previous lecture | next lecture | 5200 index
Fowles and Cassidy 3.1 to 3.3
For F(x) = -kx the solution is x(t) = Asin(w0t + f) where w0 = Sqrt(k/m) is the undamped frequency of the oscillator, A is the amplitude of the motion and f is the phase angle. The potential energy function is V(x) = 1/2 kx2, and energy conservation allows us to relate the energy of this oscillator to the amplitude of the motion, A = (2E/k)1/2. Given the initial position and velocity, we found that the amplitude and phase are given by tanf = w0x0/v0 and A2 = x02 + v02/w02.
Let's first consider the simplest example, a mass attached to a spring, able to slide across a frictionless surface [drawing]. If the spring constant is k = 3N/m, and m = 0.3kg, what is the undamped frequency?
w0 = Sqrt(k/m) = Sqrt(3/0.3) = 3.1rad/s or, f0 = w0/2p = 0.5Hz
If we pull the mass 10cm from its equilibrium position and release it from rest at time t = 0, what will be its position for later times?
The initial conditions are x0 = 0.1m and v0 = 0m/s. Therefore, A = 0.1m, f = p/2, and x(t) = (0.1m) sin{(3.1/s)t + p/2}.
Consider the same mass and spring, but this time the spring is supporting the mass in uniform gravity. First, since the weight of the mass will stretch the spring, is it still a SHO? [drawing]
The answer is yes. Basically, the constant external force changes the equilibrium position of the mass, but aside from that the motion will be the same. Let's say that the initial equilibrium position is at x = Xe (we let this be zero before, but now that the equilibrium position will change, let's write it out explicitly). If the mass is at position X (measured from the same x = 0 position as Xe) then the spring is stretched or compressed by the amount X - Xe. The spring force is then Fsp = -k(X - Xe) and the total force on the mass is F = -k(X - Xe) + mg. Check that the signs make sense, make a drawing and think about the direction of the weight and the spring force. The new equilibrium position is the place where the total force is zero: X'e = Xe + mg/k. If we now define x to be the displacement from this equilibrium position, x = X - X'e, then the total force takes on the familiar form F = -kx.
The equilibriums position moves down by an amount mg/k but once that is taken into account, the equation of motion will return to the familiar form, and we can use the earlier results.
A simple pendulum is a point mass attached to a fixed point by a rigid, massless rod. It is approximated by a so-called plumb bob, a small weight attached to one end of a string, with the other end of the string secured to a rigid support. The weight can swing freely back and forth in a uniform gravitational field, and when hanging vertically the pendulum is in equilibrium, so this is an oscillator. [drawing] There are two forces acting on the weight, the force of gravity, and the tension in the string. Decompose the force of gravity into a components parallel to and perpendicular to the string. The parallel component is balanced by the tension in the string. The perpendicular component remains, it has a magnitude of mgsinq , where q is the angle the string makes with vertical, and it points back to the equilibrium position. The displacement of the weight from the equilibrium position is the arc length, s = lq. This may seem like an odd variable to use -- why not use the horizontal position of the weight -- but this is the proper coordinate. The instantaneous speed is after all ds/dt, and NOT dx/dt.
Thus, the equation of motion is m[sddot] = ml[qddot] = -mgsinq. This is not the equation of motion of a SHO. However, if the amplitude of the oscillation is small, then the angle q will be small, and we can approximate sinq by q. Then the equation of motion becomes [qddot] = -(g/l)q and this is the equation for a SHO, but in the variable q. (By multiplying both sides by l, I can change this into an equation for s.) In the approximation of small oscillations, we can treat a simple pendulum like a SHO, and apply the results we obtained. In particular, note that in place of k/m we have the quantity g/l. The frequency of the pendulum is w0 = Sqrt(g/l) and the period is T0 = 2pSqrt(l/g).
Replacing sinq by q is a specific example of a general technique. The general technique is to approximate a complex function by an appropriate series expansion, and truncate the expansion to as few terms as possible. The above example used the Taylor series expansion sinq = q - q3/6 + ... This series was truncated at the first term, so it is limited in accuracy to angles where sinq is approximately equal to q.
Notice that the frequency and period depend only on the length of the pendulum, not on the mass. This is the effect noted by Galileo, and is why pendula make good clocks -- it is much easier to make 2 pendula with the same length than to make masses and springs that are identical.
What length pendulum has a period of 1sec? Using the relation for period given above, l = gT02/4p2 = 0.25m.
The Morse potential approximates the potential energy function for a diatomic molecule. If x represents the distance between the two atoms, x > 0, then V(x) = V0[1 - e-(x-x0)/d]2 - V0 where V0 , x0 , and d are constants. [Sketch the potential function.] It is useful to know the location of the minimum potential energy, and the behavior as x approaches 0 and infinity (we look at 0 because x is not supposed to be less than zero). The location of the minimum is found by setting the derivative to zero:
dV(x)/dx = 2V0/d [1 - e-(x-x0)/d][e-(x-x0)/d] = 0
One solution is x = infinity, but this is not a physically interesting solution, since two atoms infinitely far apart would not form a molecule. A second solution is obtained when 1 - e-(x-x0)/d = 0 for which x = x0. And V(x0) = -V0.
As x approaches infinity, V(x) approaches 0, and as x approaches 0, V(x) approaches V0[e2x0/d - 2ex0/d]. This is greater than zero as long as x0/d > ln2 = 0.69. Although this is not guaranteed, for a realistic model of a molecule, it is the case.
Notice that the potential function has the characteristics of a well (or a bowl). For energies less than zero and greater than -V0 the allowed range of the atoms is restricted. If the energy of the atoms equals -V0 then the atoms remain at a fixed separation distance x0. If the energy is a little bit greater, E = -V0 + DE, then the distance between the atoms can vary by a small amount around the equilibrium distance x0. That is, the atoms oscillate. I would like to calculate the frequency for small oscillations.
To accomplish this we need to approximate the force (or potential) near the equilibrium position. Recall from calculus, that any smooth function can be expanded in a polynomial series about some point, which we call the Taylor (or sometimes MacLauren) series expansion:
f(x) = f(x0) + df(x0)/dx [x-x0] + 1/2 d2f(x0)/dx2 [x-x0]2 + ... = f(x0) + f'(x0)[x-x0] + 1/2 f''(x0)[x-x0]2 + ...
Let's use this to approximate the potential function in the region of its minimum. We already know the first two terms, V(x0) = -V0 and dV(x0)/dx = 0. Remember that the minimum was found by setting the first derivative to zero. Calculating the second derivative gets a bit messy, but luckily you almost never need to calculate the third derivative! From above, dV(x)/dx = 2V0/d [1 - e-(x-x0)/d][e-(x-x0)/d] so differentiating once more gives
d2V/dx2 = 2V0/d2 [2e-(x-x0)/d-1][e-(x-x0)/d]
and at x = x0 this gives d2V(x0)/dx2 = 2V0/d2 . We then write the Taylor series approximation:
V(x) = -V0 + V0/d2 [x - x0]2 + ...
The force on the object is approximately
F(x) = -dV/dx = -2V0/d2 [x - x0] + ...
This expression very closely resembles F = -kx, with k replaced by 2V0/d2 . In fact, we can say that for small oscillations, the force is approximately linear, and the resulting motion is approximately simple harmonic! Immediately, we can write down the frequency for small oscillations using the SHO result, w0 = Sqrt{2V0/md2}.
This is where the real power and importance of the SHO problem lies. Many physical problems involve things that are, in some way or another, "bound" in a potential well, and it is almost always possible to approximate the potential well by a quadratic function near the minimum, and then apply the SHO result.