fibers) in the inner ear to vibrate, creating nerve signals. How energy is transferred between the vibrating fluid and the fiber depends not just on its natural frequency $\omega_0$), but on the viscosity of the fluid in which the fibers move. To understand the mechanism how this works, we'll first consider the simplest example of a damped oscillator, and then, in the follow-on, we'll explore the important phenomenon of resonance — how a damped oscillator interacts with a driving force.

## Included a resistive force

In our discussion of a simple harmonic oscillator (mass on a spring, the pendulum), 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 key conceptual 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 — when it gets back to the stable point, there is no force to stop it at 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 equations

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.

$$ma = -kx$$

$$\frac{d^2x}{dt^2} + \omega_0^2 x = 0\quad \quad \omega_0^2 = \frac{k}{m}$$

(This form, with all the derivatives written out and all the terms moved to the left is the standard form you will meet if you take a course in differential equations, of which this is, of course, an example.)

Once the units and constants are sorted out, the solution to this differential equation of motion is straightforward: $x(t) = A \cos{(ω_0 t)}$, where $A$ gives the amplitude of the oscillation, $ω_0$ 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. (See Reading the content in the harmonic oscillator solution.)

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 that is proportional to the velocity and which we will assume to be small:

$$ma = -k -bv$$

Putting in the acceleration as a second derivative and rearranging the parameters, we get

$$\frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + \omega_0^2 x = 0 \quad \quad \omega_0^2 = \frac{k}{m} \quad \gamma = \frac{b}{m}$$

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

$$x(t) = A(t) \cos{(\omega_1 t)}$$

$$A(t) = A_0 e^{-\gamma t/2}$$

$$\omega_1^2 = \omega_0^2 -\frac{\gamma^2}{4}$$

(We've left out a phase shift, $\phi$. We could include it if we wanted.)

## 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 $A_0$, a constant (since $e^0 = 1$).

The damping constant: It’s useful to make sense of the damping constant by considering its dimensionality. 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 $A(t) = A_0 e^{-t/τ}$. 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 expect our frequency to become a function of time, but it did shift a bit becoming smaller. We note that if $\gamma$ is too big (if $γ/2 > ω_0$) then $ω_1^2$ 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 $γ/2 < ω_0$, the system will oscillate. This situation is called underdamped. If $\gamma/2$ is bigger than $\omega_0$, the system is called overdamped. The boundary between the two — $\gamma/2 = \omega_0$ — is called critically damped.

These conditions make sense when they are expressed in terms of the period and the damping times:

• underdamped: damping time great than the undamped period $(\tau > T)$ so it oscillates in a shorter time than it damps;
• critically damped: damping time equal to the undamped period $(\tau = T)$ so it damps at the rate it would have oscillated - no time to oscillate;
• overdamped: damping time shorter than the undamped period $(\tau < T)$. To see picks up a more slowly damping term check the technical page!

## Overdamped and critically damped systems

When $γ/2 ≥ ω_0$ 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 boundary between the two is critically damped: $γ/2 = ω_0$. 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.

In the overdamped case $γ/2 > ω_0$. 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.

Joe Redish 3/23/12

Article 669