The Simple Pendulum

Legend has it that in 1583 a nineteen-year-old medical student named Galileo Galilei sat in the cathedral of Pisa, watching a bronze lamp swing from the ceiling on its long chain. The drafts pushed it through wide arcs and small ones — yet, timing the swings against his own pulse, he noticed something that should not have been true: every swing seemed to take the same time, big or small. That observation, the isochronism of the pendulum, became the beating heart of every precision clock for the next three centuries, and it is the first deep fact you meet in the physics of oscillation. The system itself could hardly be simpler: a compact mass on a string of length $L$ , swinging under gravity. But it hides three surprises. The period $T = 2\pi\sqrt{L/g}$ contains no mass — a lead bob and a wooden one keep identical time. It contains no amplitude — at least for small swings, which is Galileo's isochronism. And the exact equation of motion, $\ddot\theta + (g/L)\sin\theta = 0$ , is genuinely nonlinear: one of the simplest mechanical systems whose exact solution demands elliptic integrals rather than sines and cosines. You already have every tool this article needs: the simple harmonic oscillator, the idea of linearizing about a stable equilibrium, and energy conservation. The pendulum is where those tools earn their keep — and where you first watch a linear approximation fail, measurably and controllably, as the amplitude grows. Master it and you own the template for every "small oscillations" problem in physics, plus the classic instrument that measured the Earth's gravity, ran the world's clocks, and, in Foucault's hands, proved the planet turns.

Key takeaways

A simple pendulum is an idealized point mass $m$ on a massless, inextensible string of length $L$ swinging under gravity $g$ .
The exact equation of motion, $\ddot\theta + (g/L)\sin\theta = 0$ , is nonlinear; its exact period involves the complete elliptic integral $K$ .
For small angles $\sin\theta \approx \theta$ and the motion is simple harmonic: $\omega_0 = \sqrt{g/L}$ , $T = 2\pi\sqrt{L/g}$ .
The period contains neither the mass (gravity and inertia both scale with $m$ ) nor, to leading order, the amplitude — the isochronism that made pendulum clocks possible.
Energy conservation, $E = \tfrac12 mL^2\dot\theta^2 - mgL\cos\theta$ , is exact at any amplitude and organizes the phase portrait: libration inside the separatrix, full rotation outside.
At larger amplitudes the period grows, $T \approx T_0(1 + \theta_0^2/16 + \cdots)$ — about $+1\%$ at $23°$ and $+18\%$ at $90°$ .
Key idea: linearizing the gravitational restoring torque about the stable equilibrium turns the pendulum into a simple harmonic oscillator with $\omega_0 = \sqrt{g/L}$ — a frequency set purely by geometry and gravity, independent of amplitude and mass.

Where this lives today

The pendulum's day job — timekeeping — ended in the 1930s, when quartz oscillators overtook even the heroic Shortt free-pendulum clocks that had kept world time to a few seconds per year. But its second job, measuring gravity, never really stopped; it just changed hardware. The logic of Worked Example 1 — extract $g$ from a measured length and a measured time — now runs inside free-fall absolute gravimeters and atom interferometers, which drop corner-cube mirrors or clouds of cold atoms instead of swinging bobs and read $g$ to parts in $10^9$ . Surveys with such instruments map aquifer depletion, magma movement under volcanoes, and the mass loss of ice sheets: the same physics Kater's reversible pendulum served in 1817, at a billion times the precision. And the original theatre piece still runs daily. Foucault pendulums swing in the Panthéon and in hundreds of science museums, their planes turning at $\omega_\oplus \sin\lambda$ — the simple pendulum promoted into a planetary instrument (the full story lives in the Coriolis force article). Meanwhile, every seismometer and every skyscraper tuned-mass damper is a pendulum wearing work clothes: the 660-tonne sphere hanging inside Taipei 101 is a pendulum with $T \approx 7\ \mathrm{s}$ , deliberately tuned near the tower's sway period to steal energy from typhoon gusts. Four centuries on, the physics of a mass on a string is still holding buildings up and clocks — of the gravitational kind — to account.

The physical picture

Displace the bob to an angle $\theta$ and let go. Gravity pulls straight down with force $mg$ , but the string forbids any motion along its own length — the bob is trapped on a circular arc of radius $L$ . So resolve gravity along the two directions that matter: a component $mg\cos\theta$ along the string, absorbed by the tension, and a component $mg\sin\theta$ along the arc, pointing back toward the bottom. That tangential piece is the engine of the whole problem: a restoring force that grows as you stray farther, almost exactly like a stretched spring near the bottom of the swing.

