13.4 Small-Angle Approximations and Rotational Oscillations
A Clock That Keeps Time
[Video: A grandfather clock in a quiet room. The camera lingers on the brass pendulum as it swings steadily back and forth, its tick marking out even seconds. Then the camera zooms in on the bob at the bottom of the pendulum arm. As the pendulum reaches its maximum displacement, it pauses --- and then gravity pulls it back. A slow-motion overlay traces the arc. The path is not a straight line; it is a curve, a segment of a circle. The word "arc" appears along the path. Then, on screen, the exact equation of motion appears: $\frac{d^2\theta}{dt^2} = -\frac{g}{L}\sin\theta$. The $\sin\theta$ glows. A question fades in: "Is this simple harmonic motion?"]
A pendulum swings back and forth with remarkable regularity. For centuries, it was the best timekeeping mechanism humans had. Huygens built the first pendulum clock in 1656, and its descendants kept time for nearly three hundred years.
But here is a puzzle. In Sections 13.1 through 13.3, we built the theory of simple harmonic motion around a very specific equation:
$$\frac{d^2x}{dt^2} = -\omega^2 x$$
The solution is a sinusoid, the period is independent of amplitude, and everything works out cleanly. But the pendulum equation is not that. The pendulum equation is:
$$\frac{d^2\theta}{dt^2} = -\frac{g}{L}\sin\theta$$
That $\sin\theta$ makes all the difference. The SHM equation has $\theta$ on the right side. The pendulum equation has $\sin\theta$. These are not the same function. The SHM equation can be solved exactly with sines and cosines. The pendulum equation cannot --- its exact solution involves elliptic integrals, which are far beyond what we need here.
And yet, a grandfather clock works. The pendulum swings with a steady period, just like a mass on a spring. How?
The answer lies in an approximation --- one of the most useful in all of physics. For small angles, $\sin\theta \approx \theta$. Replace $\sin\theta$ with $\theta$ in the pendulum equation, and you get the SHM equation. The pendulum becomes a simple harmonic oscillator --- approximately.
But "approximately" raises a question that deserves a careful answer: how small is "small enough"?
A Prediction
Before you read on: A pendulum of length 1 meter swings with an amplitude of 5 degrees. A second identical pendulum swings with an amplitude of 30 degrees. Both are released from rest.
Do they have the same period? If not, which one has a longer period, and by roughly how much?
Commit to an answer before continuing.
Many students expect both pendulums to have the same period, since the SHM result says period is independent of amplitude. Others guess that the larger swing takes longer, but are unsure by how much.
Here is the reality. The 5-degree pendulum has a period that is almost exactly $T = 2\pi\sqrt{L/g} \approx 2.006$ seconds. The 30-degree pendulum has a period about 1.7% longer --- roughly $2.04$ seconds. That may sound small, but over an hour, the two clocks would drift apart by about a minute. Over a day, the difference would be unmistakable.
The small-angle approximation works beautifully for the 5-degree pendulum. It is already starting to show cracks at 30 degrees. Understanding why requires looking at where $\sin\theta \approx \theta$ comes from and where it breaks down.
Where Does $\sin\theta \approx \theta$ Come From?
Let's start with the mathematical fact before we apply it to physics.
The Taylor series for $\sin\theta$, expanded about $\theta = 0$, is:
$$\sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots$$
where $\theta$ is measured in radians (this is essential --- the approximation does not work in degrees).
If $\theta$ is small, then $\theta^3$ is much smaller than $\theta$, and $\theta^5$ is smaller still. For example, if $\theta = 0.1$ radians (about 5.7 degrees):
| Term | Value |
|---|---|
| $\theta$ | $0.1$ |
| $\theta^3/6$ | $0.000167$ |
| $\theta^5/120$ | $0.0000000833$ |
The first correction term, $\theta^3/6$, is less than 0.2% of the leading term. The approximation $\sin\theta \approx \theta$ keeps only the first term of the Taylor series and throws away everything else. For small $\theta$, the discarded terms are negligible.
