In short

A simple pendulum is an idealised bob of mass m on an inextensible, massless string of length L, swinging in a vertical plane under gravity. For small angles (a few degrees, so that \sin\theta \approx \theta), the tangential restoring force is proportional to the displacement, and the motion is simple harmonic. The time period is

\boxed{\; T = 2\pi\sqrt{\frac{L}{g}}. \;}

Striking feature: T depends only on the length L and on g — not on the bob's mass, and (to leading order) not on the amplitude. Double the length and the period grows by \sqrt{2}; move to the Moon where g = 1.62 m/s² and a one-metre pendulum ticks once every \approx 4.93 s instead of the Earthly \approx 2 s. Measure L and T and you have measured g — the workhorse school experiment that every Indian physics lab has repeated for a century.

Walk into the Jantar Mantar observatory at Jaipur, or look up at the long chain of a brass bell in a South Indian temple, or the ceiling-hung flywheel pendulum of a grandfather clock — and you are looking at one of the most important devices in the history of physics. A weight at the end of a string. Gravity pulls it back toward the vertical. Friction-free (more or less) contact lets it swing past. Pendulums timed the first useful clocks in the 17th century. Pendulums measured the shape of the Earth (by varying g with latitude). Pendulums, as you will see in Compound and Torsional Pendulums, generalise into the swinging of every rigid body about a pivot.

A child on a park swing in Lodi Gardens does not know the equation of her motion, but she knows this: the time between two big pumps is a little longer than the time between two small pumps — but only a little. To first approximation, small and large swings take the same time. That tiny constancy — of the period as the amplitude changes — is what Galileo is said to have noticed while watching a chandelier swing in a cathedral. It is also what this article explains: why the period of a pendulum is (nearly) independent of the size of the swing, and what exactly "nearly" means.

The machinery is: draw the free-body diagram, take the component of gravity tangential to the swing, apply Newton's second law, use the small-angle approximation, and read off the SHM equation. The conclusion — T = 2\pi\sqrt{L/g} — should emerge as inevitable. Then the going-deeper section restores the amplitude correction, T = T_0 \bigl[1 + \theta_0^2/16 + \dots\bigr], so you can see exactly how good the small-angle answer is.

Setting up the simple pendulum

A simple pendulum is the idealisation: a point mass m hanging from a rigid support by a string of length L. The string is assumed massless and inextensible (no stretching or shortening), and the motion is confined to a single vertical plane. There is no friction at the support, no air resistance.

When the string makes an angle \theta with the downward vertical, the bob is displaced by an arc length s = L\theta along the circular path. (\theta is measured in radians throughout.) As the bob swings, \theta changes with time — that is what we want to solve for.

Free-body diagram of a simple pendulum at angle θ A pendulum of length L hangs from a fixed pivot at the top. The bob is displaced to the right by an angle θ from the downward vertical. The string is drawn as a solid line from pivot to bob. Two force vectors act at the bob: the string tension T pointing up along the string toward the pivot, and gravity mg pointing straight down. The gravitational force is decomposed into components: mg cos θ along the string (balanced by tension) and mg sin θ perpendicular to the string, tangent to the swing, pointing back toward equilibrium. An arc shows the swing path; an arrow labels the arc length s = L θ. pivot L m θ arc s = L θ T (tension) mg mg sin θ (tangential) mg cos θ (radial)
Free-body diagram of a simple pendulum at angle $\theta$ from the vertical. The bob experiences tension $T$ along the string (toward the pivot) and gravity $mg$ downward. Decomposing gravity gives $mg\cos\theta$ along the string (balanced by tension up to the centripetal term) and $mg\sin\theta$ tangent to the circle — the **restoring force** that pulls the bob back toward the vertical.

Two forces act on the bob: the tension T in the string (pointing from the bob toward the pivot — along the string) and the gravitational force mg (pointing straight down). These two forces live in different directions, so you split them along natural directions of the motion: radial (along the string) and tangential (perpendicular to the string, along the direction of motion).

Along the radial direction, the net force provides the centripetal acceleration of the circular path. Along the tangential direction, the net force accelerates the bob along the arc. The tangential direction is where the oscillation lives.

