Taylor 5.2-5.4 (today) and 5.4-5.6 (Wednesday).
Now let's look at the energy of the harmonic oscillator. Take the solution x(t) = A cos(ωt - δ) and substitute into the expression for the potential energy,
Differentiate x(t) to obtain the velocity and use this to write the kinetic energy
where we've used the relation ω=k/m. The total energy is the sum
a constant as it should be since the linear restoring force is conservative.
where there are four constants to be determined by the initial conditions: Ax, Ay, δx, and δy. Four are required in general for two independent second order differential equaitons.
If ωx/ωy is a rational number, then the motion is periodic (i.e. the mass will eventually return to its initial position with its initial velocity and completely retrace the path it took to get there. If the ratio is irrational, then the motion never repeats.
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 3 cases to consider:
q² > 0 This situation is called overdamped.
q² = 0 This situation is called critically damped.
q² < 0 This situation is called underdamped.
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:
When q2 < 0, the solution to the DE is an exponential with a complex exponent. As shown in Appendix D, eiq = cosq + isinq. I will skip the derivation (you can find it in the text), and jump to the solution:
x(t) = e-gt [Csin(wdt) + Dcos(wdt)] = e-gt Acos(wdt + f)
v(t) = -Ae-gt [gsin(wdt + f) - wdcos(wdt + f)]
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 f. The frequency wd = iq = (w02 - g2)1/2 = (k/m - c2/4m2)1/2 and is called the damped frequency (w0 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 wd rather than w0 .
As when there is no damping, the energy of the oscillator at any instant in time is given by:
E = 1/2 mx'2 + 1/2 kx2 .
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:
dE/dt = mx'x" + kxx' = (mx" + kx)x' = -cx'2
since mx" + cx' + 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.
DE = ò0Td dE/dt dt = -gmw02A2e-2gt Td
where Td = 2p/wd. This result valid when the damping is small -- the weak damping limit. The energy of the oscillator at time t is E = 1/2 mA2w02e-2gt, so we find that the fraction energy loss is
|DE/E| = 2gTd .
We identify 2γ to be an inverse time such that the product is unitless. Calling the time t, then t=(2g)-1, and
|ΔE/E| = Td / τ.
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,
Q = 2πτ/ Td .
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 |
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