Physics · Oscillations and Waves

Simple pendulum

Q: How do you calculate the period of a simple pendulum?

For small angles use $T_0=2\pi\sqrt{\frac lg}$. Length l is the distance from the fixed pivot to the centre of mass, not only visible string length; g is local gravitational acceleration. Enter metres and $\frac{m}{s^{2}}$ to obtain seconds. For $l = 1.00 m$ and $g =\frac{ 9.81 m}{s^{2}}$, $T_0$ ≈ 2.006 s. One complete back-and-forth motion is a period; merely crossing the midpoint is insufficient for counting full cycles. At larger angles the ideal actual period is longer and needs an amplitude correction. Reversing $g=\frac{4\pi^2l}{T_0^2}$ checks the calculation under the same small-angle assumptions.

Q: Why does the period not depend on the mass?

Tangential weight is $-mg\sin\theta$ and tangential inertia is $ml\ddot\theta$. Both are proportional to the same mass, so m cancels from the ideal equation of motion. Equal length, equal local g and equal initial conditions therefore give the same ideal motion for different masses. This applies to the nonlinear angle equation as well as the small-angle version. It does not exclude real effects: for equal body shape, air resistance relative to inertia can differ. An extended body also needs its moment of inertia; changing mass distribution can alter physical-pendulum period even when total mass itself cancels for a fixed geometry.

Q: How do you determine g with a simple pendulum?

Measure centre-of-mass length l and time many complete periods at a small release angle. Divide total time by cycle count, then use $g\approx\frac{4\pi^2l}{T^2}$. For $l = 0.900 m$ and 30 periods in 57.12 s, $T = 1.904 s$ and $g \approx{}\frac{ 9.801 m}{s^{2}}$. Even at 2°, angle increases ideal period by roughly 0.0076%; without correction g is underestimated by about 0.0152%. That control value does not replace measurement uncertainty. Check centre-of-mass location, support stability, string extension, friction and consistent counting. Matching position and direction of motion identifies a complete cycle.

Q: What changes when you quadruple the pendulum length?

At unchanged g, quadrupling length doubles small-angle period because $T_0\propto\sqrt l$. Frequency consequently halves. Reversing $g=\frac{4\pi^2l}{T_0^2}$ gives the same value because numerator and denominator both quadruple. For the same maximum angle, maximum height $l(1-\cos\theta_0)$ quadruples and bottom speed doubles. Equal angular amplitude must not be confused with equal sideways displacement. With unchanged angle, the period scaling also holds in the ideal elliptic large-angle model; if angle changes, its amplitude factor must additionally be included. These comparisons assume the same ideal constraint and negligible losses in both configurations.

Q: When does the pendulum formula stop being valid?

$T=2\pi\sqrt{\frac lg}$ is a small-angle approximation, not an exact equation for every release angle. Acceptable angle depends on required accuracy; at 10° the leading period correction is about 0.190%, with no sudden change of motion type. At 60° the complete ideal period is about 7.318% greater than $T_0$. String elasticity, moving support, substantial friction or an extended body can also change the model. The string must stay taut: release from rest above the horizontal cannot maintain the circular path. An elliptic integral alone does not repair that missing constraint. Always separate approximation error from failure of the physical model.

Ideal small-angle pendulum period depends on centre-of-mass length l and local g. Mass cancels; at larger angles period increases with amplitude.

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Formula

Calculate

T = 2 π \frac{l}{g}

T in s · l from pivot to centre of mass in m · local g in m/s² (often approximated as 9.81 m/s²)

Variables & units

Variables & units
Symbol	Meaning	Unit
$T$	Period of one complete back-and-forth motion in the small-angle model	s
$l$	Pendulum length from fixed pivot to centre of mass	m
$g$	Local gravitational acceleration; often approximated as $\frac{9.81 m}{s^{2}}$	m/s²

LaTeX

T = 2\pi \sqrt{\frac{l}{g}}

QUANTA STUDYquanta-study.de

Diagram: a ball on a string of length l is deflected from the dashed rest position, a dashed arc shows the swing to the opposite end position, next to it an arrow g pointing down. — Small-angle schematic at ±8° about vertical equilibrium. l runs from pivot to bob center of mass; T ≈ 2π√(l/g) assumes an ideal undamped simple pendulum and small angles.