Writing Newton's second law along the arc

The component of gravity along the tangent (pointing back toward equilibrium, opposite to increasing \theta) has magnitude mg\sin\theta. The tension has no tangential component (it is entirely along the string). So the tangential equation of motion is

m a_\text{tan} = -mg \sin\theta,

where the minus sign captures "pulls back toward \theta = 0."

The tangential acceleration is the second derivative of the arc length with respect to time: a_\text{tan} = d^2 s/dt^2 = L \, d^2\theta/dt^2 = L\ddot\theta (since s = L\theta and L is fixed). Substituting:

m L \ddot\theta = -mg\sin\theta.

Cancel the mass m — and notice that it is the same m on both sides, because the "inertial mass" in Newton's second law equals the "gravitational mass" in the weight mg (this equivalence is what Einstein would later elevate into general relativity):

\boxed{\; L \ddot\theta = -g \sin\theta. \;} \tag{1}

Equation (1) is the exact equation of the simple pendulum. It describes every pendulum swing from tiny wiggles to full 180° inversions. Unfortunately, it is non-linear (because of the \sin\theta), and non-linear differential equations do not, in general, have closed-form solutions in elementary functions. The general solution of (1) is expressible only in terms of elliptic functions — not the kind of thing you want to write down for a 15-year-old.

What rescues the problem is that for small angles, \sin\theta \approx \theta, and the linear approximation turns (1) into an SHM equation you can solve in one step.

The small-angle approximation — \sin\theta \approx \theta

Expand \sin\theta in a Taylor series about \theta = 0:

\sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots = \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \dots

For \theta in radians:

\theta (°) \theta (rad) \sin\theta Relative error in \sin\theta \approx \theta
5 0.0873 0.0872 0.13%
10 0.1745 0.1736 0.51%
15 0.2618 0.2588 1.15%
20 0.3491 0.3420 2.06%
30 0.5236 0.5000 4.72%
45 0.7854 0.7071 11.1%

For angles up to about 10° — the range of a typical pendulum-clock swing — the approximation is accurate to about half a percent. For a school experiment where you time a pendulum to three significant figures, you can confidently work up to about 15° before the approximation starts to leak into the answer.

Why: the next correction term is \theta^3/6, so the relative error is approximately \theta^2/6. At \theta = 0.1 rad the error is 0.01/6 \approx 0.17\%; at \theta = 0.3 rad (\approx 17°) it has grown to \sim 1.5\%. The correction scales as amplitude squared.

With \sin\theta \approx \theta, equation (1) becomes

\boxed{\; L\ddot\theta = -g\theta, \qquad \ddot\theta = -\frac{g}{L}\theta. \;} \tag{2}

This is the SHM equation — the same \ddot x = -\omega^2 x form from Equation of SHM and Phase, now with \theta playing the role of the displacement and

\omega^2 = \frac{g}{L}, \qquad \omega = \sqrt{\frac{g}{L}}.

The solution is

\theta(t) = \theta_0 \sin(\omega t + \varphi),

where \theta_0 is the angular amplitude (the maximum swing angle) and \varphi is the phase set by the initial conditions. The time period is

\boxed{\; T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{g}}. \;} \tag{3}

Why: T = 2\pi/\omega for any SHM. Substituting \omega = \sqrt{g/L} and inverting gives T = 2\pi/\sqrt{g/L} = 2\pi\sqrt{L/g}.

Three striking consequences follow.

Three things the formula tells you before you calculate anything

1. The period is independent of the mass

Nowhere in equation (3) does m appear — it cancelled in going from mL\ddot\theta = -mg\sin\theta to L\ddot\theta = -g\sin\theta. Take two pendulums of the same length but different bob masses — a heavy brass bob and a light aluminium one — pull them both to the same small angle and release them together. They swing in perfect synchrony. The heavier bob has more inertia (harder to accelerate) but also feels a proportionally larger gravitational force — the two effects cancel exactly.

This is the simplest-possible manifestation of the equivalence of inertial and gravitational mass, which is the experimental basis for Einstein's general relativity. You can demonstrate it on a playground set with two swings of equal chain length and two kids of different weight, releasing them from the same angle.