Figure

The pendulum's anatomy — gravity resolved along the string and along the arc

Now the two famous cancellations. Why no mass? The restoring torque is proportional to $m$ (gravity pulls harder on a heavier bob), but so is the rotational inertia $mL^2$ (a heavier bob is harder to accelerate). Form Newton's law and the $m$ drops out — the pendulum is a rotational free-fall experiment, the same reason a hammer and a feather fall together. Why no amplitude? Double the swing and you double the restoring force, but you also double the distance to cover each cycle. Near the bottom these effects compensate exactly, so the round trip takes the same time. Isochronism is not magic; it is the signature of a force that is linear in displacement. That signature fails at large angles, because the true restoring force goes as $\sin\theta$ , which lags behind $\theta$ — at $90°$ the force is only $\sin(90°)/(\pi/2) \approx 64\%$ of what a linear spring would supply. A weaker-than-linear force means slower round trips: the period grows with amplitude. Here is the whole story in one curve:

Figure

Isochronism and its breakdown — the true period versus amplitude

Read the flat left end of that curve and you understand three hundred years of horology: below about $20°$ the period is amplitude-independent to better than $1\%$ , so a clock whose swing slowly decays barely changes rate. Read the right end and you see the nonlinearity taking over, all the way to the divergence at $\theta_0 = 180°$ — a pendulum released just short of the inverted position hovers near the top for an arbitrarily long time before falling. That whole story is quantified in the anharmonic pendulum article; here we will derive its leading term. Energetically, the swing is a metronome of exchange: potential energy is highest at the turning points, kinetic energy peaks at the bottom, and the total stays rigorously constant. Plotted in the $(\theta, \dot\theta)$ plane this exchange traces closed loops — and, at higher energy, curves that never close, when the pendulum stops oscillating and starts rotating over the top. We will draw that portrait after the equations are on the table.

From gravity to the equation of motion

Everything begins with one exact statement. Take torques about the pivot: tension passes through it (no torque), gravity supplies $-mgL\sin\theta$ (the minus sign says it always opposes the displacement), and the moment of inertia of a point bob is $mL^2$ . Newton's rotational law $I\ddot\theta = \tau$ then reads, after the mass cancels,

\ddot\theta + \frac{g}{L}\sin\theta = 0 \qquad (1)

where: $\theta$ — angular displacement from the downward vertical; $g$ — gravitational acceleration ( $9.81\ \mathrm{m/s^2}$ at Earth's surface); $L$ — string length. This equation is exact for the idealized pendulum at any amplitude, including over-the-top rotation. Notice what is absent: no $m$ anywhere, and the nonlinearity confined to the single $\sin\theta$ . All the richness of the pendulum — isochronism, its breakdown, the separatrix — lives in the difference between $\sin\theta$ and $\theta$ .

Derivation AEquation of motion from the torque about the pivot

We are deriving the exact pendulum equation (1) from Newton's second law for rotation, $I\ddot\theta = \tau$ , taken about the pivot point. The strategy is to choose the pivot deliberately: the string tension — an unknown, position-dependent constraint force — passes straight through it, so its torque vanishes and it never enters the problem. That single choice reduces a two-dimensional constrained-force problem to one clean scalar equation. The things to watch are the sign of the gravitational torque, which must be restoring, and the mass cancellation at the end, which is the equivalence of gravitational and inertial mass showing up in miniature.

Step 1. Compute the torque of gravity about the pivot. The bob sits at distance $L$ from the pivot, and only the component of gravity perpendicular to the string — the tangential component $mg\sin\theta$ — produces torque:

\tau = -mgL\sin\theta

Why the minus sign: for $\theta > 0$ the torque acts to decrease $\theta$ , and vice versa — gravity always steers the bob back toward the vertical. A sign error here produces exponential runaway instead of oscillation, an instant red flag.

Step 2. The moment of inertia of a point mass at distance $L$ from the axis is

I = mL^2

Why this is the whole inertia: the string is massless by assumption. A real rod or an extended bob adds its own $I$ — that generalization is precisely the physical pendulum.

