Connect pendulum length, gravity and period

Measure length from the pivot

A simple pendulum model has a point-like bob suspended from a fixed pivot by a massless, inextensible string. Its length L is the distance from the pivot to the bob’s center. The release angle theta is measured from the downward vertical, not the horizontal. The diagram keeps a fixed distance scale as you change length.

Set release angle to 10° and compare lengths 0.5 m and 2 m. Before reading the timing values, predict whether a fourfold length change makes a fourfold period change. Count a complete cycle as returning to the same release configuration and phase; travel from one turning point to the opposite turning point is only half a cycle.

Find the restoring part of gravity

Gravity acts downward with force magnitude mg, where m is bob mass and g = 9.81 m/s² in this model. The component along the circular path is −mg sin(theta). The negative sign means it restores a positive displacement toward the downward equilibrium. Tension acts along the string, so it does not contribute to the tangential restoring force.

Tangential acceleration is L times angular acceleration. Newton’s law gives mL times angular acceleration = −mg sin(theta). Mass cancels, giving angular acceleration = −(g/L) sin(theta). This cancellation is why ideal pendulum timing does not depend on bob mass. The lab fixes mass at 1 kg; it does not offer a separate mass experiment.

Use the small-angle model

For small angles measured in radians, sin(theta) is close to theta. The equation then has the harmonic form, with angular frequency √(g/L), giving Tsmall = 2π√(L/g). Tsmall is the approximate full period in seconds. L is in metres, and g is in metres per second squared. This is an approximation to the full sine law, not a general amplitude-independent rule.

At 1 m, Tsmall ≈ 2.006067 s. At 0.5 m it is about 1.418503 s; at 2 m about 2.837007 s. The length ratio between the last two cases is four, while the period ratio is two. Frequency, which is one divided by period, changes in the opposite direction.

The nonlinear reference shown beside the estimate retains sin(theta). At 10° it is slightly larger: approximately 2.009893 s at 1 m. Both models retain square-root length scaling when the release angle and gravity are fixed. Changing length does not remove the angle approximation’s fractional error.

The OpenStax pendulum reference explains the ideal force model and small-angle timing relation.

Connect geometry to energy

At release from rest, the bob is raised by h = L(1 − cos(theta)) above the lowest point. This follows from the vertical projection L cos(theta). At 1 m and 10°, h ≈ 0.015192 m. The initial gravitational energy above the bottom is mgh ≈ 0.149036 J for the 1 kg bob.

With no drag or pivot friction, this becomes kinetic energy at the bottom: mv²/2 = mgh. Thus bottom speed is v = √(2gh) ≈ 0.545960 m/s. Mass cancels again. The full geometric height is used here even when the period is approximated; do not mix a release angle in degrees directly into a formula requiring radians.

At a fixed release angle, quadrupling length quadruples height and doubles bottom speed. That higher speed does not imply a shorter period: the bob also travels along a longer arc. Use the full timing relation rather than speed alone.

Transfer using a changed gravity model

An ideal small-angle pendulum has period 2 s at a particular length and gravitational acceleration. Predict its period if the length doubles while gravity halves. Then explain whether doubling bob mass would change that prediction.

Check the pendulum transfer

The ratio L/g becomes four times larger, so period doubles to 4 s. Bob mass cancels from the ideal equation, so doubling it does not change this prediction. A real bob’s dimensions, air drag or suspension could change when mass is changed; those effects are outside the point-bob model.