In short

A compound pendulum (also called a physical pendulum) is any rigid body that oscillates about a fixed horizontal axis under gravity. Let I be the moment of inertia about that axis, m the total mass, and d the distance from the axis to the centre of mass. For small angular displacements, the time period is

T = 2\pi \sqrt{\frac{I}{m g d}}.

The equivalent simple pendulum length — the length of a simple pendulum with the same period — is L_\text{eq} = I/(m d). A torsional pendulum is a rigid body suspended by a wire or thread that twists through angle \theta against a restoring torque \tau = -\kappa\theta, where \kappa is the wire's torsional stiffness. The time period is

T = 2\pi\sqrt{\frac{I}{\kappa}}.

In both cases the formula is the rotational analogue of T = 2\pi\sqrt{m/k} from the spring-mass: moment of inertia replaces mass, and the restoring-torque coefficient (mgd or \kappa) replaces the spring constant. Master this once, and every rigid-body oscillator — grandfather clock, balance wheel, Cavendish torsion balance, Kater's reversible pendulum — collapses into the same two-line calculation.

The pendulum swinging inside your nana's old grandfather clock in Lucknow is not a mass on a massless string. It is a metre-long wooden rod with a heavy brass disc at the bottom — and every bit of that rod, not just the disc, contributes to the swinging motion. This is a compound pendulum, and its time period is not the textbook T = 2\pi\sqrt{L/g} from the simple pendulum chapter. You cannot just plug in the length of the rod.

Open up any mechanical wristwatch — a HMT Janata, a Titan, or the quartz-free movement in a Rolex — and you will see a little wheel spinning back and forth, back and forth, about four times per second. It is called a balance wheel, and it is doing the same job the pendulum does in a grandfather clock: providing a steady, repeating rhythm to divide up a second. But it is not swinging under gravity. It is twisting against a tiny hairspring wound around its axle. That is a torsional pendulum, and it obeys a different formula again.

Both systems — the compound pendulum and the torsional pendulum — are rigid-body oscillators. Instead of a point mass moving through space along a line, a whole extended body is rotating around a fixed axis. The equation of motion is not Newton's second law F = ma but its rotational form \tau = I\alpha. The moment of inertia I replaces the mass, and the restoring torque replaces the restoring force. Once you see the parallel, both formulas fall out by the same derivation as the spring-mass and the simple pendulum, just with rotational quantities instead of linear ones. By the end of this article you will be able to write down T for any such system in one line.

The compound pendulum — an extended body swinging under gravity

Take any rigid body and suspend it from a horizontal axis that does not pass through its centre of mass. Give it a small angular nudge and it will swing back and forth. The grandfather-clock pendulum is the canonical example. Other everyday examples: a wooden metre stick hanging from a nail through one end, a uniform ring balanced on a knife edge, a thin bent rod pivoted at a corner.

Compound pendulum geometryA rigid body hangs from a horizontal axis O. The centre of mass C is a distance d below O. The body is tilted by angle θ from the vertical. A gravity vector mg points straight down from C. The component mg sin θ, perpendicular to the line OC, provides the restoring torque. $O$ vertical $C$ (centre of mass) $\theta$ $d$ $mg$
A compound pendulum: a rigid body pivoted at $O$ with centre of mass $C$ a distance $d$ below $O$. When the body is tilted by angle $\theta$, gravity $mg$ acts at $C$ and exerts a restoring torque about $O$. Only the perpendicular component of gravity — $mg\sin\theta$ — contributes to the torque.

Let O be the pivot, C the centre of mass. The line OC has length d. When the body is tilted so that OC makes angle \theta with the vertical, gravity acts at C straight downward with magnitude mg. The component of gravity perpendicular to OC is mg\sin\theta, and this component provides the restoring torque about O:

\tau = -m g d \sin\theta. \tag{1}

Why: torque is force times perpendicular distance from the axis. The perpendicular component of gravity is mg\sin\theta; its line of action passes through C, a distance d from O. The minus sign captures that the torque acts to reduce \theta — it pulls the body back toward the vertical.

