The Simple Pendulum

The pendulum retired as a primary clock in 1927 when quartz arrived, but its physics never left the front line. The mirrors of the LIGO gravitational-wave detectors hang as the last stage of quadruple pendulums — above the pendulum's resonant frequency a suspended mass sits almost perfectly still, so four cascaded stages isolate the optics from ground vibration by a factor of billions; the formula being exploited is this module's $\omega = \sqrt{g/L}$. Foucault pendulums still swing in science museums on every continent, doing with a bob and wire what it otherwise takes a gyroscope to prove — the planet turns. Seismometers descend directly from the horizontal pendulum, and your own walking pace is set by leg-as-physical-pendulum mechanics: comfortable stride frequency tracks $\sqrt{g/L_{\text{leg}}}$, which is why walking on the Moon collapsed into hopping. Even the humble metronome — an inverted adjustable pendulum — keeps the world's music honest with the same square root.

Hang a small heavy bob on a light string of length $L$ and pull it aside by angle $\theta$. Two forces act: string tension along the string, and weight $mg$ straight down. Split the weight along the string and along the arc. The component along the string, $mg\cos\theta$, is cancelled and managed by tension — it bends the path into a circle but never speeds the bob up or slows it down. The component along the arc, $mg\sin\theta$, is the whole story: it always points toward the bottom of the swing, a restoring force supplied by the planet itself.

Now the two famous cancellations, in plain sight. Why no mass? The restoring pull is proportional to $m$ — gravity tugs a heavier bob harder — but the inertia resisting acceleration is the same $m$. Newton's law divides one by the other and the mass is gone: the pendulum is a slow-motion free-fall experiment, the same reason a hammer and a feather drop together in vacuum. Why no amplitude? Double the swing and you double the restoring force, but you also double the distance to cover each cycle; near the bottom the two effects compensate exactly. Isochronism is not magic — it is the signature of a restoring force linear in displacement, and it holds precisely as far as $\sin\theta$ stays close to $\theta$.

Measure the bob's position by its arc-length displacement $s = L\theta$ from the lowest point. The force along the arc is $-mg\sin\theta$; Newton's second law along the arc reads $ma = -mg\sin\theta$, so

where: the mass has already cancelled — before any approximation. Equation (1) is exact but not yet SHM: the force follows $\sin\theta$, not $\theta$. Enter the small-angle approximation: in radians, $\sin\theta \approx \theta$ when $\theta$ is small. The honesty table — $\sin 5^\circ = 0.0872$ vs $\theta = 0.0873$ (0.1% off); $\sin 15^\circ = 0.2588$ vs $0.2618$ (1.1%); $\sin 30^\circ = 0.5$ vs $0.5236$ (4.7%) — shows the approximation earning its keep to about $15^\circ$ and fraying beyond. Within it, $\sin\theta \to \theta = s/L$:

where: $s$ — displacement along the arc. This is the SHM signature $a = -\omega^2 s$ with the frequency slot filled by gravity-over-length:

where: $L$ — pivot to the centre of the bob (not to its top); $g$ — the local free-fall acceleration, $9.8\ \mathrm{m/s^2}$ standard ($10$ on most JEE/NEET papers, $9.81$ on CIE). Read the structure the way the technique module will drill it: stiffness over inertia. The effective stiffness of gravity along the arc is $mg/L$ (force $mg\theta = (mg/L)s$ per unit arc), the inertia is $m$, and $\sqrt{(mg/L)/m} = \sqrt{g/L}$ — mass cancelling in the division, exactly as advertised.

The pendulum's exam home is the lab question, and its centrepiece is a straight line. Square the period law: $T^2 = \dfrac{4\pi^2}{g}L$. Plot $T^2$ against $L$ and the data must march along a straight line through the origin with slope $4\pi^2/g$ — so the slope measures the planet:

where: the graph form beats single-point computation because it averages many measurements and exposes systematic errors (a line missing the origin betrays a length mis-measured to the bob's top instead of its centre). Timing discipline does the rest: time 20 oscillations and divide, because a $0.2$ s human reaction error splits 20 ways into $0.01$ s.

Because $T$ depends on $g$, the pendulum is also a comparator of places. In 1671 the astronomer Jean Richer carried a pendulum clock from Paris to Cayenne, near the equator, and found it losing about 2.5 minutes a day: gravity there is weaker (the Earth bulges, and spin steals some pull), and the pendulum reported it a century before anyone could compute it. The same sensitivity powers the exam's favourite clock problems: any fractional change in $L$ or $g$ moves the period by half as much, $\dfrac{\Delta T}{T} = \dfrac12\dfrac{\Delta L}{L} - \dfrac12\dfrac{\Delta g}{g}$ (the square root's fingerprint), and a slow clock loses $\left(\Delta T/T\right) \times 86{,}400$ seconds per day.