2. The period is independent of the amplitude (to first order)

To the small-angle approximation, T = 2\pi\sqrt{L/g} does not depend on \theta_0. Two pendulums of the same length, one swinging through ±3° and the other through ±10°, take the same time to complete a swing. This is what Galileo noticed watching the cathedral lamp (though his "same time" was a few-percent approximation — the amplitude correction we work out in the going-deeper section adds a small but measurable dependence at larger angles).

The independence is called isochronism. It is what made pendulum clocks possible: the clock's pendulum does not need to be kept at exactly the same amplitude to keep the same time.

3. The period depends on L and g — and on nothing else

From T = 2\pi\sqrt{L/g}, the period grows as \sqrt{L}: a 1 m pendulum has T \approx 2.006 s on Earth (g = 9.81 m/s²); a 4 m pendulum has T = 2\sqrt{4/9.81}\pi \approx 4.01 s — exactly twice as long.

The period shrinks as \sqrt{g}: take the same 1 m pendulum to the Moon (g_\text{Moon} = 1.62 m/s²) and its period becomes T = 2\pi\sqrt{1/1.62} \approx 4.93 s — almost two and a half times longer. This is why — in the Apollo 12 mission footage — the astronauts walking about take long, loping strides; the rotational pendulum of their leg (from the hip) has a period that is longer in weaker gravity.

A seconds pendulum is one whose half-period (the time for one swing from one extreme to the other) is exactly one second — so T = 2 s. From T = 2\pi\sqrt{L/g} with T = 2 and g = 9.81, you get L = gT^2/(4\pi^2) = 9.81 \times 4/(4\pi^2) \approx 0.994 m. Every grandfather clock with a tick-tock of one beat per second has its pendulum roughly that length, with a length-adjustment nut to fine-tune.

Seeing it: an animation and a parameter explorer

The animation below shows a pendulum of length L = 1 m swinging through a 15° amplitude. Period: T = 2\pi\sqrt{1/9.81} \approx 2.006 s. The animation runs for 4 seconds so you see about two full swings.

Animated: simple pendulum of length 1 m swinging at 15 degrees amplitude A red bob on a string of length 1 metre swings in a vertical plane from a fixed pivot. The amplitude is 15 degrees. The motion follows theta of t equals 0.2618 radians times sine of omega t, with omega equal to square root of g over L, giving a period of about 2 seconds. Over the 4-second animation, the bob completes about two full swings. pivot length = 1 m
Watch the 1 m pendulum swing through a 15° amplitude. The angle follows $\theta(t) = \theta_0 \sin(\omega t)$ with $\theta_0 = 15° \approx 0.2618$ rad and $\omega = \sqrt{g/L} \approx 3.13$ rad/s. The position is $(L\sin\theta, -L\cos\theta)$ measured from the pivot. In 4 seconds the bob completes about two full swings. Click replay to watch again.

To get a feel for how T depends on L, drag the length of the pendulum below. The vertical bar represents the period; watch it grow with \sqrt{L}.

Interactive: time period versus length for a simple pendulum A curve T(L) = 2 pi square root of L over 9.81 plotted from L equals 0 to L equals 4 metres. A draggable red point moves along the x axis between 0.1 and 4 metres; a vertical segment rises from the point to the curve; a horizontal dashed segment extends from the curve to the y axis. Readouts display L, T, and the frequency f. L (m) T (s) 0 1 2 3 4 1 2 3 4 T = 2π √(L/g) drag the red point along the axis
Drag the red point along the $L$-axis to change the pendulum's length. Watch how the time period $T$ follows the square-root curve. A 1 m pendulum ticks every $\approx 2.006$ s; a 4 m pendulum ticks every $\approx 4.01$ s — exactly twice as long, confirming the $\sqrt{L}$ scaling. The frequency $f = 1/T$ falls accordingly.

Using the pendulum to measure g — the laboratory experiment

Invert the formula:

g = \frac{4\pi^2 L}{T^2}.

If you can measure the length L and the period T accurately, you can extract the acceleration due to gravity. This is the oldest and most robust lab method for measuring g, and it is one of the few school experiments where the accuracy you can achieve at home rivals that of a research lab.