Step 3. Assemble Newton's rotational law and cancel:

mL^2\,\ddot\theta = -mgL\sin\theta \quad\Longrightarrow\quad \ddot\theta = -\frac{g}{L}\sin\theta

Why the mass disappears: gravitational mass drives the torque, inertial mass resists the acceleration, and they are the same $m$ . The pendulum is rotational free fall — Newton himself swung pendulums with bobs of gold, glass, and wood to test exactly this cancellation. *Dimension check: $g/L$ has units $\mathrm{m\,s^{-2}/m} = \mathrm{s^{-2}}$ , exactly right for an angular acceleration per radian — and the same result emerges from the Lagrangian route below.*

Now the payoff move — the one you will repeat in every corner of physics. Near the bottom of the swing, Taylor-expand: $\sin\theta = \theta - \theta^3/6 + \cdots$ . Keeping only the linear term turns (1) into the simple harmonic oscillator equation,

\ddot\theta + \omega_0^2\,\theta = 0, \qquad \omega_0 = \sqrt{\frac{g}{L}} \qquad (2)

where: $\omega_0$ — the natural angular frequency of small oscillations. The error of the approximation is the discarded $\theta^3/6$ term: at $\theta = 0.26\ \mathrm{rad}$ ( $15°$ ) it is about $1\%$ of the linear term, which is why "small" in practice means "under twenty degrees or so." Everything you know about simple harmonic motion now applies verbatim. The general solution is

\theta(t) = \theta_0\cos(\omega_0 t + \varphi) \qquad (3)

where: $\theta_0$ — the angular amplitude; $\varphi$ — the phase constant, both set by how you release the bob. Releasing from rest at angle $\theta_0$ gives $\varphi = 0$ ; a sharp tap at the bottom gives $\varphi = -\pi/2$ . And the period — the number Galileo timed with his pulse — is

T = \frac{2\pi}{\omega_0} = 2\pi\sqrt{\frac{L}{g}} \qquad (4)

where: $T$ — the time for one complete back-and-forth. Only $L$ and $g$ enter. Doubling the length lengthens the period by $\sqrt{2}$ ; taking the same pendulum to the Moon ( $g \to g/6$ ) slows it by $\sqrt{6} \approx 2.45$ . A one-metre pendulum on Earth swings with $T = 2\pi\sqrt{1/9.81} = 2.006\ \mathrm{s}$ — which is why the seconds pendulum, whose every one-way beat is exactly one second, has the memorable length

L_{\mathrm{sec}} = \frac{g}{4\pi^2}\,T^2 = \frac{9.81 \times 4}{4\pi^2} \approx 0.994\ \mathrm{m} \qquad (5)

where: $T = 2\ \mathrm{s}$ — one second per half-swing. That number is no accident of history: the metre itself was nearly defined as the seconds-pendulum length in the 1790s, losing out to the Earth-quadrant definition only because $g$ varies from place to place.

Which equation answers which question

you want	use	notes
exact dynamics, any amplitude	$\ddot\theta + (g/L)\sin\theta = 0$ , eq. (1)	nonlinear; elliptic solutions
small-swing motion	$\theta(t) = \theta_0\cos(\omega_0 t + \varphi)$ , eq. (3)	valid to $\sim 1\%$ below $\sim 23°$
the clock number	$T = 2\pi\sqrt{L/g}$ , eq. (4)	no mass, no amplitude
speed or tension mid-swing	energy, eq. (6)	exact at any amplitude
period at real amplitudes	$T(\theta_0)$ series, eq. (7)	first correction $\theta_0^2/16$
measuring $g$	$g = 4\pi^2 L/T^2$	time many swings, keep $\theta_0$ small

The same equation of motion drops out of the Lagrangian machinery in four lines — worth seeing side by side with the torque route, because for the pendulum's descendants (double pendulums, pendulums on carts, coupled oscillators) the torque bookkeeping becomes hopeless while the Lagrangian route scales effortlessly.

Go deeper · Derivation B · 1 minEquation of motion from the Lagrangian

Derivation BEquation of motion from the Lagrangian

We are re-deriving the pendulum equation (1) from the Euler-Lagrange equation, both to confirm the Newtonian result and to display the generalized-coordinate structure that makes this method the default for constrained systems. The strategy: pick the single angle $\theta$ as the generalized coordinate — the string constraint is then satisfied automatically, and the tension never appears at all. Build $\mathcal{L} = T - V$ , turn the crank, and compare. Watch the sign convention in the potential energy: we measure height from the pivot, so $V$ is negative below it, and only differences in $V$ matter.

Step 1. Kinetic energy. The bob moves on a circle of radius $L$ , so its speed is $v = L\dot\theta$ :

T = \tfrac12 m v^2 = \tfrac12 mL^2\dot\theta^2

Step 2. Potential energy. The bob hangs a depth $L\cos\theta$ below the pivot; taking the pivot as the zero of height,

V = -mgL\cos\theta

Why this is the right shape: $V$ is minimized at $\theta = 0$ (stable equilibrium, curvature $V''(0) = mgL > 0$ ) and maximized at $\theta = \pi$ (the inverted, unstable point). The entire phase portrait is encoded in this one cosine.