But as $\theta$ grows, the cubic term becomes significant:
| $\theta$ (degrees) | $\theta$ (radians) | $\sin\theta$ | $\theta$ (linear approx) | Relative error |
|---|---|---|---|---|
| 5 | 0.0873 | 0.0872 | 0.0873 | 0.13% |
| 15 | 0.2618 | 0.2588 | 0.2618 | 1.2% |
| 30 | 0.5236 | 0.5000 | 0.5236 | 4.7% |
| 45 | 0.7854 | 0.7071 | 0.7854 | 11% |
| 60 | 1.0472 | 0.8660 | 1.0472 | 21% |
| 90 | 1.5708 | 1.0000 | 1.5708 | 57% |
At 15 degrees, the approximation is off by about 1%. At 30 degrees, nearly 5%. By 45 degrees, the error is over 10%, and the approximation is becoming unreliable for any quantitative work.
Pause and think: Look at the table above. The error grows rapidly with angle. Why does it grow so much faster than the angle itself? (Hint: the leading correction is proportional to $\theta^3$, not $\theta$.)
The answer is that the error is dominated by the $\theta^3/6$ term. If you double $\theta$, the error roughly increases by a factor of $2^3 = 8$. That cubic growth means the approximation degrades quickly once $\theta$ gets beyond a modest range.
The Pendulum Equation
Now let's apply this to the physics. Consider a simple pendulum: a mass $m$ attached to a rigid, massless rod of length $L$, free to swing in a vertical plane. The angle $\theta$ is measured from the vertical equilibrium position.
[Video: A simple pendulum diagram. A rod of length $L$ hangs from a fixed pivot. The mass $m$ is at the bottom. The angle $\theta$ is measured from the vertical. The forces on the mass are shown: gravity $mg$ straight down, and the tension $T$ along the rod. The component of gravity tangent to the arc is $-mg\sin\theta$, which provides the restoring torque.]
The mass moves along a circular arc of radius $L$. Using Newton's second law for rotation (or equivalently, the tangential component of $F = ma$ along the arc), the equation of motion is:
$$mL\frac{d^2\theta}{dt^2} = -mg\sin\theta$$
Dividing both sides by $mL$:
$$\frac{d^2\theta}{dt^2} = -\frac{g}{L}\sin\theta$$
This is the exact pendulum equation. It is a second-order, nonlinear differential equation. The nonlinearity comes from the $\sin\theta$ --- if the right side were just proportional to $\theta$, the equation would be linear and we could solve it as SHM. But $\sin\theta$ is not a linear function of $\theta$.
Compare this to the SHM equation we know:
$$\frac{d^2\theta}{dt^2} = -\omega^2\theta$$
The pendulum equation has $\sin\theta$ where the SHM equation has $\theta$. That single difference makes the pendulum equation nonlinear and unsolvable in terms of elementary functions.
The Approximation That Makes It Solvable
Here is the key move. If $\theta$ remains small throughout the motion, then $\sin\theta \approx \theta$, and the pendulum equation becomes:
$$\frac{d^2\theta}{dt^2} \approx -\frac{g}{L}\theta$$
This is exactly the SHM equation, with $\omega^2 = g/L$. The solution is:
$$\theta(t) = \theta_0 \cos!\left(\sqrt{\frac{g}{L}}\,t + \phi\right)$$
and the period is:
$$T = 2\pi\sqrt{\frac{L}{g}}$$
This is a remarkable result. The period depends only on the length of the pendulum and the gravitational acceleration. It does not depend on the mass of the bob. It does not depend on the amplitude --- as long as the amplitude is small enough for the approximation to hold.
That last caveat is the whole point of this section. The formula $T = 2\pi\sqrt{L/g}$ is not the exact period of a pendulum. It is the period of a linearized pendulum --- a pendulum whose equation has been simplified by replacing $\sin\theta$ with $\theta$.