The procedure (as run in every Indian school physics laboratory):

  1. Tie a brass bob to one end of a thin string. Clamp the other end to a rigid support so that the pendulum hangs freely.
  2. Measure the length L from the point of suspension (the effective pivot, not where your hand is holding it) to the centre of the bob. Use a metre scale or Vernier callipers; record L \pm \Delta L with a realistic uncertainty (say, \pm 1 mm for a typical 1 m string).
  3. Displace the bob by a small angle (less than 10°) and release it without imparting any sideways kick.
  4. Use a stopwatch to time 20 complete oscillations (not just one — this reduces the relative error in the timing). Record t_{20}.
  5. The period is T = t_{20}/20.
  6. Compute g = 4\pi^2 L/T^2.
  7. Repeat the whole experiment for several different lengths (say L = 0.5, 0.75, 1.0, 1.25, 1.5 m) and either compute g from each and average, or — better — plot T^2 against L and find the slope. From T^2 = (4\pi^2/g) \cdot L, the slope of the best-fit straight line is 4\pi^2/g, so g = 4\pi^2/\text{slope}.

Error propagation. If L and T are measured with fractional errors \Delta L/L and \Delta T/T, then from g = 4\pi^2 L/T^2:

\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}.

The factor of 2 on the \Delta T/T says the period measurement is the more important one — a 1% error in T gives a 2% error in g. This is precisely why the procedure times 20 swings and not one: if a single stopwatch click has a reaction-time error of \pm 0.2 s, then timing 20 swings (total \approx 40 s) gives \Delta T/T \approx 0.2/40 = 0.5\%, while timing one swing (2 s) would give \Delta T/T \approx 10\%.

A careful measurement with L = 1.00 \pm 0.001 m and t_{20} = 40.1 \pm 0.2 s gives T = 2.005 \pm 0.01 s and

g = \frac{4\pi^2 \times 1.00}{(2.005)^2} \approx 9.82 \text{ m/s}^2 \pm 1\%,

which is well within experimental agreement of the "true" value g = 9.81 m/s² at most Indian latitudes.

Worked examples

Example 1: The seconds pendulum at Jaipur

A clockmaker in Jaipur wants to build a pendulum clock whose pendulum has a time period of exactly 2 seconds (a "seconds pendulum"). What must the pendulum length be? Take g = 9.79 m/s² (the value at Jaipur's latitude; g varies slightly with latitude).

Seconds pendulum diagram A pendulum hangs with length L from a pivot at the top. An arrow indicates the swing from one extreme to the other taking exactly one second. The total period from one extreme and back again is 2 seconds. The length L is labelled to be found. pivot half-period = 1 s L = ? T = 2 s
A seconds pendulum: a swing from one extreme to the other takes 1 s; a full period (there and back) takes 2 s. The length $L$ is chosen so that $2\pi\sqrt{L/g} = 2$ s.

Step 1. Write down the formula and what you know.

T = 2\pi\sqrt{L/g}, with T = 2 s and g = 9.79 m/s². Solve for L.

Why: the problem gives you T and g; you want L. Square the equation and isolate L.

Step 2. Square both sides.

T^2 = 4\pi^2 \frac{L}{g}.

Step 3. Solve for L.

L = \frac{g T^2}{4\pi^2} = \frac{9.79 \times (2)^2}{4 \pi^2} = \frac{9.79 \times 4}{39.478} = \frac{39.16}{39.478}.
L \approx 0.992 \text{ m}.

Why: 4\pi^2 \approx 39.478. The numerator is 9.79 \times 4 = 39.16. The two are almost identical because g \approx \pi^2 is close to true (not by coincidence — it is a consequence of the original metre being defined so that a seconds pendulum in Paris is almost exactly a metre long).

Step 4. Sanity check with g = 9.81 m/s² (standard value).

L = \frac{9.81 \times 4}{39.478} \approx 0.994 \text{ m}.

The slight change in g between Jaipur and the standard value changes L by only 2 mm. A clockmaker would fine-tune the bob's height by a nut on a threaded rod to get the rate exactly right.

Result: L \approx 0.992 m. A pendulum of just under one metre length keeps one-second time in Jaipur.

What this shows: The "one-metre-for-one-second" coincidence is the origin of the metre itself — an 18th-century attempt to define the metre as the length of a seconds pendulum. The French Revolution-era committee eventually picked the meridian-arc definition instead (one ten-millionth of the Paris meridian's quadrant), but the near-miss tells you the two values are naturally close: g \approx \pi^2 \approx 9.87 m/s² means a seconds pendulum is approximately a metre.