Step 3. The Lagrangian is the difference:

\mathcal{L} = T - V = \tfrac12 mL^2\dot\theta^2 + mgL\cos\theta

Step 4. Write the Euler-Lagrange equation for the single coordinate $\theta$ :

\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial\mathcal{L}}{\partial\dot\theta} - \frac{\partial\mathcal{L}}{\partial\theta} = 0

Step 5. Compute the two derivatives: $\partial\mathcal{L}/\partial\dot\theta = mL^2\dot\theta$ , whose time derivative is $mL^2\ddot\theta$ , and $\partial\mathcal{L}/\partial\theta = -mgL\sin\theta$ .

Step 6. Assemble and divide by $mL^2$ :

mL^2\ddot\theta + mgL\sin\theta = 0 \quad\Longrightarrow\quad \ddot\theta + \frac{g}{L}\sin\theta = 0

Why the tension never appeared: the coordinate $\theta$ moves only along the constraint surface, and constraint forces do no work along allowed displacements. That silence is the Lagrangian method's superpower — for a double pendulum the Newtonian route needs four constraint equations; the Lagrangian route needs none.

Energy and the phase portrait

Multiply equation (1) by $mL^2\dot\theta$ and integrate once, or simply add $T + V$ from the Lagrangian derivation — either way, the pendulum carries an exact first integral, conserved at every amplitude:

E = \tfrac12 mL^2\dot\theta^2 - mgL\cos\theta \qquad (6)

where: $E$ — the total mechanical energy; $\tfrac12 mL^2\dot\theta^2$ — kinetic energy (since $v = L\dot\theta$ ); $-mgL\cos\theta$ — potential energy, zero reference at pivot height. Given a release condition, (6) hands you the speed at any angle without solving any differential equation: release from rest at $\theta_0$ and the bottom-of-swing speed is $v = \sqrt{2gL(1 - \cos\theta_0)}$ . It also hands you the string tension, $T_{\mathrm{str}} = mg(3\cos\theta - 2\cos\theta_0)$ , which at the bottom of a $60°$ release reaches $2mg$ — a fact that has snapped more than one cheap string. Solving (6) for the angular velocity gives the trajectory traced in the $(\theta, \dot\theta)$ plane,

\dot\theta = \pm\sqrt{\frac{2}{mL^2}\left(E + mgL\cos\theta\right)} \qquad (7)

where: the $\pm$ selects the forward or backward half of the swing. One picture organizes every possible motion:

Figure

The pendulum phase portrait — libration inside the separatrix, rotation outside

Three energy regimes, one diagram. For $E < mgL$ the trajectories are closed loops around the origin: the pendulum librates, swinging back and forth between turning points where the loop crosses the $\theta$ -axis. For $E > mgL$ the curves never touch $\dot\theta = 0$ : the pendulum rotates, circulating over the top forever, faster at the bottom than at the top. And the boundary case $E = mgL$ is the separatrix, the trajectory of a pendulum with exactly enough energy to reach the inverted point — which it approaches asymptotically, never quite arriving. Small loops near the origin are nearly elliptical (that is simple harmonic motion); loops near the separatrix are visibly flattened, spending long times near the turning points — the geometric face of the growing period.

Beyond small angles: the exact period

How wrong is $2\pi\sqrt{L/g}$ at real amplitudes? Energy conservation answers exactly. The result of the derivation below is worth displaying first:

T(\theta_0) = 2\pi\sqrt{\frac{L}{g}}\left(1 + \frac{\theta_0^2}{16} + \frac{11\,\theta_0^4}{3072} + \cdots\right) \qquad (8)

where: $\theta_0$ — the amplitude in radians. The leading correction $\theta_0^2/16$ is the number to memorize: it says the error hits $1\%$ at $\theta_0 = 0.4\ \mathrm{rad} \approx 23°$ , and about $0.43\%$ at $15°$ . A clock regulated at small amplitude but swung at $60°$ runs slow by over $7\%$ — an hour and three quarters per day.

