Taylor 5.4-5.5 (today) and 5.5-5.6 (Friday).
What is the frequency for small oscillations of a mass m about the minimum of the potential U(r) = a/r ² - b/r ?
First, find the minimum of the potential.
thus rmin = 2a/b. The value of the potential at this minimum is U(rmin) = b ²/4a - b ²/2a = -b ²/4a.
Second, to determine the frequency for small oscillations, we need the second derivative of U at the minimum.
The second derivative is positive, so this is a position of stable equilibrium. In the small amplitude approximation, this is the equivalent of k, giving
Now let's apply a new technique to the important problem of the damped harmonic oscillator. We assume that the damping (resistance) term is linear in the velocity. This makes the problem possible to solve in closed form, but in this case, it is also an important example with interesting behavior to understand.
There are a number of ways to solve DE's, for instance direct integration, "guessing" the solution, or the method of operators that I will demonstrate here. Let D stand for the "operation" of taking a derivative (a time derivative for our purposes). Then ax = Dvx, and vz = Dz. With this notation, the above differential equations become:
Let γ = c/2m, ω0² = k/m, and divide through by m to get:
This equation looks like a second order polynomial. A second order polynomial can always be factored (allowing for the use of imaginary numbers) and it turns out that we can do the same with our operator expression because none of the terms in parenthesis (all constants) are effected by the derivative operation (we say they commute).
The order of the two terms is arbitrary. Clearly if the result of the first (rightmost) term in brackets is zero, then the equation is satisfied. But we have solved this earlier, and the solution is an exponential function. The same holds here. Let's call the quantity γ² - ω0² = q². Then there are two solutions for x:
This result may seem a bit strange at first. After all, in the undamped case, the solution was a sine or cosine function not an exponential, and letting γ-->0 takes us to the undamped case. What happens when γ=0? Then q2=-ω0² leading us to conclude that q is imaginary. And x is the exponential of an imaginary number times t, i.e. the solution becomes a sine or cosine function.
Recall that q² = γ² - ω0². Then we have 4 cases to consider:
q² > 0 This case is called overdamped.
q² = 0 This case is called critically damped.
q² < 0 This case is called underdamped.
q² = -ω0² This case corresponds to γ=0 and is called undamped. We have already solved this case and will skip it.
This is mathematically the easiest case. The solution is:
This solution corresponds to the position returning exponentially to the equilibrium position. Despite the name "damped harmonic oscillation", there is no oscillation! Well, if you arrange things right there might be one oscillation.
And note that q is always less than γ, so it is never possible for γ-q to become negative and therefore turn the A1 exponential into a positive exponential.
When q=0, the two terms of the DE are identical:
So the solution x1 = Ae-γt has only one arbitrary constant. But this is a second order DE -- there should be TWO arbitrary constants. Physically this corresponds to being able to independently specify the initial position and velocity for the oscillator. The solution to this dilema is to ask if there is a function x2(t) that yields a result of the form Ae-γt after being acted on by [D + γ]. Then the second [D + γ] guarantees that the result is 0. Mathematically this means:
The solution, which can be verified by substitution, is
So, our general solution for the critically damped case is the sum of x1 and x2:
This solution is the transition from overdamped to underdamped. It is of interest because for some devices (shock absorbers on a vehicle) this is the optimal regime.
When q² < 0, the solution to the DE is an exponential with a complex exponent. Recall that eiθ = cosθ + isinθ. The solution is then:
These two forms of the solution are equivalent. In the first case, C and D are the unknown constants while in the second case they are A and δ. The frequency ωd = iq = (ω0² - γ²)½ = (k/m - c²/4m²)½ and is called the damped frequency (ω0 is the undamped or natural frequency).
This solution is almost the same as the solution of the undamped oscillator. There are two differences:
The amplitude decays exponentially to zero, and
the frequency of oscillation is ωd rather than ω0.
As when there is no damping, the energy of the oscillator at any instant in time is given by:
When there is no damping the energy is constant. With damping, we expect that the energy will tend to decrease with time and we find that:
since ma + cv + kx = 0 is the equation of motion. The frictional constant c is positive as well as the square of the velocity, so the change of energy with time is always negative, that is, the energy is always decreasing. The instantaneous rate of energy loss varies with the velocity, but the fractional energy loss averaged over one period of oscillation is approximately constant.
where Td = 2π/ωd. This result is valid when the damping is small -- the weak damping limit. The energy of the oscillator at time t is E = ½mA²ω0²e-2γt, so we find that the fraction energy lost is
We identify 2γ to be an inverse time such that the product is unitless. Calling the time &tau, then τ=(2γ)-1, and
The quality factor, Q, is defined as 2π times the energy stored in the oscillator divided by the energy lost in a single period of oscillation. Therefore we get immediately that, in the case of weak damping,
mx[ddot] + cx[dot] + kx = 0
Three cases to consider, overdamped, critically damped, and underdamped.
| quantity | relations | comments |
|---|---|---|
| ω0 | ω02 = k/m | positive |
| ωd | ωd2 = ω02 - γ2 | positive |
| γ | γ = c/2m | positive |
| damping | γ | decay parameter |
|---|---|---|
| none | γ=0 | 0 |
| under | γ<ω0 | γ |
| critical | γ=ω0 | γ |
| over | γ>ω0 | γ - sqrt(γ² - ω0²) |
For an oscillator with a natural (undamped) frequency of ω0, let's look at how quickly the system returns to its equilibrium state as a function of the damping characterized by γ. This is determined by how quickly the exponential factor decreases to zero, and is shown in Fig. 5.13 (β = γ). The system returns to equilibrium most quickly at critical damping, making this the optimal regime for shock absorbers on vehicles.
Every real oscillator has some friction or energy loss, and so will eventually stop oscillating. It is therefore common to drive an oscillator with a periodic force. We'll investigate the behavior of an oscillator with a driving force F. We now have the solution to the homogeneous differential equation. If an additional driving force is applied, then we need to consider the equation:
In general, the force can take on any form, but a particularly interesting case is if the force is time dependent, F = F(t). Then, the differential equation is a non-homogeneous, linear, ODE, and techniques exist to solve it. We will solve the case where the force varies cosinusoidally with time: F(t) = F0cos(ωt).