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Lecture 10

Lecture 9 | Lecture 11


By Giuseppe Tenti and annotated by Douglas Wilhelm Harder.

The next three lectures will examine four 2nd-order initial-value problems with constant coefficients including:

  • Mechanical Oscillator Model
  • Electrical Oscillator Model
  • Free Oscillations
  • Forced Oscillations
  • The Schrödinger Equation
  • The Particle in a Box

The Mechanical Oscillator Model

Consider the mass attached to two equal strength springs shown in Figure 1. The two springs (elastic objects storing mechanical energy) are anchored to two walls and the mass is sitting on a surface with resistance which will oppose motion. When the mass is at position tex:$$x = 0$$, the force by the two springs is equal and therefore the mass is at equilibrium.


Figure 1. A mass at rest between two springs.

The force on the mass when it is at rest is 0, but if there is a non-zero force on the mass, the force will cause an acceleration due to Newton's law: tex:$$F = ma$$. Because acceleration is defined as the second derivative of position with respect to time, we may write tex:$$F = m\ddot{x}$$.

Suppose now that the mass is displaced to a distance tex:$$x$$ from the equilibrium point, as is shown in Figure 2.


Figure 2. The mass in Figure 1 displaced to a distance tex:$$x$$ and the resulting force.

At this point, the springs cause a force on the mass attempting to force the mass back to equilibrium. The size of the force is proportional to the stretch of the spring, and therefore we may write:

tex:$$m\ddot{x} = -kx$$

where tex:$$k$$ is the spring constant. This is called Hooke's law. If these were the only forces, our differential equation would therefore be

tex:$$m\ddot{x} + kx = 0$$

where both tex:$$m > 0$$ and tex:$$k > 0$$. By rewriting this as

tex:$$\ddot{x} + \frac{k}{m}x = 0$$

we already know that the general solution must be of the form

tex:$$x(t) = c_1\cos\left( \sqrt{\frac{k}{m}} t \right) + c_1\sin\left( \sqrt{\frac{k}{m}} t \right)$$.

This says that, if the mass is ever displaced and released, it will oscillate forever.

In reality, however, there is a friction between the mass and the surface, as is shown in Figure 3, and once the mass begins to move, the frictional force moves to oppose that movement.


Figure 3. The friction opposing the movement of the mass.

We can therefore add this mass to to our equation:

tex:$$m\ddot{x} = -kx - \gamma\dot{x}$$

or

tex:$$m\ddot{x} + \gamma\dot{x} + kx = 0$$.

Note: both of these forces are phenomenological in the sense that they are local linear approximations based on experimental data. If a spring is stretched too far, that is, it is pushed to its elastic limit, it will fail, and neither spring can be compressed beyond the wall it is attached to.

Finally, let us assume that there is a external force independent of the springs which is also placing a force on the mass. This is a force which may vary with time, and therefore we will represent it by a function tex:$$F(t)$$ as shown in Figure 4.


Figure 4. A force tex:$$F(t)$$ external to the springs and friction.

This gives us our final differential equation

tex:$$m\ddot{x} + \gamma\dot{x} + kx = F(t)$$.

Equation 1

Standard Form

Because the differential equation in Equation 1 is so common, engineers will generally rewrite this differential equation in the form

tex:$$\ddot{x} + 2\zeta\omega\dot{x} + \omega^2x = W(t)$$

where

  • tex:$$\omega$$ is the natural frequency of the oscillator,
  • tex:$$\zeta$$ is the damping parameter, and
  • tex:$$W(t)$$ is the reduced external force.

For this example of an oscillating mass, we have that:

  • tex:$$\omega = \sqrt{\frac{k}{m}}$$ is the natural frequency of the oscillator,
  • tex:$$\zeta = \frac{\gamma}{2m\omega}$$ is the damping parameter, and
  • tex:$$W(t) = \frac{F(t)}{m}$$ is the reduced external force.