The rotational form of Newton's second law says the angular acceleration is torque divided by moment of inertia about the pivot:

I \frac{d^2\theta}{dt^2} = \tau = -m g d \sin\theta,

or

\frac{d^2\theta}{dt^2} = -\frac{m g d}{I}\sin\theta. \tag{2}

So far exact. Now apply the small-angle approximation — the same move that turns the simple pendulum into SHM. For small \theta (in radians), \sin\theta \approx \theta. Equation (2) becomes

\frac{d^2\theta}{dt^2} = -\omega^2 \theta, \qquad \omega^2 = \frac{m g d}{I}. \tag{3}

This is the equation of SHM in the angular variable \theta. Its general solution is \theta(t) = \theta_0 \cos(\omega t + \varphi), and the period is

\boxed{\; T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{I}{m g d}}. \;} \tag{4}

Why: compare with the spring-mass T = 2\pi\sqrt{m/k}. The role of "mass" is played by I (rotational inertia), and the role of "spring constant" is played by mgd (the restoring-torque coefficient — the torque per unit angular displacement). Same equation, same period formula, different physical meanings.

Recovering the simple pendulum

A good formula is not trusted until you have checked it against a known case. The simple pendulum — a point mass m at the end of a massless string of length L — is a compound pendulum with I = mL^2 (all mass at one point at distance L) and d = L (the centre of mass is where the point mass is). Plug in:

T = 2\pi\sqrt{\frac{mL^2}{mg L}} = 2\pi\sqrt{\frac{L}{g}}.

The simple-pendulum formula falls out exactly. This is the sanity check: the compound-pendulum result includes the simple pendulum as a special case.

The equivalent simple pendulum length

The right-hand side of equation (4) can be rewritten:

T = 2\pi\sqrt{\frac{I}{m g d}} = 2\pi\sqrt{\frac{I/(md)}{g}} = 2\pi\sqrt{\frac{L_\text{eq}}{g}},

where

\boxed{\; L_\text{eq} = \frac{I}{m d}. \;} \tag{5}

In other words, every compound pendulum swings with the same period as a simple pendulum of length L_\text{eq} — the equivalent simple pendulum length. If you want to match a grandfather clock's swing, you do not need to know its shape or mass distribution; you need to build a simple pendulum of length L_\text{eq}, and it will keep time identically.

This is more than a trick. It tells you that the only dynamical quantity that matters is the combination I/(md). Two pendulums with the same I/(md) swing at the same rate, regardless of shape.

A uniform rod pivoted at one end

The most-worked-out case. Take a uniform rod of length L and mass m, pivoted at one end. What is the period?

Moment of inertia about the end: I_\text{end} = \tfrac{1}{3} m L^2 (standard result — see Moment of Inertia, or derive from I_\text{cm} = mL^2/12 using the parallel axis theorem). The centre of mass is at the midpoint, so d = L/2. Plug in:

T = 2\pi\sqrt{\frac{mL^2/3}{m g \cdot L/2}} = 2\pi\sqrt{\frac{2L}{3g}}. \tag{6}

The equivalent simple pendulum length is

L_\text{eq} = \frac{I}{md} = \frac{mL^2/3}{m \cdot L/2} = \frac{2L}{3}.

A uniform rod of length L hanging from one end swings with the period of a simple pendulum of length 2L/3. This is faster than a simple pendulum of length L (which would be a point mass at the bottom of the rod) — because the rod's mass is distributed along its length, some of the mass is closer to the pivot and "helps" the swing. A rod of length 1 m has T = 2\pi\sqrt{2/(3 \times 9.8)} \approx 1.64 s — considerably shorter than the 2.01 s period of a simple pendulum of length 1 m.

Animated: uniform rod pendulum with L = 1 m, amplitude 0.3 radA 1 m uniform rod pivoted at its top end swings back and forth under gravity. With θ₀ = 0.3 rad (about 17°), the small-angle SHM approximation is good to within a few percent. The rod's angular position is θ(t) = 0.3 cos(ωt) with ω = √(3g/2L) ≈ 3.83 rad/s, giving T ≈ 1.64 s. The tracked point is the bottom tip of the rod, at position (L sin θ, −L cos θ). pivot
The tip of a 1 m uniform rod pivoted at the top. Angular displacement $\theta(t) = 0.3\cos(\omega t)$ with $\omega = \sqrt{3g/2L} \approx 3.83$ rad/s. Period $T = 2\pi/\omega \approx 1.64$ s — shorter than the 2.01 s period of a simple pendulum of the same length $L$. Click replay to watch again.

Torsional pendulum — twisting against a wire

Now leave gravity behind. Take a rigid body — a disc, a wheel, a dumbbell — and suspend it from a wire or thin rod attached to its centre. Twist it slightly about the axis of the wire, and let go. The wire resists the twist with a restoring torque. The body oscillates rotationally: twisting one way, unwinding, twisting the other, unwinding, repeating.