The move we just made --- replacing a nonlinear equation with a linear one by approximating near equilibrium --- is called linearization. You saw this idea in Section 4.4, where we discussed it as a general strategy. Here, linearization turns a genuinely hard problem (the nonlinear pendulum) into an easy one (SHM).
Seeing the Approximation Break
[Interactive: Side-by-Side Pendulum Comparison. The screen shows two pendulums of the same length, side by side. The left pendulum obeys the exact equation $d^2\theta/dt^2 = -(g/L)\sin\theta$ (solved numerically). The right pendulum obeys the linearized equation $d^2\theta/dt^2 = -(g/L)\theta$. Below them, two $\theta(t)$ graphs trace the motion in real time. An amplitude slider lets students adjust the initial angle from 1 degree to 90 degrees. At small angles, the two pendulums swing in perfect sync and the graphs overlap. As the amplitude increases, the exact pendulum starts to lag behind the linear one. A readout shows: (1) the exact period, (2) the SHM period, and (3) the percentage error. Guided prompts appear: "Set the amplitude to 10 degrees. Can you see any difference?" ... "Now try 30 degrees. When do the pendulums first visibly get out of sync?" ... "At what angle does the period error first exceed 1%? 5%?"]
This interactive reveals several important things:
At small amplitudes (below about 15 degrees), the two pendulums are virtually indistinguishable. The graphs overlap, the periods match, and the SHM approximation is excellent. A clock built on these swings would keep excellent time.
At moderate amplitudes (15 to 30 degrees), the differences become visible. The exact pendulum takes slightly longer to complete each swing. The period error climbs from about 1% at 15 degrees to about 2% at 23 degrees. The two $\theta(t)$ graphs start to drift apart after a few cycles.
At large amplitudes (above 45 degrees), the approximation clearly fails. The exact pendulum's motion is noticeably non-sinusoidal --- it spends more time near the extremes of its swing, where it moves slowly, and less time near the center. The period is significantly longer than $2\pi\sqrt{L/g}$.
Before you read on: Based on what you have seen (or can reason about), why does a large-amplitude pendulum have a longer period than the SHM formula predicts? Think about what happens to the restoring force at large angles.
The answer is physical. The restoring force on the pendulum is proportional to $\sin\theta$, not $\theta$. For large $\theta$, $\sin\theta < \theta$ --- the actual restoring force is weaker than the linearized version assumes. A weaker restoring force means the pendulum accelerates less, moves more slowly, and takes longer to complete a cycle. The SHM approximation overestimates the restoring force at large angles, which makes it underestimate the period.
The Physical Pendulum and Other Rotational Oscillators
The simple pendulum is the most familiar example, but the same logic applies to any rigid body that swings about a pivot. These are called physical pendulums (or compound pendulums).
Consider a rigid body of mass $m$ pivoting about a fixed point $O$. The distance from $O$ to the center of mass is $d$. The moment of inertia about the pivot is $I_O$. When the body is displaced by angle $\theta$ from its equilibrium (hanging straight down), the gravitational torque about the pivot is:
$$\tau = -mgd\sin\theta$$
Newton's second law for rotation gives:
$$I_O \frac{d^2\theta}{dt^2} = -mgd\sin\theta$$
For small angles, $\sin\theta \approx \theta$:
$$I_O \frac{d^2\theta}{dt^2} \approx -mgd\,\theta$$
This is the SHM equation with:
$$\omega = \sqrt{\frac{mgd}{I_O}}, \qquad T = 2\pi\sqrt{\frac{I_O}{mgd}}$$
Notice the structure. The simple pendulum is a special case where all the mass is at distance $L$ from the pivot, so $I_O = mL^2$ and $d = L$, giving $\omega = \sqrt{g/L}$ as before. For any other shape --- a swinging door, a metronome arm, a baseball bat hung from one end --- you just substitute the appropriate $I_O$ and $d$.
Pause and think: A uniform rod of length $L$ and mass $m$ is pivoted at one end and allowed to swing. What are $I_O$ and $d$ for this system? What is the period of small oscillations?
