ME3120 · Dynamic Modelling
Theme 2 · Newton's method
Small-angle linearization
The pendulum's \(\sin\theta\) blocked us. For small swings we trade it for a straight line — and the equation snaps into the clean spring shape, frequency and all.
Source: course notes; the small-angle approximation (\(\sin\theta\approx\theta\)).
Before you start
What you need first
- The pendulum equation \(\ddot\theta + \tfrac{g}{L}\sin\theta = 0\) — and that the \(\sin\theta\) makes it nonlinear.
- The standard form \(\ddot{(\;)} + \omega_n^2(\;) = 0\) and reading \(\omega_n\) off it (from the spring topic).
- Radians — angles measured as arc length over radius.
What you'll be able to do
- Use \(\sin\theta \approx \theta\) and know when it is allowed.
- Turn the pendulum into a clean linear equation.
- Read off \(\omega_n=\sqrt{g/L}\) and the period \(T\).
Where we got stuck
Last topic we built the pendulum's equation of motion with Newton:
$$\ddot\theta + \frac{g}{L}\sin\theta = 0$$
The trouble is the \(\sin\theta\): the unknown \(\theta\) is trapped inside it,
so the equation is nonlinear. We cannot read a natural frequency off it the
way we did for the spring. We need a way to straighten it out.
The key move
The small-angle approximation
Linearization means: near a point of interest, replace a curved function with the straight line that just touches it there (its tangent).
For \(\sin\theta\) near \(\theta = 0\), that tangent line is simply \(\theta\) itself. So, for small angles measured in radians:
$$\sin\theta \approx \theta \qquad(\theta \text{ in radians})$$
The graph shows why: near zero the curve \(y=\sin\theta\) and the line
\(y=\theta\) sit right on top of each other. They only peel apart as \(\theta\) grows.
Radians are essential. \(\sin\theta\approx\theta\) is
only true in radians. In degrees it is false (e.g. \(\sin 10^\circ = 0.174\), not \(10\)).
Near \(\theta=0\) the line \(y=\theta\) is the tangent to
\(y=\sin\theta\); they agree closely in the small-angle region and drift apart later.
When is it allowed?
"Small" is not a feeling — we can measure the error. Here is \(\theta\) (in radians) against the true \(\sin\theta\):
| Angle | \(\theta\) (rad) | \(\sin\theta\) | Error of \(\theta\approx\sin\theta\) |
|---|---|---|---|
| \(10^\circ\) | 0.1745 | 0.1736 | ≈ 0.5 % |
| \(15^\circ\) | 0.2618 | 0.2588 | ≈ 1.2 % |
| \(20^\circ\) | 0.3491 | 0.3420 | ≈ 2.1 % |
| \(30^\circ\) | 0.5236 | 0.5000 | ≈ 4.7 % |
Rule of thumb: below about \(10^\circ\) the error is under
\(1\%\) (excellent); up to \(\sim 20^\circ\) it is still only a few percent (usually fine).
Past that, the \(\sin\theta\) really matters and the approximation breaks down.
Linearize
Apply it to the pendulum
Take the pendulum equation and swap \(\sin\theta\) for \(\theta\):
1Start from the nonlinear equation:
$$\ddot\theta + \frac{g}{L}\sin\theta = 0$$
2Use \(\sin\theta \approx \theta\) (small swings):
$$\ddot\theta + \frac{g}{L}\,\theta = 0$$
The curve is gone. This is now linear in \(\theta\) — the
same friendly shape as the spring's \(\ddot x + \tfrac{k}{m}x = 0\).
A small swing: the bob barely leaves straight-down, so
\(\sin\theta\approx\theta\) is excellent here.
Read off the natural frequency and period
Any free linear oscillation has the standard form
\(\ \ddot{(\;)} + \omega_n^2(\;) = 0\). Match our linearized pendulum to it and read
\(\omega_n\) straight off.
