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.