(You should find $I_O = \frac{1}{3}mL^2$ and $d = L/2$, giving $T = 2\pi\sqrt{2L/(3g)}$.)
This is a powerful pattern: any rotational system with a gravitational restoring torque becomes a simple harmonic oscillator in the small-angle limit. The details of the body's shape and mass distribution are all encoded in $I_O$ and $d$. The small-angle approximation does the same work here as it did for the simple pendulum --- it replaces $\sin\theta$ with $\theta$ and turns a nonlinear equation into a linear one.
Torsional Oscillations: Springs Without Gravity
Not all rotational oscillators rely on gravity. A torsional oscillator consists of an object attached to a wire or spring that exerts a restoring torque proportional to the angular displacement:
$$\tau = -\kappa\theta$$
where $\kappa$ (the Greek letter kappa) is the torsional spring constant, measured in N$\cdot$m/rad.
[Video: A disk suspended from a thin wire. The disk is rotated slightly and released. It twists back and forth about the vertical axis. A slow-motion replay shows the smooth oscillation. On screen, the restoring torque $\tau = -\kappa\theta$ is displayed, and the analogy to $F = -kx$ for a mass on a spring is drawn explicitly.]
Newton's second law for rotation gives:
$$I\frac{d^2\theta}{dt^2} = -\kappa\theta$$
This is already the SHM equation --- no small-angle approximation needed. The restoring torque is exactly proportional to $\theta$, not to $\sin\theta$. The angular frequency is:
$$\omega = \sqrt{\frac{\kappa}{I}}, \qquad T = 2\pi\sqrt{\frac{I}{\kappa}}$$
Compare this to the linear spring: $\omega = \sqrt{k/m}$. The correspondence is exact: $\kappa$ plays the role of $k$, and $I$ plays the role of $m$. Torsional oscillators show up in many precision instruments --- the balance wheel of a mechanical watch, the torsion balance used by Cavendish to measure the gravitational constant, and the galvanometer that first detected tiny electric currents.
The key distinction: a torsional oscillator with $\tau = -\kappa\theta$ is exactly SHM, while a pendulum with $\tau = -mgd\sin\theta$ is only approximately SHM (for small angles). The small-angle approximation is needed when the restoring torque involves $\sin\theta$ rather than $\theta$.
The Logic of Approximation
Let's step back and look at what we have done, because the reasoning pattern here is one of the most important in physics.
-
Start with the exact equation. For the pendulum: $\frac{d^2\theta}{dt^2} = -\frac{g}{L}\sin\theta$.
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Identify the regime where the equation simplifies. When $\theta$ is small, $\sin\theta \approx \theta$.
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Make the substitution. Replace $\sin\theta$ with $\theta$ to get: $\frac{d^2\theta}{dt^2} = -\frac{g}{L}\theta$.
-
Solve the simplified equation. This is SHM, and we know the solution.
-
Assess the validity. The solution is accurate as long as $\theta$ remains small enough that $\sin\theta \approx \theta$ holds to the precision you need.
This is exactly the linearization strategy from Section 4.4, applied to a specific physical system. In Section 13.5, you will see this generalized even further: any stable equilibrium point, for any potential energy function, produces SHM when the system is displaced slightly. The small-angle pendulum is one instance of a universal pattern.
But this section emphasizes something that the general theory can obscure: you should always know how good your approximation is. The formula $T = 2\pi\sqrt{L/g}$ is not universally true for pendulums. It is the answer to a simpler, approximate problem. Knowing when it is accurate and when it is not is part of understanding the physics.
How Good Is "Good Enough"?