The physical pendulum, in one paragraph. Real swinging objects — a metre rule on a nail, your leg from the hip, a church bell — are not point bobs on strings. For any rigid body pivoted a distance $d$ from its centre of mass, with moment of inertia $I$ about the pivot, the same small-angle argument gives $T = 2\pi\sqrt{I/(mgd)}$ — inertia over stiffness once more, with rotational inertia upstairs and the gravity lever $mgd$ downstairs. A metre rule swinging about its end ($I = \tfrac13 mL^2$, $d = L/2$) ticks like a simple pendulum of length $2L/3 \approx 67$ cm. Boards want state-and-use at most; the full treatment is the university deep dive.

1 · Board level — design a seconds pendulum, then take it to the Moon (CBSE/ISC pattern)

A seconds pendulum beats seconds: each one-way swing takes $1\ \mathrm{s}$, so $T = 2\ \mathrm{s}$. (a) Find its length on Earth ($g = 9.8\ \mathrm{m/s^2}$). (b) What length gives $T = 2\ \mathrm{s}$ on the Moon ($g = 1.62\ \mathrm{m/s^2}$)? (c) If the Earth pendulum is taken to the Moon unchanged, what is its period there?

Step 1. Invert the period law: $L = \dfrac{gT^2}{4\pi^2} = \dfrac{9.8 \times 4}{39.48} = \boxed{0.993\ \mathrm{m}}$ — almost exactly a metre, a coincidence so convenient that an early definition proposed for the metre was the seconds pendulum.

Step 2. Moon design: $L_{\text{moon}} = \dfrac{1.62 \times 4}{39.48} = \boxed{0.164\ \mathrm{m}}$ — a hand-span pendulum, because weak gravity is a weak spring.

Step 3. Earth pendulum on the Moon: $T' = T\sqrt{g_E/g_M} = 2\sqrt{9.8/1.62} = 2 \times 2.46 = \boxed{4.92\ \mathrm{s}}$. A clock regulated by it would lose about 14.6 hours per day.

2 · Board level — the lab measurement of g (A-Level/IB practical pattern)

A student sets a pendulum of length $1.000\ \mathrm{m}$ (pivot to bob centre) swinging at small amplitude and times 20 complete oscillations: $40.2\ \mathrm{s}$. Find $g$, and state why timing 20 oscillations beats timing one.

Step 1. Period: $T = 40.2/20 = 2.01\ \mathrm{s}$.

Step 2. Invert: $g = \dfrac{4\pi^2 L}{T^2} = \dfrac{39.48 \times 1.000}{(2.01)^2} = \dfrac{39.48}{4.040} = \boxed{9.77\ \mathrm{m/s^2}}$.

Step 3. Why 20: the stopwatch error ($\sim 0.2$ s of human reaction) attaches to the whole measurement; divided across 20 cycles it contributes only $0.01$ s to $T$ — a 0.5% period error instead of 10%. Since $g \propto 1/T^2$, every percent of period error becomes two percent in $g$; timing many swings is not neatness, it is the experiment.

3 · Competitive level — the drifting clock (JEE-Main/NEET pattern, ~2 minutes)

A pendulum clock keeps perfect time at sea level. It is moved to a mountain laboratory where $g$ is $0.2\%$ smaller. How much time does it lose per day?

Step 1. Half-rule: $T \propto g^{-1/2}$, so $\dfrac{\Delta T}{T} = -\dfrac12\dfrac{\Delta g}{g} = +\dfrac12(0.2\%) = 0.1\%$ — the period lengthens by one part in a thousand.

Step 2. A clock whose pendulum runs 0.1% slow loses 0.1% of a day, every day: $0.001 \times 86{,}400 = \boxed{86.4\ \mathrm{s} \approx 1.4\ \text{minutes per day}}$.

Step 3. Direction audit: weaker $g$ = weaker restoring = slower swings = clock loses. (Heat has the same signature via expansion: a summer pendulum is longer, so grandfather clocks run slow in July — same half-rule, applied to $\Delta L/L$.) Timing note: ~90 s. The entire question is the half-rule $\Delta T/T = \tfrac12 \Delta L/L - \tfrac12 \Delta g/g$ and the 86,400. Distractor watch: the full 0.2% (172.8 s) for the student who forgot the square root, and a "gains time" option for the sign-blind.

4 · Competitive level — JEE-Advanced stretch: the string's worst moment (~3.5 minutes)

A $0.30\ \mathrm{kg}$ bob on a $0.75\ \mathrm{m}$ string is released from rest at $60^\circ$ to the vertical. Find (a) the bob's speed at the lowest point, (b) the string tension there. (c) If the string snaps at tension $6.0\ \mathrm{N}$, does it survive the swing? ($g = 9.8\ \mathrm{m/s^2}$.)