1Compare \(\ddot\theta + \dfrac{g}{L}\theta = 0\) with \(\ddot\theta + \omega_n^2\theta = 0\), so \(\omega_n^2 = g/L\). Take the root:
$$\omega_n = \sqrt{\dfrac{g}{L}}\,$$natural frequency (rad/s)
| Symbol | Meaning | SI unit |
|---|---|---|
| \(\omega_n\) | natural (angular) frequency | rad/s |
| g | acceleration due to gravity (\(9.81\)) | m/s² |
| L | rod length | m |
2The period (time for one full swing) is \(T = 2\pi/\omega_n\):
$$T = 2\pi\sqrt{\dfrac{L}{g}}$$
Two famous facts fall out. The period depends only on
\(L\) and \(g\) — not on the mass, and (for small swings) not on how far
you pull it. A longer pendulum is always slower.
Same shape as the spring
| Mass–spring | Pendulum (small angle) | |
|---|---|---|
| Equation | \(\ddot x + \tfrac{k}{m}x = 0\) | \(\ddot\theta + \tfrac{g}{L}\theta = 0\) |
| "Stiffness/inertia" | \(k/m\) | \(g/L\) |
| Natural frequency | \(\omega_n=\sqrt{k/m}\) | \(\omega_n=\sqrt{g/L}\) |
🔭 Looking ahead: straightening a curve near an operating point is the
single most-used trick in modelling and control. Later we linearize a whole robot the
same way — about a chosen pose — to study how it behaves nearby.
Frequency and period of a 1-metre pendulum
A simple pendulum has a rod length \(L = 1.0\ \text{m}\) and swings through small angles. Take \(g = 9.81\ \text{m/s}^2\). Find its natural frequency \(\omega_n\) and period \(T\).
The worked-example pendulum: \(L=1.0\ \text{m}\), small swings.
1Natural frequency from \(\omega_n=\sqrt{g/L}\):
$$\omega_n = \sqrt{\frac{9.81}{1.0}} = \sqrt{9.81} = 3.13\ \text{rad/s}$$
2Period from \(T = 2\pi/\omega_n\):
$$T = \frac{2\pi}{3.13} \approx 2.01\ \text{s}$$
A one-metre pendulum takes about two seconds per full swing — and that is true
whether the bob is heavy or light, pulled out \(5^\circ\) or \(10^\circ\).
✏️ Try it yourself
A simple pendulum has rod length \(L = 0.25\ \text{m}\) and swings through small angles. Take \(g = 9.81\ \text{m/s}^2\). Find \(\omega_n\) and the period \(T\).
Step 1.
\(\omega_n = \sqrt{g/L} = \sqrt{9.81/0.25} = \sqrt{39.24} = 6.26\ \text{rad/s}\)
Step 2.
\(T = \dfrac{2\pi}{6.26} \approx 1.00\ \text{s}\)
Read it:
a quarter the length of the 1-m pendulum, so \(\omega_n\) doubles and the period halves
(\(2.01\to 1.00\) s) — because \(\omega_n\propto 1/\sqrt{L}\).
Common mistakes to avoid
| Mistake | Fix |
|---|---|
| Using \(\sin\theta\approx\theta\) in degrees | It only holds in radians. Convert first (\(180^\circ = \pi\) rad). |
| Linearizing at large angles | Past \(\sim 20^\circ\) the error grows fast — keep the full \(\sin\theta\) (or simulate it). |
| Thinking the period depends on mass | \(T = 2\pi\sqrt{L/g}\) has no \(m\) — mass cancelled back when we built the model. |
| Forgetting \(\sqrt{\;}\) when reading \(\omega_n\) | The standard form gives \(\omega_n^2 = g/L\); take the square root for \(\omega_n\). |
| Calling the approximation "exact" | It is a tangent-line model, valid only for small swings — always state that assumption. |
Recap — the whole topic on one screen
$$\sin\theta\approx\theta \;\Rightarrow\; \ddot\theta + \frac{g}{L}\theta = 0 \;\Rightarrow\; \omega_n=\sqrt{\tfrac{g}{L}},\quad T=2\pi\sqrt{\tfrac{L}{g}}$$
| Idea | What you own now |
|---|---|
| The trick | Near \(\theta=0\), replace the curve \(\sin\theta\) with its tangent \(\theta\) |
| When it's valid | Small angles in radians (\(\lesssim 20^\circ\), error a few %) |
| The linear model | \(\ddot\theta + \tfrac{g}{L}\theta = 0\) — same shape as the spring |
| The results | \(\omega_n=\sqrt{g/L}\), \(T=2\pi\sqrt{L/g}\) — no mass, no amplitude |