Simple pendulum period calculator

Find the period, length or gravitational acceleration of a simple pendulum swinging through small angles.

Find

Pendulum length l

Distance from the pivot to the centre of the bob, e.g. 1 m.

Gravitational acceleration g

Local gravitational acceleration, 9.81 m/s² on Earth, about 1.62 m/s² on the Moon.

m/s²

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for T, substitute, evaluate

1. Rearrange for T
$T=2\pi\sqrt{\frac lg}$
2. Substitute values (in SI units)
$T=2\pi\sqrt{\frac{1}{9.81}}=2.00607$
3. Result
T = 2.006 s

Result: Period T

T = 2.006 s

With l = 1 m and g = 9.81 m/s², T = 2.006 s.

Quick tips

Four times the length, twice the period: T grows with √l. The mass of the bob does not matter.
Estimate on Earth: T ≈ 2 s · √(l in m). A 1 m pendulum takes about 2 s, a 25 cm one about 1 s.

Assumptions and limits

Full oscillation period, pivot-to-centre-of-mass length, l > 0 and g > 0. Small angles, fixed pivot, taut massless string and negligible damping. This is not the exact period at large angles.

Derivation & background

Tangentially, $ml\ddot\theta=-mg\sin\theta$ . Mass cancels and $\sin\theta\approx\theta$ for small angles in radians gives $\ddot\theta+\frac gl\theta=0$ . Therefore $\omega_0=\sqrt{\frac gl}$ and $T_0=2\pi\sqrt{\frac lg}$ . There is no sharp cutoff at 10°: leading relative period correction is $\frac{\theta_0^2}{16}$ , about 0.190% at 10°. Gravity measurements also involve length and timing uncertainties and friction.

Exam blueprint

Validity range

Small-angle approximation for a point mass on a massless, inextensible, taut string with a fixed pivot and approximately constant local g. Neglect friction and driving. There is no sharp 10° cutoff: acceptable amplitude depends on required accuracy.

Derivation steps

Tangential weight first produces a nonlinear equation; the small-angle approximation gives the simple period.

1Define θ from vertical equilibrium and l from the fixed pivot to the centre of mass. The string remains taut.
2Tangential weight is $-mg\sin\theta$ ; with $s=l\theta$ , $ml\ddot\theta=-mg\sin\theta$ .
3Cancel m: $\ddot\theta+\frac gl\sin\theta=0$ . This is the exact ideal nonlinear equation.
4For small angles in radians replace $\sin\theta$ by θ, giving $\ddot\theta+\frac gl\theta=0$ .
5Compare with $\theta=\theta_0\cos(\omega_0t+\varphi)$ : differentiating twice gives $\omega_0^2=\frac gl$ .
6A full phase advance of 2π gives $T_0=\frac{2\pi}{\omega_0}=2\pi\sqrt{\frac lg}$ ; larger angles require nonlinear period correction.

Rearrangements

Pendulum length

l = \frac{g \cdot T^2}{4\pi^2}

A seconds pendulum has full small-angle period 2 s. At $g=\frac{9.81 m}{s^{2}}$ its model length is approximately 0.994 m.

Gravitational acceleration

g = \frac{4\pi^2 \cdot l}{T^2}

Standard lab method to determine g.

Task variant

An ideal pendulum of length $l=0.8\,\mathrm m$ has full period $T=1.8\,\mathrm s$ at small angles. Find local $g$ , neglecting friction and driving.

$g=\frac{4\pi^2l}{T^2}\approx\frac{39.48\cdot0.8}{3.24}\,\mathrm{m\,s^{-2}}\approx9.75\,\mathrm{m\,s^{-2}}$ .

How long must a pendulum be for $T = 1 s$ ?