This differential equation is a linear 2nd-order inhomogeneous differential equation with constant coefficients. If the reduced external force is zero, the differential equation is homogeneous.

Science and engineering use the following terminology:

The differential equation tex:$$\ddot{x} + 2\zeta\omega\dot{x} + \omega^2x = W(t)$$ describes a forced oscillator and tex:$$W(t)$$ is the forcing function.

The differential equation tex:$$\ddot{x} + 2\zeta\omega\dot{x} + \omega^2x = 0$$ describes a free oscillator or an unforced oscillator.

The Electrical Oscillator Model

As you have seen in your circuit theory courses, a circuit containing a capacitor and an inductor have two energy-storage elements that cannot replaced by a single storage element. (You will recall that resistors, capacitors, and inductors in parallel or series may normally be replaced by a single element of the same kind while two such elements in series may be switched.) Such a circuit is called a linear second-order circuit as it is described by a linear second-order differential equation.

While there are arbitrarily many different circuits, we will focus on the simple RLC circuit shown in Figure 5.


Figure 5. An RLC circuit with a switch.

If there is no charge on the capacitor and no current passing through the inductor, the circuit is said to be quiescent. At a moment in time, usually at time tex:$$t = 0$$, the switch is closed and the current begins to flow through the wire—the circuit is now active.

To determine the activity in the circuit, it is necessary to recall that the voltage drop across

  • the inductor is tex:$$L \frac{di}{dt}$$,
  • the resistor is tex:$$R i$$, and
  • the capacitor is tex:$$\int \frac{i}{C} dt$$.

Therefore, according to Kirchoff's Voltage Law (KVL or the principle of conservation of charge) we have

tex:$$L\frac{di}{dt} + Ri + \frac{1}{C} \int i dt = V(t)$$

If our goal is to find the current, we may differentiate the equation:

tex:$$L\frac{d^2i}{dt^2} + R\frac{di}{dt} + \frac{i}{C} = \dot{V}(t)$$

or, if we are interested in finding the charge, we can substitute tex:$$i = \dot{q}$$ to get

tex:$$L\ddot{q} + R\dot{q} + \frac{q}{C} = V(t)$$.

Both of these differential equations are mathematically equivalent. The derivative of a solution to the first differential equation is a solution to the second, but also, both have the same natural frequency (tex:$$\omega = \frac{1}{\sqrt{LC}}$$) and the same damping parameter (tex:$$\zeta = \frac{R}{2L\omega}}$$)—only the reduced external force (tex:$$W = \frac{\dot{V}(t)}{L}$$ or tex:$$W = \frac{V(t)}{L}$$, respectively) is different.

We will write this common mathematical structure as

tex:$$\ddot{\phi} + 2\zeta\omega\dot{\phi} + \omega^2\phi = f(t)$$

and we will then only have to specify the variables tex:$$\phi$$, tex:$$\zeta$$, tex:$$\omega$$, and tex:$$f$$ as well as the necessary initial conditions to get the desired solution.

Finding the Standard Form in General

To convert the differential equation

tex:$$a\ddot{x} + b\dot{x} + cx = F(t)$$

into the standard form, divide by tex:$$a$$ to get

tex:$$\ddot{x} + \frac{b}{a}\dot{x} + \frac{c}{a}x = \frac{F(t)}{a}$$

Now, in general for reasons of stability, the ratio tex:$$\frac{c}{a} \ge 0$$. Thus, solve tex:$$\omega^2 = \frac{c}{a}$$ and for convenience, choose the positive root. We can do this because tex:$$\sin(-\omega t) = -\sin(\omega t)$$ and therefore it is simply convenient to choose the positive frequency.

Next, solve tex:$$2 \zeta \omega = 2 \zeta \sqrt{\frac{c}{a}} = \frac{b}{a}$$ to get tex:$$\zeta = \frac{b}{2 a \omega} = \frac{b}{2 \sqrt{ac}}$$.