17.2 Modeling a Spring (Harmonic Oscillator)

A spring is a device that you can expand or contract, but when you do so it tries to get back to its equilibrium position. Suppose an object with weight $$m$$ is attached to the end of a spring of negligible weight, and its equilibrium position is at $$x=0$$. Then the force on the spring when the weight is at position $$x$$ is $$-kx$$, where $$k$$ is what is called the "spring constant" of the device.

The equation of motion of the system is then $$mx'' = -kx$$, or $$mx'' + kx = 0$$.

We know the general solution to this equation, because we can recognize it as the equation satisfied by $$\sin\omega t$$ and $$\cos\omega t$$ when $$\omega = \left(\frac{k}{m}\right)^{\frac{1}{2}}$$. Here the ‘frequency’ of the oscillation is $$\frac{\omega}{2\pi}$$, since sines and cosines repeat as a function of their argument with period $$2\pi$$. (We are using radians as our angle measure.)

Exercise 17.1: Differentiate the function below and show that it it is the general solution to the frictionless spring equation:

$x(t) = x(t_0)\cos(\omega(t-t_0)) + x'(t_0) \frac{\sin(\omega(t-t_0))}{\omega}$

This general solution can also be written as the sum of exponentials; ($$a e^{i \omega t} + be^{-i \omega t}$$), for appropriate $$a$$ and $$b$$.

This solution has the spring oscillating on forever.

In reality there is also friction in the motion which as in the previous section can be modeled by adding a term to the force of the form $$-fx'$$.

The equation of motion then becomes

$mx'' + fx' + kx = 0$

We can solve this equation by looking for solutions of the form $$e^{zt}$$. On substituting this form into the equation for $$x$$ we get:

$mz^2 + fz + k = 0$

which quadratic function of $$z$$ has solutions $$z = \frac{-f \pm (f^2 - 4km)^{\frac{1}{2}}}{2m}$$. The two solutions for $$z$$ here take the place of the two exponentials $$i\omega$$ and $$-i\omega$$ that appear in the frictionless problem. The first term in these solutions, $$-\frac{f}{2m}$$, produces an exponential damping factor in the solutions of $$e^{-\frac{f}{2m}(t-t_0)}$$

As long as $$f^2$$ is less than $$4km$$, the second terms in these solutions for $$z$$ are imaginary so that they give rise to sinusoidal behavior with reduced frequency compared to $$\omega$$. $$X$$ therefore dies off exponentially according to the factor discussed in the previous paragraph, and oscillates according to this factor as it does so.

When $$f^2$$ is $$4km$$ or greater, the solutions for $$z$$ are real numbers and the spring is said to be critically damped. There is no oscillation at all, only exponential decay in the displacement from equilibrium $$x$$, as a function of time $$t$$.

The models described above are useful, but not terribly exciting. More interesting results can be obtained when we consider a spring subject to external stimuli.