$l=\frac{gT^2}{4\pi^2}\approx\frac{9.81\cdot1}{39.48}\,\mathrm m\approx0.248\,\mathrm m$ , about 25 cm.

Common mistakes

Including the bob mass in the calculation.

In the ideal small-angle model period depends only on length and local g. Equal length and conditions give different point masses equal periods; real friction or extended bobs can alter this.

Timing half a swing (one way) as a full period.

T is the time for there and back; in experiments time 10 periods and divide.

Applying the small-angle formula without checking required accuracy.

There is no sharp cutoff at $10^\circ$ . The leading relative period correction is $\frac{\theta_0^2}{16}$ , with angle in radians. Larger angles lengthen the ideal period; use the complete calculation if the deviation is too large for the task.

Measuring l to the end of the string instead of the centre of mass of the bob.

The pendulum length runs from the pivot to the centre of mass of the bob.

Exam context

Common: determining g from measurement series, computing the length change for a desired period or explaining why mass and (small) amplitude play no role.

Formula cluster

Harmonic oscillations

Mechanical counterpart of the spring pendulum and the LC circuit.

T = 2π·√(m/k)

Spring Pendulum (Oscillation Period)

T = 2π·√(L·C)

Thomson formula for an LC circuit

Worked example

For $l = 1.00 m$ and $g =\frac{ 9.81 m}{s^{2}}$ the small-angle model gives $T_0=2\pi\sqrt{\frac lg}\approx2.00607\,\mathrm s$ . A full period of 1.00 s requires rearranging to $l=\frac{gT_0^2}{4\pi^2}\approx0.248490\,\mathrm m$ , approximately 0.25 m.

Applications

Small-angle measurements of local g and initial time-scale estimates for pendulum clocks or swings. Extended bodies, large angles and spatial Foucault motion require additional models.

How do you calculate with Simple Pendulum (Period)?

Here is how to work through a typical Simple Pendulum (Period) (T = 2π·√(l/g)) task step by step:

1
Task
An ideal pendulum of length $l=0.8\,\mathrm m$ has full period $T=1.8\,\mathrm s$ at small angles. Find local $g$ , neglecting friction and driving.
Solution path
$g=\frac{4\pi^2l}{T^2}\approx\frac{39.48\cdot0.8}{3.24}\,\mathrm{m\,s^{-2}}\approx9.75\,\mathrm{m\,s^{-2}}$ .
2
Task
How long must a pendulum be for $T = 1 s$ ?
Solution path
$l=\frac{gT^2}{4\pi^2}\approx\frac{9.81\cdot1}{39.48}\,\mathrm m\approx0.248\,\mathrm m$ , about 25 cm.

Understand and apply the equation

Pendulum period increases with length and, for small angles, does not depend on mass. The simple square-root expression is an approximation whose accuracy can be checked against amplitude. Timing, large angles, and a half-period mistake show the consequences.

A complete return journey counts as one period

A released pendulum bob moves towards the lowest point, passes it, and returns. A full period ends only when it reaches the same position with the same direction of motion again. One crossing of the lowest point, or travel from one turning point to the other, is not a full cycle.[1]

You need weight, an angle measured from downward vertical, and the idea of a restoring force. Length runs from the suspension point to the small bob’s centre of mass. The familiar square-root expression describes small angles. The ideal pendulum model can still describe larger angles, but its period must then be calculated differently.[1]

Length, angle and the constraint

The model uses a point mass m, a massless inextensible string of length l and a fixed pivot. The string remains taut and motion stays in a vertical plane. θ is the signed angle from the vertical equilibrium position; positive and negative angles lie on opposite sides. Arc coordinate is $s=l\theta$ when θ is in radians. Pendulum length reaches the centre of mass, not merely the top of the bob.[1]

String tension acts radially and therefore does not directly supply tangential acceleration. The tangential weight component is $-mg\sin\theta$ . Radially, tension must also maintain circular motion. A slack string violates the circular-path assumption; an extended rigid body requires a physical-pendulum model with moment of inertia.

How does the small-angle period arise, and why does mass cancel?