Example 2: A pendulum in a lift (elevator) accelerating upward

A simple pendulum of length L = 0.50 m is placed in a lift that accelerates upward at a = 2.0 m/s². Find the new time period of small oscillations. Take g = 9.8 m/s².

Pendulum in an upward-accelerating lift A pendulum of length L hangs inside a lift cabin. The lift accelerates upward with acceleration a equal to 2 metres per second squared, so inside the cabin the effective gravity felt by the bob is g plus a, pointing downward. The pendulum swings at a shorter period because the effective gravity is larger. a = 2 m/s² (lift accelerating up) g + a effective gravity
A pendulum in a lift accelerating upward at $a$. Inside the lift, the bob feels an effective gravitational acceleration $g_\text{eff} = g + a$, leading to a shorter period than when the lift is at rest.

Step 1. Identify the effective gravitational acceleration.

Inside the lift's (non-inertial) frame, every object feels an additional pseudo-force equal to its mass times the lift's acceleration, pointing opposite to the lift's acceleration. The lift accelerates upward, so the pseudo-force on the bob points downward — the same direction as gravity. Effectively, the bob feels gravity of magnitude g + a pointing downward.

Why: in a non-inertial frame accelerating with \vec a_\text{frame}, any body behaves as if there were an extra force -m\vec a_\text{frame} on it. For an upward-accelerating lift, this adds mg equivalent of downward force to the bob. Equivalently, stand in an upward-accelerating lift and you feel "heavier" — the floor pushes up on you with force m(g+a).

Step 2. The pendulum equation becomes L\ddot\theta = -g_\text{eff}\sin\theta \approx -(g+a)\theta.

The angular frequency is

\omega' = \sqrt{\frac{g+a}{L}}

and the new period is

T' = 2\pi\sqrt{\frac{L}{g+a}}.

Step 3. Plug in numbers.

T' = 2\pi\sqrt{\frac{0.50}{9.8 + 2.0}} = 2\pi\sqrt{\frac{0.50}{11.8}} = 2\pi\sqrt{0.04237}.
T' = 2\pi \times 0.2058 \approx 1.29 \text{ s}.

Why: the effective gravity of 11.8 m/s² is larger than the standard 9.8, so the restoring force is larger, \omega is larger, and the period is shorter. Compare to the stationary-lift period T = 2\pi\sqrt{0.5/9.8} \approx 1.42 s.

Step 4. Check extremes.

If the lift were in free fall (a = -g, so g_\text{eff} = 0), the pendulum would have T' \to \infty — the bob would not swing at all, because there is no effective restoring force. This is exactly what astronauts experience on the International Space Station: a pendulum up there is a bob on the end of a loose string, floating, going nowhere.

If the lift accelerated downward at a = 5 m/s², the effective gravity would be 9.8 - 5 = 4.8 m/s² and the period would grow to T' = 2\pi\sqrt{0.5/4.8} \approx 2.03 s.

Result: T' \approx 1.29 s in the upward-accelerating lift, down from \approx 1.42 s at rest.

What this shows: The simple pendulum is a local gravimeter. Take it in a lift, up a mountain, to the Moon — wherever the effective gravitational acceleration is, the pendulum reports it through its period. This is the principle behind instruments that measure variations in g across the Earth's surface, used in oil prospecting and geodesy.

Common confusions

  • "The period depends on the mass of the bob." It does not — provided the string is massless and air resistance is negligible. Every simple pendulum of length L in a given g has the same period, regardless of what hangs from it. A cannonball and a tennis ball on identical 1 m strings swing together.

  • "The small-angle approximation means the pendulum won't work at large angles." It will work — but the period depends slightly on amplitude, and the motion is no longer strictly sinusoidal. See the going-deeper section for the first correction, T(\theta_0) \approx T_0(1 + \theta_0^2/16).

  • "The pendulum formula doesn't apply to a stick or a ceiling fan." Correct — the formula T = 2\pi\sqrt{L/g} is only for a simple pendulum (point mass on massless string). An extended rigid body swinging about a pivot is called a compound pendulum, and its period is T = 2\pi\sqrt{I/(mgd)} where I is the moment of inertia about the pivot and d is the distance from pivot to centre of mass. See Compound and Torsional Pendulums.