Torsional pendulum geometryA disc suspended by a thin wire from a fixed ceiling. The disc is rotated by angle θ about the vertical axis of the wire. The wire exerts a restoring torque τ = −κθ that tries to unwind the disc back to its equilibrium orientation. wire, stiffness $\kappa$ $\theta$ $\tau = -\kappa\theta$ disc ($I$)
A torsional pendulum: a disc of moment of inertia $I$ hangs from a wire whose torsional stiffness is $\kappa$ (units: N·m per radian). Twisting the disc through angle $\theta$ produces a restoring torque $\tau = -\kappa\theta$. The disc oscillates about its equilibrium angular position.

The defining relation is Hooke's law for torsion. For a thin elastic wire twisted through a small angle \theta, the restoring torque is proportional to \theta:

\tau = -\kappa \theta. \tag{7}

Here \kappa (the Greek letter "kappa") is the torsional stiffness of the wire, in units of N·m per radian. \kappa depends on the wire's material and its geometry — for a cylindrical wire of length \ell, radius r, and shear modulus G, \kappa = \pi G r^4 / (2\ell). But for this chapter we just treat \kappa as given; the system obeys equation (7) exactly, not approximately. There is no "small angle" required — the linearity of Hooke's law for elastic torsion holds for any twist within the elastic limit (often many full turns for a thin metal wire).

Newton's second law (rotational form):

I \frac{d^2\theta}{dt^2} = -\kappa \theta,

or

\frac{d^2\theta}{dt^2} = -\omega^2 \theta, \qquad \omega^2 = \frac{\kappa}{I}. \tag{8}

This is SHM, exactly. The period is

\boxed{\; T = 2\pi\sqrt{\frac{I}{\kappa}}. \;} \tag{9}

Why: same structure as the compound pendulum, with mgd (the gravitational restoring-torque coefficient) replaced by \kappa (the elastic one). The torsional pendulum is cleaner in one sense — no small-angle approximation is needed — because Hooke's law for torsion is already linear.

Why is the torsional formula exact while the compound-pendulum formula is approximate?

Because the two restoring torques are different shapes. The gravitational restoring torque on a compound pendulum is -m g d \sin\theta — nonlinear in \theta. You have to assume small angles to replace \sin\theta by \theta and get SHM. For large amplitudes the gravitational pendulum's period grows slightly with amplitude (see simple-pendulum for the exact expression using elliptic integrals).

The torsional restoring torque is -\kappa\theta — already linear. The wire obeys Hooke's law across a huge range of angles. So the torsional pendulum's period is exact, independent of amplitude, for any twist that stays within the wire's elastic limit. This is what makes the balance wheel of a mechanical watch so accurate: the period does not drift with how much the mainspring is wound, because the torsional restoring law is linear to very high precision.

Worked examples

Example 1: A metre stick pivoted at one end

A uniform wooden metre stick (length L = 1.00 m, mass m = 0.15 kg) is pivoted at one end on a horizontal nail so it hangs vertically and can swing freely. Find:

(a) the moment of inertia about the pivot, (b) the period of small oscillations about the vertical, (c) the equivalent simple pendulum length.

Metre stick hanging from a nail, pivoted at one endA vertical uniform rod of length 1 m hanging from a nail at the top. The centre of mass is marked at 0.5 m from the top. The rod is shown at a small angular displacement to illustrate the swing. pivot centre of mass (at $L/2$) $L = 1.00$ m $m = 0.15$ kg
A uniform metre stick pivoted at one end. Its centre of mass is at the midpoint, so $d = L/2 = 0.50$ m.

Step 1. Compute the moment of inertia about the pivot.

For a uniform rod pivoted at one end,

I = \frac{1}{3} m L^2 = \frac{1}{3} \times 0.15 \times (1.00)^2 = 0.0500 \text{ kg·m}^2.

Why: I = mL^2/3 is the standard moment of inertia of a uniform rod about one end. It can be derived from I_\text{cm} = mL^2/12 about the centre, using the parallel axis theorem with offset d = L/2: I = mL^2/12 + m(L/2)^2 = mL^2(1/12 + 1/4) = mL^2/3.

Step 2. Identify d.

d = L/2 = 0.50 m. The centre of mass is at the midpoint of a uniform rod.

Step 3. Compute the period.