For the ideal pendulum bob at small angular amplitudes, $T_0=2\pi\sqrt{\frac{l}{g}}$ . Measure $l$ from pivot to centre of mass. Weight and inertia are both proportional to mass, which cancels in the tangential equation of motion.[1]

\begin{aligned}ml\ddot\theta&=-mg\sin\theta\\\ddot\theta+\frac gl\sin\theta&=0\end{aligned}

Mass cancels from this model’s exact equation of motion.[1]

For small angles in radians, $\sin\theta$ is approximately θ. The equation becomes $\ddot\theta+\frac gl\theta=0$ . Differentiating θ₀cos(ω₀t + φ) twice gives −ω₀²θ, so $\omega_0^2=\frac gl$ . A full oscillation advances phase by 2π. Crossing the midpoint alone is not a full period: the same position must be reached with the same direction of motion. Centripetal force instead concerns the radial resultant: it keeps the mass on its circular path but does not supply the tangential restoring component.

\begin{aligned}\omega_0&=\sqrt{\frac{g}{l}}\\{} T_0&=2\pi\sqrt{\frac{l}{g}}\\{} f_0&=\frac{1}{T_0}\end{aligned}

Subscript 0 denotes the small-angle period here.

How small must the release angle be for the required accuracy?

There is no sharp cutoff at 10°. Accuracy decreases continuously as maximum angle θ₀ grows. For ideally constrained motion, the leading period correction is $\frac T{T_0}\approx1+\frac{\theta_0^2}{16}$ , with θ₀ in radians. At 8° this leading term is about 0.122%, and at 10° about 0.190%. This is period error, not the relative error of $\sin\theta$ ≈ θ; they are different quantities.[4]

\frac{T}{T_0}=1+\frac{\theta_0^2}{16}+\frac{11\theta_0^4}{3072}+\cdots

The series is a small-amplitude approximation; convert degrees to radians first.[4]

When deriving g from T, period error enters quadratically: $g_{\mathrm{estimated}}=\frac{4\pi^2l}{T^2}$ . A longer actual period therefore produces an underestimated g when using the small-angle equation. Errors in centre-of-mass length, timing, pivot friction and string elasticity add to this. A small angle alone does not guarantee a precise measurement.

Larger amplitude lengthens the period

QUANTA STUDYquanta-study.de

T_0

is the small-angle period. The solid curve uses the complete elliptic integral; the dashed curve uses only

1+\frac{\theta_0^2}{16}

, with angle in radians. Release from rest with a taut string is consistent over the range up to

60^\circ

.[2][3]

Free with a Quanta account. Your download stays ready after registration.

Explore further: the exact large-angle period

For release from rest at θ₀, energy conservation gives $\frac12ml^2\dot\theta^2+mgl(1-\cos\theta)=mgl(1-\cos\theta_0)$ . Integrating gives travel time between turning point and midpoint. Four equal quarter motions give the period. Substitution $\sin\frac\theta2=\kappa\sin u$ with $\kappa=\sin\frac{\theta_0}2$ produces the complete elliptic integral of the first kind.[2][3]

\begin{aligned}\dot\theta^2&=\frac{2g}l(\cos\theta-\cos\theta_0)\\\frac T4&=\sqrt{\frac l{2g}}\int_0^{\theta_0}\frac{\mathrm d\theta}{\sqrt{\cos\theta-\cos\theta_0}}\end{aligned}

The positive square root gives elapsed time for a quarter path.

\begin{aligned}T&=4\sqrt{\frac lg}\,K(\kappa)\\{} K(\kappa)&=\int_0^{\frac{\pi}{2}}\frac{du}{\sqrt{1-\kappa^2\sin^2u}}\\{} \kappa&=\sin(\frac{\theta_0}{2})\end{aligned}

κ is the modulus; some software instead expects parameter κ².[2][3]

As θ₀ → 0, K → π/2 and the small-angle equation returns. Period increases with angle. The mathematical limit T → ∞ as θ₀ → π assumes a maintained rigid circular constraint. A simple string cannot push a body released from rest above the horizontal outward and becomes slack there. That range must not be presented as an ordinary string-pendulum oscillation.