Go deeper · Derivation C · 2 minThe exact period from energy conservation

Derivation CThe exact period from energy conservation

We are deriving the exact period of the pendulum at arbitrary amplitude $\theta_0$ , ending in the complete elliptic integral $K$ and the practical series (8). The strategy is the standard one for any one-dimensional conservative system: use energy conservation to solve for $\dot\theta$ as a function of $\theta$ , separate variables, and integrate time over one quarter of the cycle — from the bottom of the swing to the turning point — then multiply by four, since the four quarter-swings are congruent. Watch two things: the turning-point condition that fixes $E$ in terms of $\theta_0$ , and the change of variables that tames the integrable square-root singularity at the turning point.

Step 1. Fix the energy by the turning point. At maximum swing $\theta = \theta_0$ the bob is momentarily at rest, $\dot\theta = 0$ , so equation (6) gives

E = -mgL\cos\theta_0

Step 2. Solve the energy equation for the angular speed at a general angle $\theta$ :

\tfrac12 mL^2\dot\theta^2 = mgL(\cos\theta - \cos\theta_0) \quad\Longrightarrow\quad \dot\theta = \sqrt{\frac{2g}{L}(\cos\theta - \cos\theta_0)}

Why the argument is positive: during libration $|\theta| < \theta_0$ , and the cosine is decreasing on $[0, \pi]$ , so $\cos\theta > \cos\theta_0$ everywhere inside the swing.

Step 3. Separate variables and integrate over a quarter period, from $\theta = 0$ to $\theta = \theta_0$ :

\frac{T}{4} = \sqrt{\frac{L}{2g}}\int_0^{\theta_0}\frac{\mathrm{d}\theta}{\sqrt{\cos\theta - \cos\theta_0}}

Why a quarter: bottom-to-turning-point is one fourth of the full cycle by the symmetry of the potential; the integrand's divergence at $\theta = \theta_0$ is an integrable square-root singularity — the bob lingers at the turning point but not infinitely long.

Step 4. Convert to standard form with the substitution $\sin(\theta/2) = k\sin\phi$ , where $k = \sin(\theta_0/2)$ . The half-angle identity $\cos\theta - \cos\theta_0 = 2(k^2 - \sin^2(\theta/2))$ and a few lines of algebra give

T = 4\sqrt{\frac{L}{g}}\,K(k), \qquad K(k) = \int_0^{\pi/2}\frac{\mathrm{d}\phi}{\sqrt{1 - k^2\sin^2\phi}}

Why this substitution: it maps the swing $[0, \theta_0]$ onto the fixed interval $[0, \pi/2]$ and absorbs the turning-point singularity, leaving the complete elliptic integral of the first kind — the pendulum is the reason this special function was invented.

Step 5. Expand for small amplitude. With $K(k) = \tfrac{\pi}{2}\left(1 + \tfrac14 k^2 + \tfrac{9}{64}k^4 + \cdots\right)$ and $k = \sin(\theta_0/2) = \theta_0/2 - \theta_0^3/48 + \cdots$ , collect powers of $\theta_0$ :

T = 2\pi\sqrt{\frac{L}{g}}\left(1 + \frac{\theta_0^2}{16} + \frac{11\,\theta_0^4}{3072} + \cdots\right)

Why the limit checks out: as $\theta_0 \to 0$ , $K \to \pi/2$ and $T \to 2\pi\sqrt{L/g}$ — amplitude has dropped out exactly, recovering isochronism as the leading term of a controlled expansion rather than an article of faith. *Numerical check: at $\theta_0 = 60°$ the exact elliptic result gives $T/T_0 = 1.0732$ while the two-term series gives $1.0728$ — agreement to four figures, confirming that the series is already excellent well before $90°$ .*

Limiting cases

Case	Result	Meaning
small amplitude ( $\theta_0 \lesssim 20°$ )	$T \to 2\pi\sqrt{L/g}$	linear SHM regime; period independent of amplitude and mass
moderate amplitude	$T \approx T_0\left(1 + \theta_0^2/16\right)$	first anharmonic correction; $+1\%$ at $23°$ , $+18\%$ at $90°$
$\theta_0 \to \pi$	$T \to \infty$ (logarithmically)	the bob hovers near the inverted point; the separatrix in phase space
$E > mgL$	no turning points	rotation, not oscillation: the pendulum circulates over the top
$g \to 0$ (free fall / orbit)	$T \to \infty$	a pendulum in a freely falling lab does not swing — no restoring torque
$L \to \infty$ at fixed swing arc	$T \propto \sqrt{L}$	Foucault's 67 m Panthéon pendulum: stately 16.4 s period

Common trapThe period is independent of amplitude — only in the small-angle fiction

