The mass-spring oscillator is easy to analyse. If the displacement of the mass away from equilibrium is denoted $x(t)$, then Newton’s law requires
$$m \dfrac{d^2x}{dt^2} + kx=0 \tag{1}$$
for free motion of the mass, with no external force applied. If a force $f(t)$ were to be applied, it would simply replace the zero on the right-hand side of this equation. The general solution of the equation for free motion can be written
$$x(t)=A \cos(\Omega t) + B \sin(\Omega t) \tag{2}$$
where $A$ and $B$ are arbitrary constants, and
$$\Omega^2=\dfrac{k}{m}. \tag{3}$$
$\Omega$ is the oscillation frequency in radians per second: it is related to the frequency $f$ in Hz by $\Omega = 2 \pi f \tag{4}$.
This solution can be written in other forms. The sin and cos terms can be combined into a single sine wave:
$$x(t) = R \cos(\Omega t + \phi) \tag{5}$$
where the amplitude
$$R = \sqrt{A^2 + B^2} \tag{6}$$
and the phase shift $\phi$ satisfies
$$\tan{\phi} = -B/A. \tag{7}$$
The form (5) makes it particularly clear that the most general free motion of the oscillator consists of sinusoidal oscillation at (radian) frequency $\Omega$, with arbitrary amplitude and phase.
The solution can also be written compactly in terms of the complex representation (look back at section 2.1.2 if you are a bit hazy about complex numbers in this application):
$$x(t)= C e^{i \Omega t} \tag{8}$$
where as usual we implicitly assume that the real part of this complex expression is to be taken. The complex amplitude $C$ is given by
$$C=R e^{i \phi}. \tag{9}$$
Another important aspect of the behaviour of the simple oscillator is the response to a sinusoidal applied force. The easiest way to analyse this case is to use the complex representation, and assume an input $Fe^{i \omega t}$ at some chosen frequency $\omega$. From the “sine wave in, sine wave out” property, we know that the response $x(t)$ will also vary sinusoidally at the same frequency, so write it in the form $X e^{i \omega t}$. Equation (1) then requires
$$m \dfrac{d^2}{dt^2}\left[X e^{i \omega t} \right] + kX e^{i \omega t}=Fe^{i \omega t} . \tag{10}$$
Thus
$$(- \omega^2 m+k)X=F \tag{11}$$
so that the frequency response function of the oscillator, $X/F$, is given by
$$\dfrac{X}{F} = \dfrac{1}{k-\omega^2 m}=\dfrac{1/m}{\Omega^2 – \omega^2} \tag{12}. $$
For this particular linear system, the frequency response function turns out not to be complex, but it does vary with frequency $\omega$. For low frequencies $\omega < \Omega$, the result is positive so that the response is in phase with the force. For higher frequencies $\omega > \Omega$, the value is negative so that the response in the opposite phase to the force. When $\omega$ is close to the natural frequency $\Omega$, the response is very large: this is the phenomenon of resonance. When the two frequencies are exactly equal, this model predicts response which is infinite. That is obviously non-physical: we will find out how to fix this problem shortly (see sections 2.2.7 and 2.2.8).