How do you find pendulum length or g from the measured period?

\begin{aligned}g&=\frac{4\pi^2l}{T_0^2}\\{} l&=\frac{gT_0^2}{4\pi^2}\\{} [\frac{l}{g}]&=\mathrm{s^2}\end{aligned}

Use metres for l, m/s² for g and seconds for T.

g is local gravitational acceleration; 9.81 m/s² is a common approximation, not a universal constant. Quadrupling length doubles the small-angle period. Mass cancels because weight and inertia are both proportional to m. Real aerodynamic effects can nevertheless vary with mass for the same geometry. When timing many periods, count full cycles and divide total time by their number; using half cycles systematically distorts the result.

How do you determine g from 30 complete oscillations?

The pendulum measurement determines g from length and mean period. Free fall offers another ideal measurement approach using displacement and elapsed time; each method needs its own initial conditions and error checks. The comparison below remains a pendulum measurement.[1][5]

Evaluating a small-angle measurement

Given: Centre-of-mass length l = 0.900 m; 30 periods in 57.12 s; initial angle 2°.
Find: Estimate g and assess the angle contribution.

Approach: The timing series gives a full period; then invert the small-angle formula and estimate amplitude bias.
$\begin{aligned}T&=\frac{t_{\mathrm{series}}}{N}\\g_{\mathrm{est}}&=\frac{4\pi^2l}{T^2}\\\frac{T-T_0}{T_0}&\approx\frac{\theta_0^2}{16}\end{aligned}$
Divide elapsed time by 30.
$T=\frac{57.12}{30}=1.904\,\mathrm s$
Apply the small-angle equation.
$g_{\mathrm{est}}=\frac{4\pi^2\cdot0.900}{1.904^2}=9.80095\,\mathrm{m/s^2}$
Estimate the leading angle contribution to period.
$\begin{aligned}\theta_0&=\frac{2\pi}{180}\\{} \frac{T-T_0}{T_0}\approx\frac{\theta_0^2}{16}&=0.00007615\end{aligned}$

Result: The measurement gives about 9.80 m/s². Angle increases ideal period by about 0.0076% and lowers the uncorrected g estimate by roughly 0.0152%.

Check: Substituting 9.80095 m/s² and 0.900 m returns 1.904 s. Extra digits are calculation checks; without length and timing uncertainties they do not establish matching measurement accuracy.

How much does a 60-degree amplitude lengthen the ideal period?

Period and speed at a large release angle

Given: l = 1.20 m, g = 9.81 m/s², θ₀ = 60°, released from rest; ideal taut string pendulum without losses.
Find: Compare exact model period with $T_0$ and find bottom speed.

Approach: Use the exact ideal period at large initial angle. Find speed independently from energy conservation.
$\begin{aligned}\kappa&=\sin\frac{\theta_0}{2}\\T&=4\sqrt{\frac lg}\,K(\kappa)\\v_{\mathrm{bottom}}&=\sqrt{2gl(1-\cos\theta_0)}\end{aligned}$
Calculate the small-angle reference and elliptic modulus.
$\begin{aligned}T_0&=2\pi\sqrt{\frac{1.20}{9.81}}=2.197536\,\mathrm s\\{} \kappa&=\sin30^\circ=0.5\end{aligned}$
Evaluate K for modulus 0.5.
$\begin{aligned}K(0.5)\approx1.68575035\\{} T&=4\sqrt{\frac{1.20}{9.81}}\,K(0.5)=2.358356\,\mathrm s\end{aligned}$
Convert height loss into kinetic energy.
$v_{\mathrm{bottom}}=\sqrt{2gl(1-\cos\theta_0)}=\sqrt{11.772}=3.43103\,\mathrm{m/s}$

Result: T ≈ 2.358 s is 7.318% above $T_0$ . Bottom speed is about 3.431 m/s.

