The Simple Pendulum and Effective Gravity

A pendulum follows an arc, but its angular motion can be harmonic

A stone or compact bob hung from a thread provides an accessible oscillator. With a length near one metre, a small complete swing takes about two seconds. Galileo’s reported comparison of a chandelier’s swings using his pulse illustrates how a repeated rhythm can provide a clock. To investigate, hang a small mass from a fixed support using an inextensible light thread, keep the angular release small and release without pushing. Time several complete cycles between the same extreme and divide by the number of cycles. Changing the length reveals a systematic period change; using a large amplitude tests the limits of the simple model.

An ideal simple pendulum consists of a point-like bob of mass m on a massless, inextensible string of length L from a fixed rigid support. L is the distance from pivot to the bob’s centre, not just the visible free thread length. Motion is confined to a vertical plane, air resistance and pivot friction are neglected, and gravity g is uniform over the path. Choose the angular displacement θ from the downward vertical, positive toward the displaced side. The bob moves on a circular arc because the string fixes its radius, but its speed along that arc is not constant.

The two actual forces are gravity mg downward and string tension S along the string toward the pivot. Resolve gravity into a radial component mg cos θ away from the pivot and a tangential component −mg sin θ toward the lowest point. The radial net force S − mg cos θ supplies the centripetal acceleration v²/L. The tangential component changes the speed. Thus S − mg cos θ = mv²/L; tension is generally not mg cos θ. The radial acceleration is Lθ̇², involving the bob’s changing angular speed θ̇, rather than a constant SHM frequency ω squared times L.

Torque gives a direct route to angular acceleration

Tension acts along the radius, so its moment arm about the pivot is zero. Gravity’s tangential component has moment arm L and produces restoring torque τ = −mgL sin θ. Newton’s rotational equation is τ = Iα, where I is the moment of inertia about the pivot and α = d²θ/dt². Therefore α = −(mgL/I) sin θ. For the point bob and massless string, I = mL², giving the exact ideal angular equation θ″ = −(g/L) sin θ. This is restoring, but not exactly linear in θ.

To connect this equation with SHM, use the expansion sin θ = θ − θ³/3! + θ⁵/5! − ⋯, where θ is in radians and 3! = 3 × 2 × 1. When |θ| is small compared with one radian, the correction θ³/6 is small compared with θ, so sin θ ≈ θ. Substitution makes angular acceleration proportional to −θ. This is why a small pendulum swing is approximately SHM and a large swing is not. Replacing a degree number directly into sin θ ≈ θ would be wrong: 10° means about 0.1745 rad, not 10 rad.

The two numerical columns are close at small angles, but the approximation has a measurable error: at 20° the force ratio sin θ/θ is about 0.980, a difference of about 2%. This is a comparison of the restoring terms at that displacement, not the same percentage error in the complete period. There is no universal sharp “small-angle cutoff”; choose amplitudes according to the accuracy required. The ideal small-angle period’s independence of amplitude is an approximation for a pendulum.

The bob mass cancels because gravity and rotational inertia both grow in proportion to m. A longer pendulum has a longer period, scaling as √L. A larger g gives a shorter period, scaling as 1/√g. The arc displacement s = Lθ obeys the same linear equation s″ ≈ −(g/L)s, so the small oscillation is harmonic in arc length as well as angle. The bob’s full vector acceleration still includes a radial part; it is the tangential or angular equation that identifies the SHM.

Effective gravity in a turning car

Consider a car travelling at steady speed V around a circle of radius R. Its inward centripetal acceleration is a₀ = V²/R. In the car’s local frame, an approximately uniform outward inertial acceleration accompanies ordinary downward gravity. Their perpendicular vector sum defines an effective gravity with magnitude geff = √(g² + a₀²), directed downward and outward. The pendulum’s new equilibrium string aligns with that direction, tilted outward by angle β satisfying tan β = a₀/g. Effective gravity is larger in magnitude than g; it is not g − V²/R because the two contributions are perpendicular.

For small oscillations in the vertical radial plane about that tilted equilibrium, use η as the small angle from the effective vertical. The effective restoring torque is −mgeff L sin η ≈ −mgeff Lη. With I = mL², η″ ≈ −(geff/L)η. Thus the period is 2π√(L/geff). This is the local effective-gravity approximation intended by the circular-car problem: the pendulum length is small compared with R and the car’s angular rotation is slow enough compared with the pendulum oscillation that frame-rotation corrections can be neglected. The approximation should be stated rather than presented as an exact theorem for any turning rate.

Check the formula in limits. As V tends to zero, geff tends to g and the ordinary pendulum period is recovered. As the steady horizontal acceleration increases within the approximation, the resultant restoring field strengthens and the period decreases. Both g² and V⁴/R² have units m² s⁻⁴; after the inner square root geff has units m s⁻², so the outer square root has units seconds. A formula using a scalar subtraction would fail the vector geometry and could predict an imaginary period unnecessarily.

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