  • "If you double the length, the period doubles." No — it grows by \sqrt{2} \approx 1.414, because T \propto \sqrt{L}. To double the period you need to quadruple the length.

  • "The pendulum bob swings along a straight line." It does not — the bob moves along an arc of a circle. The approximation s = L\theta (arc length) and the tangential force mg\sin\theta are what the derivation uses; nothing in the derivation assumes straight-line motion. What the small-angle approximation gives you is not that the path is straight, but that the restoring force is linear in the displacement \theta.

  • "Gravity is the only force doing work on the pendulum." True — and this is why mechanical energy is conserved for a simple pendulum (in the absence of friction). The tension does no work because it is always perpendicular to the bob's velocity (which is tangent to the arc). This observation gives you an alternative derivation of the period using energy methods, of the kind sketched in Energy in SHM.

If you came here to understand what makes the simple pendulum an SHM, use the period formula, and do JEE-level problems, you have what you need. What follows is for readers who want the amplitude correction, the energy-method derivation, and the behaviour at large angles.

The amplitude correction — T at second order in \theta_0

The small-angle approximation \sin\theta \approx \theta drops the cubic term -\theta^3/6 from the Taylor expansion. Restoring that term makes the equation of motion non-linear, and the period picks up a dependence on the amplitude \theta_0 that the linear theory misses. A more careful analysis (which you can do using perturbation theory or energy conservation with an elliptic integral) gives

T(\theta_0) = T_0\left[1 + \frac{\theta_0^2}{16} + \frac{11\theta_0^4}{3072} + \dots\right],

where T_0 = 2\pi\sqrt{L/g} is the small-angle period.

Here is a derivation via energy conservation. The full (non-approximated) equation of motion is L\ddot\theta = -g\sin\theta. Multiply both sides by \dot\theta and integrate once:

L\dot\theta \ddot\theta = -g\sin\theta \cdot \dot\theta,
\frac{d}{dt}\Bigl[\tfrac{1}{2} L \dot\theta^2\Bigr] = \frac{d}{dt}\Bigl[g\cos\theta\Bigr],
\tfrac{1}{2} L \dot\theta^2 - g\cos\theta = \text{constant}.

At the extreme (\theta = \theta_0, \dot\theta = 0), the constant equals -g\cos\theta_0. So

\tfrac{1}{2} L \dot\theta^2 = g(\cos\theta - \cos\theta_0),
\dot\theta = \sqrt{\frac{2g}{L}}\sqrt{\cos\theta - \cos\theta_0}.

The time to go from \theta = 0 to \theta = \theta_0 is a quarter-period, so

\frac{T}{4} = \int_0^{\theta_0} \frac{d\theta}{\sqrt{(2g/L)(\cos\theta - \cos\theta_0)}}.

Use the identity \cos\theta - \cos\theta_0 = 2[\sin^2(\theta_0/2) - \sin^2(\theta/2)] and substitute \sin(\theta/2) = \sin(\theta_0/2)\sin\phi. After simplification,

T = 4\sqrt{\frac{L}{g}} \int_0^{\pi/2} \frac{d\phi}{\sqrt{1 - k^2\sin^2\phi}}, \qquad k = \sin(\theta_0/2).

The remaining integral is the complete elliptic integral of the first kind, K(k). Expanding in powers of k^2:

K(k) = \frac{\pi}{2}\left[1 + \frac{1}{4}k^2 + \frac{9}{64}k^4 + \dots\right].

With k^2 = \sin^2(\theta_0/2) \approx \theta_0^2/4 for small \theta_0:

T = 4\sqrt{\frac{L}{g}} \cdot \frac{\pi}{2}\left[1 + \frac{1}{4}\cdot\frac{\theta_0^2}{4} + \dots\right] = 2\pi\sqrt{\frac{L}{g}}\left[1 + \frac{\theta_0^2}{16} + \dots\right].

Numerical example. For \theta_0 = 15° \approx 0.2618 rad, the correction is \theta_0^2/16 \approx 0.00428, or about 0.43%. A pendulum clock swinging through 15° runs about 0.43% slower than the linear theory predicts — over a day (86400 s), the accumulated error is \approx 370 seconds, or 6 minutes. That is why serious pendulum clocks keep the amplitude small and constant — not because larger amplitudes fail, but because they change the period.