Check: Kinetic energy per kg is $\frac{v^2}{2}=5.886\,\mathrm{J\,kg^{-1}}$ , equal to $gl(1-\cos60^\circ)$ . Tension is $0.5mg$ at release and $2mg$ at the bottom, both positive, so the circular constraint is consistent.

Model limits and real pendulums

The simple period equation needs small angles, constant string length, a fixed support and weak friction. An extended body changes rotational dynamics: a physical pendulum has small-angle period $2\pi\sqrt{\frac{I}{mgd}}$ , where I is about the pivot axis and d is centre-of-mass distance. A moving support can drive the pendulum; string elasticity and spatial motion add degrees of freedom. For a linear spring oscillator, natural period instead depends on the ratio of effective mass to spring stiffness. Its mass does not cancel as it does for an ideal simple pendulum.[1][6]

The large-angle calculation is exact within the ideal nonlinear model, not automatically for a real experiment. For a string released from rest, tension stays positive for 0 < θ₀ < 90°. At 90° initial tension is zero; above it the required tension is negative and the string cannot hold the assumed path. Energy conservation alone does not replace this constraint.

Why do successive zero crossings span only half a period?

Timing must match both position and direction of motion. This example shows why successive crossings of the midpoint must not be counted as full periods.[1]

Two zero crossings need not make a full period

Given: An ideal small-angle pendulum crosses the lowest point at 0.00 s moving right, at 0.70 s moving left, and at 1.40 s moving right again. Local g=9.81 m/s². Assume small angles, negligible damping, a point bob, and a massless taut string.
Find: Determine full period and pendulum length; compare an analysis that uses the half-period.

Approach: the same position and direction of motion mark a complete cycle. Rearrange the small-angle period for length.
$\begin{aligned}T_0&=t_{\mathrm{right,2}}-t_{\mathrm{right,1}}\\l&=g\left(\frac{T_0}{2\pi}\right)^2\end{aligned}$
The full period is twice the spacing of successive zero crossings.
$\begin{aligned}T_0&=1.40\,\mathrm s\\l&=9.81\left(\frac{1.40}{2\pi}\right)^2\,\mathrm m\\&\approx0.487041\,\mathrm m\\l_{\mathrm{wrong}}&=9.81\left(\frac{0.70}{2\pi}\right)^2\,\mathrm m\\&\approx0.121760\,\mathrm m=\frac l4\end{aligned}$

Result: The full period is 1.40 s and model length approximately 0.487 m. Using 0.70 s as a full period gives only one quarter of that length.

Check: The correct length gives $2\pi\sqrt{\frac l{9.81\,\mathrm{m/s^2}}}=1.40\,\mathrm s$ . Direction at the middle crossing is opposite; this identifies the half-period mistake without additional instruments.

Five solved exercises

Calculate · Foundation

l = 0.490 m and g = 9.81 m/s²; small angle. Find $T_0$ .

Show hint

Use length to the centre of mass.

Learning goal: Calculate small-angle period from length.

Show solution and calculation

$T_0=2\pi\sqrt{\frac{0.490}{9.81}}\,\mathrm s\approx1.40425\,\mathrm s$ .
Reversing $g=\frac{4\pi^2l}{T_0^2}$ gives $9.81\,\mathrm{m\,s^{-2}}$ .

\begin{aligned}T_0&=2\pi\sqrt{\frac lg}\\&=2\pi\sqrt{\frac{0.490}{9.81}}\,\mathrm s\approx1.40425\,\mathrm s\end{aligned}

Rearrange · Foundation

What length gives a small-angle period of 1.00 s at g = 9.81 m/s²?

Show hint

Solve for l; do not use a half period.

Learning goal: Infer pendulum length from full period.

Show solution and calculation

$l=\frac{9.81}{4\pi^2}\,\mathrm m\approx0.248490\,\mathrm m$ .
One quarter metre is a reasonable rough approximation, not the exact model value.

\begin{aligned}T_0^2&=\frac{4\pi^2l}g\\l&=\frac{gT_0^2}{4\pi^2}=\frac{9.81\cdot1.00^2}{4\pi^2}\,\mathrm m\\&\approx0.248490\,\mathrm m\end{aligned}

Calculate · Application

Estimate the leading relative period correction at 8° release.