T = 2\pi\sqrt{\frac{I}{m g d}} = 2\pi\sqrt{\frac{0.0500}{0.15 \times 9.8 \times 0.50}} = 2\pi\sqrt{\frac{0.0500}{0.735}} = 2\pi\sqrt{0.0680}.

\sqrt{0.0680} = 0.2608, so

T = 2\pi \times 0.2608 \approx 1.64 \text{ s}.

Why: direct substitution. Note the m cancels in I/(mgd) = (mL^2/3)/(mgL/2) = 2L/(3g) — the period of a uniform rod does not depend on its mass, only on its length. T = 2\pi\sqrt{2L/(3g)} = 2\pi\sqrt{2(1)/(3 \times 9.8)} \approx 1.64 s.

Step 4. Compute the equivalent simple pendulum length.

L_\text{eq} = \frac{I}{m d} = \frac{0.0500}{0.15 \times 0.50} = \frac{0.0500}{0.075} = 0.667 \text{ m} = \frac{2L}{3}.

Result: I = 0.0500 kg·m², T \approx 1.64 s, L_\text{eq} = 2L/3 \approx 0.67 m.

What this shows: A metre stick does not swing with the period of a 1 m simple pendulum. It swings with the period of a 0.67 m simple pendulum — because its mass is spread all the way up to the pivot, and that mass contributes less to the swinging motion than if it were concentrated at the far end. The mass distribution matters, and it matters in a way that speeds up the swing.

Example 2: A thin ring pivoted on a peg

A thin metal ring (a bangle) of radius R = 4.0 cm and mass m hangs from a horizontal peg that passes through a hole at one point on its rim. Find the period of small oscillations.

Thin ring hanging from a pegA circular ring of radius R hanging from a horizontal peg that passes through its top rim. The centre of the ring is a distance R below the peg. The rim is shown in red. The dashed vertical line through the peg shows the equilibrium orientation. peg centre of mass $d = R$ $R = 4.0$ cm
A ring pivoted at one point on its rim. The centre of mass is at the geometric centre of the ring, a distance $R$ below the peg.

Step 1. Find d.

The centre of mass of a uniform ring is at its geometric centre. When the ring hangs from a point on the rim, the centre of mass is at the centre, a distance d = R below the peg.

Step 2. Find I about the peg.

The moment of inertia of a thin ring about an axis through its centre, perpendicular to its plane, is I_\text{cm} = m R^2. But the ring swings in its own plane, about the peg on its rim. The moment of inertia about the peg comes from the parallel axis theorem:

I = I_\text{cm} + m d^2 = m R^2 + m R^2 = 2 m R^2.

Why: the parallel axis theorem says I_\text{about-axis} = I_\text{about-parallel-axis-through-CM} + m(\text{offset})^2. For the ring, I_\text{cm} = mR^2 (all mass at distance R from the centre), and the offset is R. So I = mR^2 + mR^2 = 2mR^2.

Step 3. Apply the compound-pendulum formula.

T = 2\pi\sqrt{\frac{I}{m g d}} = 2\pi\sqrt{\frac{2 m R^2}{m g R}} = 2\pi\sqrt{\frac{2R}{g}}.

Plug in R = 0.04 m and g = 9.8 m/s²:

T = 2\pi\sqrt{\frac{2 \times 0.04}{9.8}} = 2\pi\sqrt{0.00816} = 2\pi \times 0.0903 \approx 0.568 \text{ s}.

Why: the mass m cancels, as it always does in gravitational pendulums. The equivalent simple pendulum length is L_\text{eq} = I/(md) = 2mR^2/(mR) = 2R = 8 cm — twice the radius.

Result: T \approx 0.57 s, L_\text{eq} = 2R = 8 cm.

What this shows: A bangle hanging on a peg swings like a simple pendulum of length 2R — its own diameter. This is a surprisingly clean result and a memorable one: no matter the ring's mass, its radius alone determines the period. A physics-lab experiment that measures the period of a ring on a peg, combined with a measurement of the radius, gives you g via g = 8\pi^2 R / T^2.

Example 3: A balance wheel in a mechanical watch

The balance wheel of a mechanical wristwatch is a thin circular disc of mass m = 1.20 g and radius R = 4.5 mm, suspended by a hairspring whose torsional stiffness is \kappa = 9.6 \times 10^{-6} N·m/rad. The wheel oscillates about its own axis. Find its period and frequency.

