The Classical Harmonic Oscillator

One of the first problems that beginning students of physics get to solve simply asks the students to describe the motion of a simple harmonic oscillator. In that simple system we find a small body of mass m attached to one end of a spring of force constant k with the other end of the spring attached to a suitably massive body that effectively does not move when the spring exerts a force upon it. We constrain the system to move in only one direction, conventionally along part of the x-axis of our coordinate system, and note that if we give the small body a certain amount of kinetic energy, it will move along a line segment in a manner that corresponds to a sinusoidal function of elapsed time.

For the force exerted by a spring we have F=-kx, in which equation x represents the distance we pull or push the free end of the spring away from the neutral point, the point that the free end occupies when the spring exerts no force at all. We imagine stretching or compressing the spring and we integrate that force law to calculate the potential energy U that we thus store in the system,

(Eq’n 1)

If we represent the total energy that we put into the system, either through an impact or by stretching/compressing the spring then letting it go free, by E, then we calculate the kinetic energy of the small body at any instant as T=E-U. Substituting the appropriate algebraic descriptions of the body’s kinetic energy and potential energy, we can calculate the body’s velocity at any instant as

(Eq’n 2)

Note that I have tacitly assumed that the spring has negligible mass relative to the mass of the small body, so all of the system’s kinetic energy goes into the body while all of the system’s potential energy goes into the spring. Defining an alpha term as α=2E/k, we can rearrange that equation into

(Eq’n 3)

We then apply the process of integration to get

(Eq’n 4)

in which we define the angular frequency of the system by writing

(Eq’n 5)

Taking the antilogarithm of Equation 4 gives us

(Eq’n 6)

in which the constant of integration C has become the multiplicative factor A
corresponding to the amplitude of the oscillation represented by the imaginary
time exponential. Because x^{2} will always have a value smaller than
α,
the second term on the left side of Equation 6 always gives us an imaginary
number and the two terms taken together represent the motion of a point on a
circle in the complex plane in accordance with the imaginary exponential on the
right side of the equality sign. We want only the real values for the location
of the small body, so we drop the imaginary terms and Equation 6 becomes

(Eq’n 7)

Now we want to add a resistive force to our oscillator and devise a more generalized description of things that oscillate. We start with the generalized oscillator equation,

(Eq’n 8)

in which α,
β,
and γ
represent constants and X represents distance moved by a mass on a spring or the
electric charge moving in a circuit. In that equation alpha can represent the
mass of a body attached to a spring or it can represent the inductance in an
electrical circuit; beta can represent friction acting on the mass/spring system
or it can represent electrical resistance in accordance with Ohm’s law; and
gamma can represent the stiffness of the spring or it can represent electrical
capacitance. Nicely enough, the operation of multiplication by those constants
commutes with the time-derivative operator. That fact lets us represent Equation
8 as a quadratic equation with the operator d_{t}=d/dt as the unknown
quantity.

Using the classical quadratic formula, which we all remember from our Algebra I class (Thank you, Mr. Disbrow), we define two new operators,

(Eq’n 9)

and

(Eq’n 10)

in which we have

(Eq’n 11)

and

(Eq’n 12)

With those operators we can rewrite Equation 8 as

(Eq’n 13)

The order in which we apply the tau-operators to X makes no difference in the calculation.

Apply the tau-operator to the left side of Equation 13 and put that equation into the form

(Eq’n 14)

The right side of Equation 13 also tells us that the combination ττ* gives us one form of the harmonic oscillator equation’s operator. Equation 14, interpreted through that fact, means that τX represents a solution of the harmonic oscillator equation, just as much as X does. That fact means that we can replace τX by X in Equation 13, thereby getting

(Eq’n 15)

That equation has a simple solution that we can obtain by taking five easy steps: (1) subtract Q*X from both sides of the equation, (2) divide the equation by X, (3) multiply the equation by dt, (4) integrate the equation, and (5) extract the antilogarithm of the integral to get

(Eq’n 16)

in which the amplitude of the solution, X_{0}, comes from the
antilogarithm of the constant of integration.

The same reasoning allows us to take

(Eq’n 17)

and get

(Eq’n 18)

The full solution of Equation 8 consists of a linear superposition of the functions in Equations 16 and 18.

If we substitute from Equation 11 into Equation 18, we get

(Eq’n 19)

in which the constants α,
β,
and γ
represent positive real numbers. Thus, the first term, -βt/2á,
specifies an exponential decay of X with the elapse of time. If 4αγ>β^{2},
the second term in the argument of the exponential function calculates out as an
imaginary number, which imposes a sinusoidal vibration on the value of X. And if
β=0,
we get a pure endless vibration. Thus we have the solution of the harmonic
oscillator equation.

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