# 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.