Isochronism is the leading term of an approximation, not a law. The true period grows with amplitude: $+0.4\%$ at $15°$ , $+1\%$ at $23°$ , $+7.3\%$ at $60°$ , $+18\%$ at $90°$ , diverging as $\theta_0 \to 180°$ . Two sibling traps: (i) the mass-independence follows from the equality of gravitational and inertial mass — the pendulum tests the equivalence principle rather than deriving it, which is exactly why Newton swung bobs of gold, glass, and wood looking for a difference; (ii) $T = 2\pi\sqrt{L/g}$ uses $L$ measured from the pivot to the center of mass of the bob, and assumes the string is massless and the bob pointlike — a fat bob or heavy cable shifts the period, which is the subject of the physical pendulum.

Worked examples

1 · Design a seconds pendulum, then take it to the Moon (confidence builder)

What length must a pendulum have so that each one-way beat takes exactly $1\ \mathrm{s}$ (period $T = 2\ \mathrm{s}$ ) at $g = 9.81\ \mathrm{m/s^2}$ ? A student then times $50$ full swings of a $1.000\ \mathrm{m}$ pendulum at small amplitude and gets $100.3\ \mathrm{s}$ — what $g$ does that measure? Finally, what is the seconds pendulum's period on the Moon ( $g_{\mathrm{Moon}} = 1.62\ \mathrm{m/s^2}$ )?

Step 1. Invert equation (4) for the length: $L = gT^2/4\pi^2 = 9.81 \times 4 / 39.48 = \boxed{0.994\ \mathrm{m}}$ . This is why grandfather clocks are roughly a metre tall inside: their pendulums beat seconds.

Step 2. The timing measurement: $T = 100.3/50 = 2.006\ \mathrm{s}$ . Then $g = 4\pi^2 L/T^2 = 39.48 \times 1.000/(2.006)^2 = \boxed{9.81\ \mathrm{m/s^2}}$ . Timing $50$ swings instead of one divides the stopwatch error per period by $50$ — the trick that makes a $0.1\%$ measurement of $g$ routine in a first-year lab.

Step 3. On the Moon, $T \propto 1/\sqrt{g}$ , so $T_{\mathrm{Moon}} = 2\ \mathrm{s} \times \sqrt{9.81/1.62} = 2\sqrt{6.06} = \boxed{4.92\ \mathrm{s}}$ . The same hardware, retimed by the planet you stand on: a pendulum is a gravimeter with a dial made of time.

2 · A 60° release: exact period, speed, and the string's worst moment (exam level)

A bob of mass $m = 0.30\ \mathrm{kg}$ on a string of length $L = 0.75\ \mathrm{m}$ is released from rest at $\theta_0 = 60°$ . Find (a) the small-angle period and the true period, (b) the speed at the bottom, (c) the string tension at the bottom, and (d) how much time a clock using this pendulum would lose per day if its gearing assumed the small-angle period.

Step 1. Small-angle period: $T_0 = 2\pi\sqrt{0.75/9.81} = 2\pi\sqrt{0.07645} = \boxed{1.737\ \mathrm{s}}$ .

Step 2. True period from the series (8) with $\theta_0 = \pi/3 = 1.047\ \mathrm{rad}$ : the correction factor is $1 + (1.047)^2/16 + 11(1.047)^4/3072 = 1 + 0.0685 + 0.0043 = 1.0728$ , so $T = 1.737 \times 1.0728 = \boxed{1.864\ \mathrm{s}}$ . (The exact elliptic answer is $1.8644\ \mathrm{s}$ — the two-term series is already good to four figures.)

Step 3. Speed at the bottom from energy conservation, eq. (6). The bob falls a height $h = L(1 - \cos 60°) = 0.75 \times 0.5 = 0.375\ \mathrm{m}$ : $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.375} = \sqrt{7.36} = \boxed{2.71\ \mathrm{m/s}}$ .

Step 4. Tension at the bottom. The bob moves on a circle, so the net upward force supplies the centripetal acceleration: $T_{\mathrm{str}} - mg = mv^2/L$ . Then $T_{\mathrm{str}} = m(g + v^2/L) = 0.30 \times (9.81 + 7.36/0.75) = 0.30 \times 19.62 = \boxed{5.89\ \mathrm{N}}$ — exactly $2mg$ , the general result $mg(3 - 2\cos\theta_0)$ at $\theta_0 = 60°$ . The string's worst moment is always the bottom of the swing, where it must both hold the weight and bend the trajectory.