For \theta_0 = 60° (a hefty swing), the correction is \theta_0^2/16 \approx 0.0685 or nearly 7% — and at this amplitude the higher-order terms in the series can no longer be safely ignored either.

The energy method for the period — bypassing Newton's second law

For a simple pendulum, the total mechanical energy (taking the lowest point \theta = 0 as the reference for PE) is

E = \tfrac{1}{2} m L^2 \dot\theta^2 + mgL(1 - \cos\theta).

The kinetic-energy term uses the fact that the speed of the bob is v = L\dot\theta, so \tfrac{1}{2} m v^2 = \tfrac{1}{2} m L^2 \dot\theta^2. The potential-energy term uses the height the bob has risen above its lowest point, h = L(1 - \cos\theta).

For small angles, 1 - \cos\theta \approx \theta^2/2, so

E \approx \tfrac{1}{2} m L^2 \dot\theta^2 + \tfrac{1}{2} m g L \theta^2.

Matching to the general SHM energy E = \tfrac{1}{2} I_\text{eff} \dot q^2 + \tfrac{1}{2} k_\text{eff} q^2 with q = \theta:

I_\text{eff} = mL^2, \qquad k_\text{eff} = mgL,
\omega = \sqrt{\frac{k_\text{eff}}{I_\text{eff}}} = \sqrt{\frac{mgL}{mL^2}} = \sqrt{\frac{g}{L}}.

The mass cancels, the period is T = 2\pi\sqrt{L/g}, and you never had to write Newton's second law. This is the energy method advertised in Energy in SHM, applied to the pendulum in three lines.

Why g \approx \pi^2 in SI units — a near-coincidence of history

The metre was originally defined so that a seconds pendulum would be almost exactly one metre long. From T = 2\pi\sqrt{L/g} with T = 2 s and L = 1 m, you need g = \pi^2 \approx 9.87 m/s². The actual value of g (varying from \approx 9.78 at the equator to \approx 9.83 at the poles) is within 1% of \pi^2.

This is not a law of nature; it is a fossilised piece of 18th-century metrological history. When the French Academy of Sciences redefined the metre as one ten-millionth of the meridian arc from the equator to the North Pole through Paris, the length turned out to be only 0.6% different from the seconds-pendulum length. The modern SI metre (defined via the speed of light) inherits this near-match.

The practical consequence: whenever you need a quick estimate and g is involved, replacing g by \pi^2 gives an answer good to 1%.

The phase portrait — large-amplitude pendulum behaviour

If you plot the pendulum's state (\theta, \dot\theta) in a phase-space diagram for different total energies, you get a family of curves.

  • For small energies (E < 2mgL), the curves are closed loops around the origin — libration (back-and-forth oscillation, as treated in the main article).
  • For exactly E = 2mgL (the energy needed to just barely reach \theta = \pi, the inverted position), the curve is a separatrix — the pendulum takes infinite time to reach the top.
  • For E > 2mgL, the curves no longer close — the pendulum has enough energy to go completely over the top, and it rotates continuously (think of a children's Ferris-wheel seat cranked hard enough to loop).

The separatrix marks the boundary between oscillation and rotation — the pendulum's equivalent of the "escape velocity" of orbital mechanics. It is a beautiful example of how a single parameter (the total energy) can change the qualitative character of a system's motion, and it is one of the simplest examples of the rich behaviour of non-linear dynamical systems.

Where this leads next

  • Equation of SHM and Phase — the general framework of simple harmonic motion, which the small-angle pendulum is a special case of.
  • Energy in SHM — the energy picture of every oscillator, including the energy-method derivation of the pendulum period.
  • Compound and Torsional Pendulums — the generalisation to an extended rigid body pivoting about a fixed axis, and the torsional pendulum (a wire twisting instead of a string swinging).
  • Damped Oscillations — what happens to the pendulum when air resistance finally starts to matter: amplitude decays exponentially and the period shifts slightly.
  • Spring-Mass Systems — the other archetypal SHM, and a useful point of comparison for what is and is not common to all oscillators.