The exact period of a simple pendulum, swinging with amplitude $\theta_0$, can be written as a series:
$$T_{\text{exact}} = 2\pi\sqrt{\frac{L}{g}}\left(1 + \frac{1}{4}\sin^2!\frac{\theta_0}{2} + \frac{9}{64}\sin^4!\frac{\theta_0}{2} + \cdots\right)$$
The leading correction term is $\frac{1}{4}\sin^2(\theta_0/2) \approx \frac{\theta_0^2}{16}$ for small $\theta_0$. This tells us the fractional error in the SHM period scales as $\theta_0^2$ --- if you double the amplitude, the error roughly quadruples.
| Amplitude $\theta_0$ | Fractional period error |
|---|---|
| 5 degrees | 0.05% |
| 10 degrees | 0.19% |
| 15 degrees | 0.43% |
| 20 degrees | 0.76% |
| 30 degrees | 1.7% |
| 45 degrees | 3.8% |
| 60 degrees | 7.3% |
| 90 degrees | 18% |
For most laboratory and engineering purposes, errors below 1% are acceptable, which means the SHM approximation is safe for amplitudes up to roughly 20--25 degrees. For a grandfather clock, whose pendulum swings only a few degrees, the approximation is superb.
Practice
Layer 1: Concrete
A simple pendulum has a length of $L = 0.80$ m. Using the small-angle approximation, find the period of oscillation. Take $g = 9.80$ m/s$^2$.
Check your answer
Using the SHM period formula: $$T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{0.80}{9.80}}$$ $$T = 2\pi\sqrt{0.08163} = 2\pi \times 0.2857 = 1.80 \text{ s}$$ The period is approximately $1.80$ seconds. This result is valid as long as the pendulum swings through a small angle --- say, less than about 20 degrees. Note what the formula does *not* depend on: the mass of the bob and (within the small-angle regime) the amplitude. A heavier bob on the same string has the same period. A wider swing (as long as it stays small) has the same period.Layer 2: Pattern
For each system below, determine the small-angle frequency $\omega$ and estimate the maximum amplitude (in degrees) for which the SHM approximation gives a period accurate to within 1%.
(a) A simple pendulum of length $L = 2.0$ m on Earth ($g = 9.80$ m/s$^2$).
(b) The same pendulum on the Moon ($g = 1.62$ m/s$^2$).
(c) A uniform thin rod of length $L = 1.0$ m pivoted at one end on Earth.
Check your answer
**(a)** Simple pendulum on Earth: $$\omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.80}{2.0}} = \sqrt{4.9} = 2.21 \text{ rad/s}$$ The period is $T = 2\pi/\omega = 2.84$ s. **(b)** Same pendulum on the Moon: $$\omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{1.62}{2.0}} = \sqrt{0.81} = 0.90 \text{ rad/s}$$ The period is $T = 2\pi/\omega = 6.98$ s. The pendulum swings much more slowly on the Moon because gravity is weaker. **(c)** Uniform rod pivoted at one end. Here $I_O = \frac{1}{3}mL^2$ and $d = L/2$: $$\omega = \sqrt{\frac{mgd}{I_O}} = \sqrt{\frac{mg(L/2)}{\frac{1}{3}mL^2}} = \sqrt{\frac{3g}{2L}} = \sqrt{\frac{3 \times 9.80}{2 \times 1.0}} = \sqrt{14.7} = 3.83 \text{ rad/s}$$ The period is $T = 2\pi/\omega = 1.64$ s. **Valid range for all three:** The 1% period accuracy threshold depends on $\theta_0$, not on $L$, $g$, or the shape of the object. From the error table, the period error is about 1.7% at 30 degrees and about 0.76% at 20 degrees. So for 1% accuracy, the amplitude should be kept below roughly **23 degrees** (about 0.4 radians). This is the same for all three systems because the quality of the $\sin\theta \approx \theta$ approximation depends only on $\theta_0$.Layer 3: Structure
Why does $\sin\theta \approx \theta$ only work for small $\theta$? What is the next correction term, and what does it tell you about the direction of the error?