Step 5. Clock error. Each true swing takes a factor $1.0728$ longer than the gearing assumes, so the clock loses a fraction $1 - 1/1.0728 = 0.0679$ of each day: $0.0679 \times 86400 \approx \boxed{5.9 \times 10^3\ \mathrm{s} \approx 98\ \mathrm{min}}$ per day. This is why every real pendulum clock keeps its amplitude small — and constant.

Origins · 1 minGalileo Galilei and how this was discovered

Galileo Galilei

1564–1642 · Pisa & Padua

Galileo met the pendulum as a teenage medical student — reputedly timing the swinging lamp of Pisa cathedral against his pulse, the only stopwatch he owned — and returned to it his whole life. He established (by experiment, his signature move) that the period depends on the length and not on the mass, and claimed it independent of amplitude; the claim is exactly true only for small swings, but it was true enough to matter. He grasped the two applications that would dominate the next two centuries: the pendulum as a timekeeper — he sketched a pendulum-clock escapement in his blind old age, dictating the design to his son — and as a measuring instrument for gravity itself. In the Discorsi (1638) the pendulum appears as evidence in his grand argument that all bodies fall alike: a lead bob and a cork bob on equal strings keep the same beat. The pendulum clock itself arrived fourteen years after his death, built by Christiaan Huygens — who also worked out what Galileo could not: the exact mathematics of the swing.

History

1583

— A young Galileo, watching a lamp swing in Pisa cathedral, times it against his pulse and notices the period seems independent of amplitude.

1638

— Galileo's Discorsi presents the pendulum as evidence that all bodies fall at the same rate, mass cancelling from the swing.

1656

— Christiaan Huygens builds the first pendulum clock, improving timekeeping accuracy from minutes to seconds per day.

1673

— Huygens' Horologium Oscillatorium gives the exact theory: the period formula, the amplitude problem, and the cycloidal pendulum that is truly isochronous.

1851

— Léon Foucault hangs a 67 m pendulum in the Panthéon; the slowly turning swing plane gives the first direct mechanical proof of Earth's rotation.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 4 (Oscillations) · Goldstein, Classical Mechanics, 3rd ed., Ch. 1 & 6 · Landau & Lifshitz, Mechanics (Vol. 1), §§5, 11 · Taylor, Classical Mechanics, Ch. 5 Key numerical anchors used throughout this article: $\omega_0=\sqrt{g/L}$ , $T=2\pi\sqrt{L/g}$ ; seconds-pendulum length $L=g(2)^2/(4\pi^2)=9.81\times4/39.478\approx0.994\,\mathrm{m}$ ; amplitude series $T=2\pi\sqrt{L/g}(1+\theta_0^2/16+11\theta_0^4/3072+\cdots)$ , consistent with the expansion of $4\sqrt{L/g}\,K(\sin(\theta_0/2))$ ; Moon-period factor $\sqrt{9.81/1.62}=\sqrt{6.06}\approx2.449$ with $g_\mathrm{Moon}\approx1.62\,\mathrm{m/s^2}$ ; $g=9.81\,\mathrm{m/s^2}$ used as the standard surface value. Worked-example results ( $T_0 = 1.737\ \mathrm{s}$ , $T = 1.864\ \mathrm{s}$ , $v = 2.71\ \mathrm{m/s}$ , $T_{\mathrm{str}} = 5.89\ \mathrm{N} = 2mg$ at $60°$ ; $g = 9.81\ \mathrm{m/s^2}$ from $T = 2.006\ \mathrm{s}$ at $L = 1.000\ \mathrm{m}$ ) follow directly from the boxed formulas.

Builds on: Simple Harmonic Motion · Small Oscillations About Equilibrium · Amplitude, Phase, and Angular Frequency. Leads to: The Physical Pendulum · Anharmonic Pendulum and Amplitude Dependence · Energy in Simple Harmonic Motion · Damped Harmonic Oscillator.

Why doesn't the mass appear anywhere in the period?

Because it enters the problem twice and cancels. Gravity's restoring torque is proportional to $m$ (heavier bob, stronger pull back), but the resistance to angular acceleration — the moment of inertia $mL^2$ — is also proportional to $m$ . Newton's law divides one by the other, and the mass is gone. This is the same cancellation that makes a hammer and a feather fall together in vacuum: the pendulum is a slow-motion free-fall experiment, which is exactly why Newton used pendulums with bobs of different materials to test the equality of gravitational and inertial mass to about one part in a thousand.

If a bigger swing travels farther, why doesn't it take longer?

