5.2 Free, Undamped Motion

We start with the idealized case: no damping and no external force. Real systems always lose some energy, but many lose it slowly, and over a few oscillations the undamped model is an excellent approximation. More importantly, it introduces the natural frequency, the single most important number of any vibrating system. Every later section measures things against it.

The equation and its general solution

With $c=0$ and $F(t)=0$, Theorem 5.1.1 becomes

Divide by $m$ and give the positive constant $k/m$ a name.

The units work out: $k/m$ has units $(\text{N/m})/\text{kg}=(\text{kg}\cdot {\text{m/s}}^2)/(\text{m}\cdot \text{kg})=1/{\text{s}}^2$, so ${\omega }_0$ has units $1/\text{s}$, which we write as rad/s.

Now solve with Section 4.2. The characteristic equation is $r^2+{\omega }_0^2=0$, with roots $r=\pm i{\omega }_0$: a complex pair with real part $0$. Hence

Fitting the initial conditions

Suppose $x(0)=x_0$ and $x^{\prime }(0)=v_0$. Differentiate:

At $t=0$ we have $\cos 0=1$ and $\sin 0=0$, so $x(0)=c_1$ and $x^{\prime }(0)=c_2{\omega }_0$. Therefore

This is worth remembering: the coefficient of the cosine is the initial position, and the coefficient of the sine is the initial velocity divided by ${\omega }_0$.

One cosine instead of two: the amplitude–phase form

A sum $c_1\cos {\omega }_{0}t+c_2\sin {\omega }_{0}t$ does not show at a glance how far the mass travels or when it is lowest. A single shifted cosine does. We want to write

with $A>0$. Here is how, step by step (assume $c_1$, $c_2$ are not both zero).

1. Factor out a length. Let $A=\sqrt{c_1^2+c_2^2}>0$. Then $$c_1\cos {\omega }_{0}t+c_2\sin {\omega }_{0}t=A\left(\frac{c_1}{A}\cos {\omega }_{0}t+\frac{c_2}{A}\sin {\omega }_{0}t\right).$$
2. Recognize a point on the unit circle. The numbers $c_1/A$ and $c_2/A$ satisfy $(c_1/A{)}^2+(c_2/A{)}^2=1$. So there is an angle $\phi$ with $$\cos \phi =\frac{c_1}{A},\sin \phi =\frac{c_2}{A}.$$ Geometrically, $\phi$ is the polar angle of the point $(c_1,c_2)$ and $A$ is its distance from the origin.
3. Use the difference formula. Since $\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta$, $$A(\cos \phi \cos {\omega }_{0}t+\sin \phi \sin {\omega }_{0}t)=A\cos ({\omega }_{0}t-\phi ).$$

Reading the motion: amplitude, period, frequency, phase

Since the cosine oscillates between $-1$ and $1$, the mass oscillates between $x=-A$ and $x=A$, forever, with the same pattern repeating. Each part of $A\cos ({\omega }_{0}t-\phi )$ has a name.

Do not confuse ${\omega }_0$ (rad/s) with $f_0$ (Hz). They differ by the factor $2\pi$. Engineers quote natural frequencies in hertz; the formulas are cleaner in rad/s.

Three facts worth remembering

1. The period does not depend on how the motion starts. $T=2\pi \sqrt{m/k}$ involves only $m$ and $k$. Pulling the mass down further gives a larger amplitude, not a slower oscillation (Figure 5.2.3). This is why the system has a natural frequency.
2. Stiffer is faster, heavier is slower. Quadrupling $k$ doubles $f_0$; quadrupling $m$ halves it.
3. The natural frequency can be read from the static deflection. Since $k{s}_0=mg$, we have $k/m=g/s_0$, so $${\omega }_0=\sqrt{\frac{g}{s_0}},f_0=\frac{1}{2\pi }\sqrt{\frac{g}{s_0}}.$$ A softly mounted machine (large sag $s_0$) has a low natural frequency. You can estimate $f_0$ with a ruler, without knowing $m$ or $k$.

Velocity, acceleration and energy

Differentiating $x=A\cos ({\omega }_{0}t-\phi )$,

So the largest speed is $A{\omega }_0$, reached while passing through equilibrium, and the largest acceleration is $A{\omega }_0^2$, reached at the turning points. For the energy (Section 5.1, with $c=0$, $F=0$), substitute and use $k=m{\omega }_0^2$:

The energy is constant: it moves back and forth between the spring and the mass but is never lost. That is why undamped motion never dies out. It also gives a quick way to find $A$ from the position and velocity at any single instant: $k{A}^2=k{x}^2+m{x}^{\prime 2}$.

Try it with the simulator below. Set the damping $c$ to $0$ and no force ($F_0=0$); change $m$, $k$, $x_0$ and $v_0$ and watch the amplitude–phase form and the graph. Check in particular that changing $x_0$ or $v_0$ never changes the period.

Worked examples