Step 1. This is not a small-angle problem — no SHM formulas allowed. Energy from release to bottom: the bob drops $h = L(1 - \cos 60^\circ) = 0.75 \times 0.5 = 0.375$ m, so $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.375} = \sqrt{7.35} = \boxed{2.71\ \mathrm{m/s}}$.

Step 2. At the bottom the bob moves on a circle of radius $L$; the net upward force must supply the centripetal demand: $T_{\text{str}} - mg = \dfrac{mv^2}{L}$. So $T_{\text{str}} = m\left(g + \dfrac{v^2}{L}\right) = 0.30\left(9.8 + \dfrac{7.35}{0.75}\right) = 0.30 \times 19.6 = \boxed{5.88\ \mathrm{N}}$.

Step 3. Recognise the general result: $T_{\text{str}} = mg(3 - 2\cos\theta_0)$, which at $\theta_0 = 60^\circ$ gives exactly $2mg = 5.88$ N ✓. Against the 6.0 N rating: survives, with 2% to spare — the paper's designed sweat.

Step 4. The string's worst moment is always the bottom of the swing, where it must both hold the weight and bend the path. At the release extreme, tension is only $mg\cos\theta_0 = 1.47$ N. Timing note: ~3–3.5 min. The reflexes being tested: large angle ⇒ energy method, not SHM; tension ⇒ centripetal accounting at the bottom; and the quotable $mg(3-2\cos\theta_0)$. Distractor watch: using $T_{\text{str}} = mg$ (statics), or $h = L\sin\theta_0$ instead of $L(1-\cos\theta_0)$ — the ballistic-pendulum geometry error, recycled.

Degrees in the small-angle machine. $\sin\theta \approx \theta$ works in radians only. Any formula with a bare $\theta$ — the SHM equation, the amplitude correction $\theta_0^2/16$, arc length $s = L\theta$ — silently demands radians; feeding degrees inflates everything by 57.3.

$L$ is pivot-to-centre. Measured to the top of the bob, the length is short by the bob's radius and the $T^2$–$L$ lab graph grows a tell-tale intercept. Examiners write the trap into data tables; the graph question expects you to name it.

The square-root half-rule. Every fractional change in $L$ or $g$ moves $T$ by half as much. Clock problems ($\Delta T/T$, seconds lost per day $= 86{,}400 \times \Delta T/T$) are half-rule questions in costume; forgetting the half doubles your answer, and the doubled answer is always among the options.

Seconds pendulum has $T = 2$ s. Each swing (one way) is a second; the full cycle is two. Setting $T = 1$ s gives $L = 25$ cm and a lost mark.

Mass distractors. Nothing at this module's level depends on the bob's mass — not the period, not the isochronism, not the lab slope. But tension does ($T_{\text{str}} = mg(3 - 2\cos\theta_0)$), which is exactly why Example 4 hands you a mass while Examples 1–3 don't need one.

Big-swing problems are energy problems. Beyond $\sim 15^\circ$, drop SHM entirely: use energy conservation for speeds and centripetal accounting for tensions. The SHM period formula silently degrades — it runs about 0.4% short at $15^\circ$, 1.7% at $30^\circ$ — and JEE-Advanced has asked precisely "longer or shorter than $2\pi\sqrt{L/g}$?" (Longer.)

Which $g$? CBSE/JEE/NEET usually declare $g = 10\ \mathrm{m/s^2}$ (seconds pendulum then comes out $L = 1.013$ m); CIE uses 9.81; IB 9.8. In lab questions, $g$ is the answer — quote it to the precision your timing supports and compare with the local value, not the textbook one.

Texts. NCERT Physics Class 11, Ch. 14 (simple pendulum section + derivation) · HC Verma, Concepts of Physics Vol. 1, Ch. 12 · Resnick–Halliday–Walker Ch. 15 · The Simple Pendulum (UG deep dive — carries the premium video, the exact large-angle period and the phase portrait).

Provenance. All results are standard. A verifier should recompute: $T(1\ \mathrm{m}, 9.8) = 2.007$ s; Example 1's $0.993$ m / $0.164$ m / $4.92$ s chain; Example 2's $g = 9.77\ \mathrm{m/s^2}$ from $T = 2.01$ s; Example 3's 86.4 s/day from the half-rule; Example 4's $v = 2.71$ m/s and $T_{\text{str}} = 5.88\ \mathrm{N} = 2mg$ against $mg(3 - 2\cos 60^\circ)$.

Prerequisites. Periodic Motion & the Definition of SHM · SHM Kinematics · The Spring–Block Oscillator. Leads to. Pendulum in Lift, Cart & on Incline (archetype) · Damped & Forced Oscillations · The Simple Pendulum (UG deep dive — premium video).