Balance wheel of a mechanical watchA small circular wheel viewed face-on, with a spiral hairspring attached to its hub, extending outward in several turns. The wheel's axis is perpendicular to the page. Arrows indicate the back-and-forth rotation about that axis. balance wheel ($I$) hairspring ($\kappa$) oscillation
The balance wheel of a mechanical watch: a circular wheel suspended by a flat hairspring coiled around its hub. The spring provides a linear restoring torque $\tau = -\kappa\theta$ for any twist $\theta$.

Step 1. Compute the moment of inertia.

Model the balance wheel as a uniform solid disc about its central axis. Then I = \tfrac{1}{2} m R^2.

I = \tfrac{1}{2} \times (1.20 \times 10^{-3}) \times (4.5 \times 10^{-3})^2 = \tfrac{1}{2} \times 1.20 \times 10^{-3} \times 2.025 \times 10^{-5}.
I = 1.215 \times 10^{-8} \text{ kg·m}^2.

Why: I = \tfrac{1}{2} m R^2 for a disc rotating about its central axis is a standard moment of inertia result. A real balance wheel has screws on its rim to tune the moment of inertia — which is how watchmakers regulate timekeeping — but for a first estimate the uniform-disc approximation is good.

Step 2. Apply the torsional-pendulum formula.

T = 2\pi\sqrt{\frac{I}{\kappa}} = 2\pi\sqrt{\frac{1.215 \times 10^{-8}}{9.6 \times 10^{-6}}} = 2\pi\sqrt{1.266 \times 10^{-3}}.

\sqrt{1.266 \times 10^{-3}} = 0.03558, so

T = 2\pi \times 0.03558 \approx 0.2235 \text{ s}.

Why: direct plug-in. The tiny value of I (because the wheel is both small and light) and the modest \kappa of the hairspring combine to give a quarter-second period — about four full oscillations per second, which matches the "tick, tick, tick" you hear in a typical mechanical watch.

Step 3. Compute the frequency.

f = \frac{1}{T} = \frac{1}{0.2235} \approx 4.47 \text{ Hz}.

Result: I \approx 1.22 \times 10^{-8} kg·m², T \approx 0.22 s, f \approx 4.5 Hz.

What this shows: A balance wheel's frequency is set entirely by its I and \kappa. Watchmakers regulate the watch by changing I (the tiny screws on the rim of the wheel) rather than changing \kappa (which would require replacing the hairspring). Move a screw inward and I decreases; the watch runs faster. Move it outward and I increases; the watch runs slower. Unlike a pendulum clock, the watch works in any orientation — in your pocket, on your wrist, upside down — because its restoring torque comes from the hairspring, not from gravity. This is why mechanical watches replaced pendulum clocks for portable timekeeping: gravity-based clocks fail the moment you put them in a pocket; hairspring-based clocks do not care.

Common confusions

If you came to understand compound and torsional pendulums well enough to work out periods for common shapes, you have what you need. What follows covers three deeper topics: the optimum pivot distance that minimises the period, Kater's reversible pendulum (the 19th-century gold standard for measuring g), and a first pass at what happens if the hairspring's torque is slightly nonlinear.

The minimum-period pivot and the "centre of percussion"

For a given rigid body, the period T = 2\pi\sqrt{I/(mgd)} depends on where you pivot it. Pivot too close to the centre of mass (d very small) and T blows up — there is almost no restoring torque. Pivot very far from the centre of mass (large d) and I grows roughly as d^2 (parallel axis theorem), making the denominator mgd grow only linearly — so T also grows. Somewhere between these extremes, T is minimised.

Use the parallel axis theorem. Let I_\text{cm} be the moment of inertia about the centre of mass, and define the radius of gyration k by I_\text{cm} = m k^2. Then

I_\text{pivot} = m k^2 + m d^2 = m(k^2 + d^2),

and

T(d) = 2\pi\sqrt{\frac{k^2 + d^2}{g d}}.

Why: substitute I = m(k^2 + d^2) into the compound-pendulum formula and the m's cancel. The period depends only on d and on k (a property of the body's shape).

To minimise T, minimise the expression under the square root. Take the derivative with respect to d:

\frac{d}{dd}\left(\frac{k^2 + d^2}{d}\right) = \frac{2d \cdot d - (k^2 + d^2)}{d^2} = \frac{d^2 - k^2}{d^2}.

