18 Jun 2025
Simple, Compound, and Torsional Pendulums
Small-oscillation equations and periods of three mechanical systems executing SHM.
The following systems execute simple harmonic motion only when their restoring force or torque is linear in the displacement.
Simple pendulum
Consider a point mass $m$ suspended from a fixed support by a light, inextensible string of length $l$. Let $\theta$ be the angular displacement from the downward vertical. The tangential component of gravity is opposite to increasing $\theta$:
\[F_\theta=-mg\sin\theta.\]The tangential acceleration is $a_\theta=l\ddot\theta$, so
\[ml\ddot\theta=-mg\sin\theta.\]After cancellation of $m$ and division by $l$, the exact equation is
\[\ddot\theta+\frac gl\sin\theta=0.\]For a small angular amplitude measured in radians, $\lvert\theta\rvert\ll1$ and $\sin\theta\simeq\theta$. The equation then becomes
\[\ddot\theta+\frac gl\theta=0.\]Comparison with $\ddot\theta+\omega_0^2\theta=0$ gives
\[\omega_0=\sqrt{\frac gl}, \qquad \boxed{T=2\pi\sqrt{\frac lg}}.\]The finite-amplitude pendulum is periodic but is not exactly simple harmonic; the boxed period is the small-angle result.
Compound pendulum
A compound pendulum is a rigid body of mass $m$ oscillating about a fixed horizontal axis. Let its centre of mass be a distance $d>0$ from the axis, and let $I$ be its moment of inertia about that axis. Gravity produces the restoring torque
\[\tau=-mgd\sin\theta.\]Using $I\ddot\theta=\tau$,
\[I\ddot\theta=-mgd\sin\theta.\]For $\lvert\theta\rvert\ll1$,
\[\ddot\theta+\frac{mgd}{I}\theta=0.\]Therefore
\[\omega_0=\sqrt{\frac{mgd}{I}}, \qquad \boxed{T=2\pi\sqrt{\frac{I}{mgd}}}.\]If $I_{\rm cm}=mk_g^2$, where $k_g$ is the radius of gyration about the centre-of-mass axis parallel to the pivot axis, the parallel-axis theorem gives
\[I=I_{\rm cm}+md^2=m(k_g^2+d^2).\]Substitution into the period yields
\[T=2\pi\sqrt{\frac{k_g^2+d^2}{gd}}.\]Define the equivalent simple-pendulum length $l_{\rm eq}$ by $T=2\pi\sqrt{l_{\rm eq}/g}$. Equating the two expressions for $T^2$ gives
\[\frac{l_{\rm eq}}g=\frac{I}{mgd},\]and hence
\[\boxed{l_{\rm eq}=\frac{I}{md}=d+\frac{k_g^2}{d}}.\]Torsional pendulum
Let a body of moment of inertia $I$ be suspended by a wire. Within the wire’s linear elastic range, a small rotation $\theta$ produces the restoring torque
\[\tau=-\kappa\theta,\]where $\kappa>0$ is the torsional constant. The rotational equation of motion is
\[I\ddot\theta=-\kappa\theta,\]or
\[\ddot\theta+\frac\kappa I\theta=0.\]Thus
\[\omega_0=\sqrt{\frac\kappa I}, \qquad \boxed{T=2\pi\sqrt{\frac I\kappa}}.\]
The three small-oscillation equations, periods, and compound-pendulum equivalent length are verified in the Maxima worksheet; every printed residual is zero.
Discussion