Show hint

Use $8^\circ=\frac{8\pi}{180}\,\mathrm{rad}$ .

Learning goal: Estimate leading amplitude correction using radians.

Show solution and calculation

$\frac{\theta_0^2}{16}\approx0.00121847=0.121847\%$ .
The fourth-order term contributes only about 0.000136%.
There is no sharp validity jump at 10°.

\begin{aligned}\theta_0&=8\frac\pi{180}\,\mathrm{rad}\\\frac{\Delta T}{T_0}&\approx\frac{\theta_0^2}{16}\\&\approx0.00121847=0.121847\%\end{aligned}

Interpret · Foundation

Three ideal string pendulums swing at small angles at the same location. A:l=0.25 m, m=50 g; B:l=0.25 m, m=200 g; C:l=1.00 m, m=50 g. Compare full periods without substituting a numerical g.

Show hint

Mass cancels from the angular equation; length appears under a square root.

Learning goal: Distinguish mass independence from square-root length scaling.

Show solution and calculation

Approach: A and B have equal length and therefore equal period despite fourfold mass.
C has fourfold length and twice the period.

Check: the squared period ratio equals length ratio 4. This assumes common local gravity, small angles and ideal string models; differing real air damping is outside the comparison.

\begin{aligned}\frac{T_B}{T_A}&=1\\\frac{T_C}{T_A}&=\sqrt{\frac{l_C}{l_A}}=\sqrt4=2\end{aligned}

[1]

Error analysis · Challenge

An ideal taut string pendulum is released from rest at 60°. The exact nonlinear calculation gives $\frac{T}{T_0}=1.073182007$ . An analysis inserts measured T into $g_{\mathrm{meas}}=\frac{4\pi^2l}{T^2}$ and claims true g. Find $\frac{g_{\mathrm{meas}}}{g}$ and the error direction.

Show hint

Measured period exceeds the small-angle period.

Learning goal: Avoid using a large-amplitude period as $T_0$ when measuring g.

Show solution and calculation

Approach: true $g=\frac{4\pi^2l}{T_0^2}$ , so $\frac{g_{\mathrm{meas}}}{g}=\left(\frac{T_0}{T}\right)^2\approx0.868267$ .
Gravity is underestimated by about 13.1733%.

Check: the supplied longer period necessarily lowers its reciprocal square. Analysis must include amplitude correction or measure smaller angles. The given ratio uses the elliptic model rather than only the leading series term.

\frac{g_{\mathrm{meas}}}{g}=\frac{1}{(T/T_0)^2}=\frac{1}{1.073182007^2}\approx0.868267

[2][3][1]

In brief

Measure a complete cycle and length to the centre of mass. The square-root expression applies at small angles; its accuracy depends on actual amplitude. At larger angles the ideal period increases. String tautness, pivot, air, and bob dimensions limit the point-pendulum model.[1]

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Clarified small-angle period, centre-of-mass length and mass cancellation; exposed existing measurement, large-angle and half-period examples and distinguished radial force, free-fall measurement and the spring oscillator.
October 1, 2026Clarified complete cycles and accuracy conditions; added a half-period-error example and an exact amplitude–period curve. Preserved measurement and large-angle examples and all five exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Simple Pendulum (Period). Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/fadenpendel

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Frequently asked questions about Simple Pendulum (Period)

How do you calculate the period of a simple pendulum?+

For small angles use $T_0=2\pi\sqrt{\frac lg}$ . Length l is the distance from the fixed pivot to the centre of mass, not only visible string length; g is local gravitational acceleration. Enter metres and $\frac{m}{s^{2}}$ to obtain seconds. For $l = 1.00 m$ and $g =\frac{ 9.81 m}{s^{2}}$ , $T_0$ ≈ 2.006 s. One complete back-and-forth motion is a period; merely crossing the midpoint is insufficient for counting full cycles. At larger angles the ideal actual period is longer and needs an amplitude correction. Reversing $g=\frac{4\pi^2l}{T_0^2}$ checks the calculation under the same small-angle assumptions.

