18 Jun 2025

Simple, Compound, and Torsional Pendulums

Small-oscillation equations and periods of three mechanical systems executing SHM.

oscillations simple-pendulum compound-pendulum torsional-pendulum

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

\[\boxed{l_{\rm eq}=\frac{I}{md}=d+\frac{k_g^2}{d}}.\]

For a body with fixed $k_g$, the equivalent length is least when

\[\frac{d l_{\rm eq}}{d\,d} =1-\frac{k_g^2}{d^2}=0.\]

Since $d>0$, this gives $d=k_g$. Moreover,

\[\frac{d^2 l_{\rm eq}}{d\,d^2}=\frac{2k_g^2}{d^3}>0,\]

so the stationary value is a minimum. Therefore

\[\boxed{l_{\rm eq,min}=2k_g}, \qquad \boxed{T_{\min}=2\pi\sqrt{\frac{2k_g}{g}}}.\]

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

\[\omega_0=\sqrt{\frac\kappa I}, \qquad \boxed{T=2\pi\sqrt{\frac I\kappa}}.\]

Solved Problems

1. Uniform rod as a compound pendulum

Problem. A uniform rod of length $1.20\ \mathrm{m}$ is pivoted about a horizontal axis through a point $0.20\ \mathrm{m}$ from one end. Find its equivalent length and small-oscillation period using $g=9.80\ \mathrm{m\,s^{-2}}$.

Solution. The centre of mass is $0.60\ \mathrm{m}$ from the same end, so

\[d=0.60-0.20=0.40\ \mathrm{m}.\]

For a uniform rod,

\[I_{\rm cm}=\frac{mL^2}{12}.\]

The parallel-axis theorem gives

\[I=m\left(\frac{L^2}{12}+d^2\right) =m(0.120+0.160) =0.280m.\]

Hence

\[l_{\rm eq}=\frac{I}{md} =\frac{0.280}{0.40} =0.700\ \mathrm{m},\]

and

\[T=2\pi\sqrt{\frac{l_{\rm eq}}{g}} =2\pi\sqrt{\frac{0.700}{9.80}} =1.679\ \mathrm{s}.\]

Checks. The mass cancels, and $I/(md)$ has units of length. Here $d>0$, so the gravitational torque is restoring for small positive $\theta$. The equivalent length lies within the rod but need not equal the pivot-to-centre-of-mass distance. As $d\to0^+$, the restoring torque vanishes and the period tends to infinity.

2. Added ring on a torsional platform

Problem. A torsional platform has moment of inertia $I_0=0.020\ \mathrm{kg\,m^2}$ and period $T_0=1.00\ \mathrm{s}$. A thin ring of mass $1.00\ \mathrm{kg}$ and radius $0.200\ \mathrm{m}$ is placed concentrically on it. Find the torsional constant and the new period.

Solution. From $T_0=2\pi\sqrt{I_0/\kappa}$,

\[\kappa=\frac{4\pi^2I_0}{T_0^2} =4\pi^2(0.020) =0.7896\ \mathrm{N\,m\,rad^{-1}}.\]

The ring’s moment of inertia about its symmetry axis is

\[I_{\rm ring}=MR^2 =(1.00)(0.200)^2 =0.040\ \mathrm{kg\,m^2}.\]

Thus $I=I_0+I_{\rm ring}=0.060\ \mathrm{kg\,m^2}$, and

\[T=2\pi\sqrt{\frac{I}{\kappa}} =T_0\sqrt{\frac{I}{I_0}} =\sqrt3\ \mathrm{s} =1.732\ \mathrm{s}.\]

Checks. $\kappa$ has units $\mathrm{N\,m\,rad^{-1}}$, so $I/\kappa$ has units $\mathrm{s^2}$. The torque $-\kappa\theta$ opposes the angular displacement. Adding positive moment of inertia must lengthen the period, as the result does. If the ring mass tends to zero, $I\to I_0$ and $T\to T_0$.

Descriptive Questions

1. Derive the exact equation of a simple pendulum and obtain its small-angle period, stating precisely where the SHM approximation enters.
2. Derive the period and equivalent length of a compound pendulum using the parallel-axis theorem.
3. Show that the compound-pendulum period is minimum when the pivot-to-centre-of-mass distance equals the radius of gyration.
4. Derive the equation and period of a torsional pendulum and explain the physical meaning and SI units of the torsional constant.

Numerical Problems

1. Find the small-angle period of a simple pendulum of length $0.810\ \mathrm{m}$ at $g=9.80\ \mathrm{m\,s^{-2}}$.

   Final answer: $T=1.806\ \mathrm{s}$.

2. A simple pendulum has period $1.50\ \mathrm{s}$ where $g=9.80\ \mathrm{m\,s^{-2}}$. Find its length.

   Final answer: $l=g(T/2\pi)^2=0.5585\ \mathrm{m}$.

3. A compound pendulum has $k_g=0.300\ \mathrm{m}$ and period $T$ when its pivot is $d_1=0.200\ \mathrm{m}$ from the centre of mass. Find the other positive pivot distance $d_2\ne d_1$ that gives the same small-oscillation period, and find that period at $g=9.80\ \mathrm{m\,s^{-2}}$.

   Final answer: Equal equivalent lengths give $d_1d_2=k_g^2$, so $d_2=0.450\ \mathrm{m}$; $l_{\rm eq}=0.650\ \mathrm{m}$ and $T=\pi\sqrt{13}/7=1.618\ \mathrm{s}$.

4. A rigid body has $k_g=0.300\ \mathrm{m}$. Find the pivot distance for minimum period and the minimum period at $g=9.81\ \mathrm{m\,s^{-2}}$.

   Final answer: $d=0.300\ \mathrm{m}$ and $T_{\min}=1.554\ \mathrm{s}$.

5. Find the period of a torsional pendulum with $I=0.012\ \mathrm{kg\,m^2}$ and $\kappa=0.300\ \mathrm{N\,m\,rad^{-1}}$.

   Final answer: $T=1.257\ \mathrm{s}$.

6. A torsional pendulum has $I=0.010\ \mathrm{kg\,m^2}$, $\kappa=0.400\ \mathrm{N\,m\,rad^{-1}}$, and angular amplitude $0.100\ \mathrm{rad}$. Find its maximum angular speed.

   Final answer: $\dot\theta_{\max}=\theta_0\sqrt{\kappa/I}=0.6325\ \mathrm{rad\,s^{-1}}$.

The small-oscillation identities and all eight worked answers are independently checked in the Maxima worksheet; every printed residual or check is zero.

References

1. Pendulum — Wikipedia.
2. OpenStax, University Physics Volume 1, Section 15.4: Pendulums.
3. John R. Taylor, Classical Mechanics, University Science Books (2005), Chapter 5.