Setting this to zero gives d^2 = k^2, i.e., d = k. Plug back in:

T_\text{min} = 2\pi\sqrt{\frac{k^2 + k^2}{g k}} = 2\pi\sqrt{\frac{2 k}{g}}. \tag{10}

The minimum period is achieved when the pivot is at distance d = k (the radius of gyration) from the centre of mass. For a uniform rod of length L, k = L/\sqrt{12}, so the fastest swing comes from pivoting at d = L/(2\sqrt{3}) \approx 0.289 L from the centre. This is the centre of oscillation — the point such that the pendulum has the shortest period.

A curious consequence: there are always two pivot points that give the same period T, symmetric about the minimum. If you plot T(d) versus d, it has a single minimum and two branches going to infinity on either side. For any T larger than the minimum, there are exactly two distances d_1, d_2 that produce it. Kater's reversible pendulum exploits this.

Kater's reversible pendulum — measuring g very precisely

The equation T = 2\pi\sqrt{(k^2 + d^2)/(gd)} can be rearranged to find g:

g = \frac{4\pi^2 (k^2 + d^2)}{T^2 d}.

But k^2 (the radius of gyration of the body) is hard to measure directly. Kater's trick: find two pivot points, at distances d_1 and d_2 from the centre of mass, such that the pendulum has the same period when pivoted from either. Call this common period T. Then

k^2 + d_1^2 = T^2 g d_1 / (4\pi^2), \qquad k^2 + d_2^2 = T^2 g d_2 / (4\pi^2).

Subtract the first from the second:

d_2^2 - d_1^2 = \frac{T^2 g}{4\pi^2}(d_2 - d_1).

Factor d_2^2 - d_1^2 = (d_2 - d_1)(d_2 + d_1), divide by (d_2 - d_1), and solve for g:

g = \frac{4 \pi^2 (d_1 + d_2)}{T^2}.

Why: the radius of gyration k — which depends on the detailed mass distribution and is hard to measure accurately — has dropped out. You only need T (measured by timing many swings and dividing) and the distance between the two pivots d_1 + d_2 (measured with a ruler).

This was the gold standard for measuring g from roughly 1817 through 1900. Captain Henry Kater's original design had two precisely-spaced knife-edge pivots in a single rigid pendulum. He would swing it on one edge, swing it on the other, adjust the mass distribution until the periods matched, and then read off g = 4\pi^2(d_1 + d_2)/T^2. Kater's reversible pendulum gave g to about four significant figures — a precision that stood until gravity metres were invented. The method is still taught in Indian physics labs because it gives a satisfying demonstration of why the equivalent-pendulum-length idea is not mere pedagogical rearrangement: it is the key to a practical measurement.

What if the hairspring is slightly nonlinear?

Real elastic wires obey Hooke's law \tau = -\kappa\theta to excellent approximation, but not exactly. A more realistic restoring torque might be

\tau = -\kappa \theta - \beta \theta^3

with \beta a small coefficient. The equation of motion becomes

I \ddot\theta = -\kappa\theta - \beta\theta^3,

which is the Duffing equation. Its solution is no longer a pure sinusoid — the higher harmonics 3\omega, 5\omega, \ldots appear as small corrections — and the period depends on amplitude. For small oscillations this amplitude-dependence is the origin of a watch's tiny drift when its mainspring unwinds (the balance wheel's amplitude slowly changes during the day) and why high-quality watches use materials like Nivarox, whose \beta is extraordinarily close to zero.

A clean way to see the amplitude dependence: if \beta > 0 (hard spring), larger amplitudes "see" a higher effective \kappa (because \kappa\theta + \beta\theta^3 > \kappa\theta for \theta > 0), so the period shortens with amplitude. If \beta < 0 (soft spring), the period lengthens with amplitude. A pure Hookean spring with \beta = 0 is isochronous — same period at every amplitude — which is the ideal a watchmaker pursues.

Why the compound and torsional formulas look alike

Both formulas have the structure

T = 2\pi\sqrt{\frac{I}{\text{torque coefficient}}}.

For the compound pendulum the torque coefficient is mgd (from gravity), derived from the perpendicular-component geometry. For the torsional pendulum it is \kappa (from the elastic wire), derived from Hooke's law for torsion. The underlying reason they look alike is that both reduce to the equation I\ddot\theta = -C\theta for some coefficient C — and any such equation is SHM with period 2\pi\sqrt{I/C}. This is the rotational version of the spring-mass T = 2\pi\sqrt{m/k}, and it is a universal template: any time a rigid body oscillates with a linear restoring torque, just identify I and the torque coefficient, and the period writes itself.

Where this leads next