Why does the period not depend on the mass?+

Tangential weight is $-mg\sin\theta$ and tangential inertia is $ml\ddot\theta$ . Both are proportional to the same mass, so m cancels from the ideal equation of motion. Equal length, equal local g and equal initial conditions therefore give the same ideal motion for different masses. This applies to the nonlinear angle equation as well as the small-angle version. It does not exclude real effects: for equal body shape, air resistance relative to inertia can differ. An extended body also needs its moment of inertia; changing mass distribution can alter physical-pendulum period even when total mass itself cancels for a fixed geometry.

How do you determine g with a simple pendulum?+

Measure centre-of-mass length l and time many complete periods at a small release angle. Divide total time by cycle count, then use $g\approx\frac{4\pi^2l}{T^2}$ . For $l = 0.900 m$ and 30 periods in 57.12 s, $T = 1.904 s$ and $g \approx{}\frac{ 9.801 m}{s^{2}}$ . Even at 2°, angle increases ideal period by roughly 0.0076%; without correction g is underestimated by about 0.0152%. That control value does not replace measurement uncertainty. Check centre-of-mass location, support stability, string extension, friction and consistent counting. Matching position and direction of motion identifies a complete cycle.

What changes when you quadruple the pendulum length?+

At unchanged g, quadrupling length doubles small-angle period because $T_0\propto\sqrt l$ . Frequency consequently halves. Reversing $g=\frac{4\pi^2l}{T_0^2}$ gives the same value because numerator and denominator both quadruple. For the same maximum angle, maximum height $l(1-\cos\theta_0)$ quadruples and bottom speed doubles. Equal angular amplitude must not be confused with equal sideways displacement. With unchanged angle, the period scaling also holds in the ideal elliptic large-angle model; if angle changes, its amplitude factor must additionally be included. These comparisons assume the same ideal constraint and negligible losses in both configurations.

When does the pendulum formula stop being valid?+

$T=2\pi\sqrt{\frac lg}$ is a small-angle approximation, not an exact equation for every release angle. Acceptable angle depends on required accuracy; at 10° the leading period correction is about 0.190%, with no sudden change of motion type. At 60° the complete ideal period is about 7.318% greater than $T_0$ . String elasticity, moving support, substantial friction or an extended body can also change the model. The string must stay taut: release from rest above the horizontal cannot maintain the circular path. An elliptic integral alone does not repair that missing constraint. Always separate approximation error from failure of the physical model.

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[1]OpenStax, University Physics Vol. 1, §15.4 Pendulums

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Simple pendulum

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Simple pendulum period calculator

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How do you calculate with Simple Pendulum (Period)?

A complete return journey counts as one period

Length, angle and the constraint

How does the small-angle period arise, and why does mass cancel?

How small must the release angle be for the required accuracy?

Larger amplitude lengthens the period

Explore further: the exact large-angle period

How do you find pendulum length or g from the measured period?

How do you determine g from 30 complete oscillations?

Evaluating a small-angle measurement

How much does a 60-degree amplitude lengthen the ideal period?

Period and speed at a large release angle

Model limits and real pendulums

Why do successive zero crossings span only half a period?

Two zero crossings need not make a full period

Five solved exercises

l = 0.490 m and g = 9.81 m/s2; small angle. Find T0T_0T0​.

What length gives a small-angle period of 1.00 s at g = 9.81 m/s2?

Estimate the leading relative period correction at 8° release.

Three ideal string pendulums swing at small angles at the same location. A:l=0.25 m, m=50 g; B:l=0.25 m, m=200 g; C:l=1.00 m, m=50 g. Compare full periods without substituting a numerical g.

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Retain Simple Pendulum (Period) for exams

l = 0.490 m and g = 9.81 m/s²; small angle. Find $T_0$ .

What length gives a small-angle period of 1.00 s at g = 9.81 m/s²?