Check your answer
The Taylor series is $\sin\theta = \theta - \frac{\theta^3}{6} + \cdots$. The approximation $\sin\theta \approx \theta$ works when $\frac{\theta^3}{6}$ is negligible compared to $\theta$ --- that is, when $\frac{\theta^2}{6} \ll 1$, or $\theta \ll \sqrt{6} \approx 2.4$ radians. In practice, the approximation is useful for $\theta$ up to about 0.3--0.4 radians (15--23 degrees), depending on how much error you can tolerate. The next correction term is $-\frac{\theta^3}{6}$, which is **negative**. This means $\sin\theta < \theta$ for all $\theta > 0$ (within the first half-period of sine). Physically, this tells us that the actual restoring torque $-mg\sin\theta$ is *weaker* than the linearized version $-mg\theta$ assumes. The approximation overestimates the restoring force. Since the restoring force is weaker than the linearized model predicts, the actual pendulum accelerates less and takes *longer* to complete a swing. The exact period is always *longer* than $2\pi\sqrt{L/g}$ --- never shorter. The SHM formula gives a lower bound on the true period. This is not a coincidence. The sign of the correction term tells you the direction of the error, which is a powerful consistency check.Layer 4: Debug
A student is asked to find the period of a simple pendulum of length 1.5 m swinging with an amplitude of 60 degrees. The student writes:
$$T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{1.5}{9.80}} = 2.46 \text{ s}$$
and reports the answer as 2.46 s. How far off is this from the actual period, and what went wrong in the student's reasoning?
Check your answer
The student used the SHM formula $T = 2\pi\sqrt{L/g}$, which is only valid for small oscillations. At 60 degrees, this formula significantly underestimates the period. From the exact correction series, the fractional period error at 60 degrees is about 7.3%. So the actual period is approximately: $$T_{\text{exact}} \approx 2.46 \times 1.073 = 2.64 \text{ s}$$ The student's answer is too short by about 0.18 seconds per cycle, which is a 7% error. What went wrong: the student applied the small-angle formula without checking whether the angle is actually small. At 60 degrees ($\theta_0 \approx 1.05$ radians), $\sin\theta$ differs from $\theta$ by over 17%. The approximation $\sin\theta \approx \theta$ has completely broken down at this angle. The fix: the student should either (1) note that the SHM formula is not valid for this amplitude and use the correction series, or (2) solve the exact equation numerically. At minimum, the student should state that the answer is an approximation that is only accurate for small swings, and 60 degrees does not qualify as "small." This is a good example of a broader lesson: **using a formula without checking its assumptions can give a confident wrong answer.** The formula itself is correct. The error is in applying it outside its domain of validity.Reflection
Think about the role of approximation in what you have learned.
When is an approximate answer "good enough"? The SHM formula for the pendulum period is never exactly right (unless the amplitude is truly zero). But for small swings, it is very close. At what point does "very close" stop being acceptable?
There is no single answer. It depends on what you need the result for. A grandfather clock needs its period accurate to fractions of a percent --- small amplitudes only. A child on a playground swing, pushed to 30 or 40 degrees, does not care about the exact period.
Consider: when you use any physics formula, you are almost certainly using an approximation of some kind. Constant gravity is an approximation. Ignoring air resistance is an approximation. Treating objects as particles is an approximation. The skill is not in avoiding approximations --- it is in knowing which approximations you are making and whether they are justified for the question you are trying to answer.
Looking Ahead
In this section, you saw a specific case of a powerful strategy: take a nonlinear equation, approximate it near equilibrium, and get SHM. The pendulum equation became solvable because $\sin\theta \approx \theta$ for small angles.
But the pendulum is just one system. In the next section, you will see that this same strategy works for any potential energy function near a stable equilibrium. The potential energy curve, whatever its shape, looks like a parabola when you zoom in close enough to a minimum. A parabolic potential means a linear restoring force, which means SHM. The pendulum is one example of a universal principle: every stable equilibrium is a simple harmonic oscillator in disguise.
The question will shift from "does this particular system oscillate?" to "what is the effective spring constant for any system near equilibrium?" That is the power of linearization --- and it is the subject of Section 13.5.