For small swings, the restoring force grows in exact proportion to the displacement: swing twice as far and you are pulled back twice as hard, move twice as fast at the bottom, and cover the doubled distance in the same time. The compensation is exact only while $\sin\theta \approx \theta$ . At larger angles the true force $mg\sin\theta$ falls below the linear value $mg\theta$ — at $90°$ it is only about $64\%$ of it — so the compensation fails in the slow direction and the period grows: $+1\%$ at $23°$ , $+18\%$ at $90°$ .

What is the string tension actually doing, if it never shows up in the equation of motion?

It enforces the constraint — keeping the bob on the circle — while doing zero work, since it is always perpendicular to the velocity. It vanishes from the torque equation because it passes through the pivot. But it is far from constant: at the turning point of a $\theta_0$ release it is $mg\cos\theta_0$ , and at the bottom it must both support the weight and bend the trajectory, reaching $mg(3 - 2\cos\theta_0)$ — twice the weight for a $60°$ swing. If you need the tension, energy conservation plus the centripetal equation delivers it in two lines.

What happens if I give the pendulum enough energy to go over the top?

The motion changes character entirely. Below $E = mgL$ (energy measured from the pivot-height zero of eq. 6) the pendulum librates — swings back and forth between turning points. Above it, there are no turning points: $\dot\theta$ never vanishes and the pendulum rotates over the top, round and round, moving fastest at the bottom. Exactly at $E = mgL$ sits the separatrix: the pendulum creeps asymptotically toward the inverted position, taking infinite time to arrive. Real pendulums given "exactly" critical energy fall unpredictably to one side or the other — a first taste of sensitive dependence on initial conditions.

Where does the 2π in the period formula actually come from?

From the geometry of a full cycle. The linearized solution is $\theta = \theta_0\cos(\omega_0 t + \varphi)$ , and the cosine repeats when its argument advances by $2\pi$ radians — one full turn of the phase. The time for that is $T = 2\pi/\omega_0$ . So the $2\pi$ is not pendulum physics at all; it is the circumference-to-radius ratio of the circle that the oscillation's phase traces. Any simple harmonic oscillator — mass on a spring, LC circuit — carries the same factor for the same reason.

My lab pendulum gave g = 9.6 instead of 9.8 — what likely went wrong?

The classic culprits, in order of frequency. (1) Length measured to the top or bottom of the bob instead of its center — a 2 cm error on a 1 m pendulum shifts $g$ by 2%. (2) Amplitude too large: at $30°$ the period runs 1.7% long, dragging your inferred $g$ down by 3.4% (it enters as $T^2$ ). (3) Timing single swings instead of many: a 0.2 s reaction error on one 2 s period is 10%; on fifty periods it is 0.2%. (4) Counting swings wrong by half a period — surprisingly common. Fix all four and a stopwatch-and-metre-stick measurement lands within 0.1–0.3% of your local $g$ .

If the amplitude of a clock's pendulum slowly decays, why doesn't the clock drift?

Two protections stack. First, isochronism itself: below $\sim 5°$ the period varies only in the fourth decimal place as the amplitude sags, because the correction goes as $\theta_0^2/16$ . Second, real clocks don't let the amplitude sag: the escapement injects a tiny push each swing, precisely replacing the energy lost to friction and air drag, so the pendulum runs at constant small amplitude. The combination — a flat period-versus-amplitude curve, operated at a fixed point on it — is what kept pendulum clocks the world's best timekeepers for nearly three centuries.

Does a pendulum's period depend on where on Earth it swings?

Yes — through $g$ , and measurably. Gravity ranges from about $9.780\ \mathrm{m/s^2}$ at the equator to $9.832\ \mathrm{m/s^2}$ at the poles (rotation plus the equatorial bulge), so a clock regulated in Paris loses about 2 minutes per day in Quito: a seconds pendulum must be about 3 mm shorter there. Far from being a nuisance, this was an instrument: eighteenth- and nineteenth-century expeditions carried pendulums to map $g$ worldwide, and Kater's reversible physical pendulum made the method absolute. It is also why the metre was ultimately not defined as the seconds-pendulum length — the "constant" in the definition varies with latitude.

The Simple Pendulum

Where this lives today

The physical picture

From gravity to the equation of motion

Which equation answers which question

Energy and the phase portrait

Beyond small angles: the exact period

Limiting cases

Worked examples

1 · Design a seconds pendulum, then take it to the Moon (confidence builder)

2 · A 60° release: exact period, speed, and the string's worst moment (exam level)

Going further

Before this

Where this leads