Visual walkthrough — Simple pendulum — small angle approximation, T = 2π√(L - g) derivation

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Step 1 — Draw the pendulum and name every part

WHAT. A pendulum is a small heavy ball (the bob) hanging from a fixed point (the pivot) by a string. Let:

• $L$ = the length of the string, from pivot to the middle of the bob.
• $\theta$ (the Greek letter "theta") = the angle the string makes with the straight-down vertical line. When the bob hangs at rest, $\theta =0$.

WHY name these first. Every equation below is a sentence made of these symbols. If you don't picture them, the equations are noise. $\theta$ is our single "how far from home" number — the whole story is how $\theta$ changes in time.

PICTURE. The red bob hangs on a black string. The dashed vertical line is "home" ($\theta =0$); the angle between it and the string is $\theta$.

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Step 2 — The two forces pulling on the bob

WHAT. Only two things touch/pull the bob:

1. Gravity, pulling straight down. Its strength is $mg$, where $m$ is the bob's mass and $g\approx 9.8{\text{m/s}}^2$ is Earth's gravitational pull per kilogram.
2. Tension $T_{\text{string}}$, pulling along the string toward the pivot.

WHY only these two matter. A "simple" pendulum has no friction and no air drag (we removed them on purpose). So these two arrows are the entire cast.

PICTURE. The red arrow is gravity $mg$ (down). The black arrow along the string is tension. Notice they point in different directions — that difference is what makes the bob swing rather than fall straight or hang still.

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Step 3 — Split gravity into "along the string" and "along the swing"

WHAT. Gravity points straight down, but the bob is forced to move on a curved arc (the string keeps it a fixed distance $L$ from the pivot). So we split the gravity arrow into two useful directions:

• Radial (along the string): $mg\cos \theta$. This just fights the tension.
• Tangential (along the arc, the direction the bob can actually move): $mg\sin \theta$.

WHY split it this way. Tension is along the string, so tension can never push the bob along its arc. Only the tangential slice of gravity speeds the bob up or slows it down. That slice is the engine of the swing — so it is the only piece we chase.

WHY does $\sin \theta$ appear, and not $\cos \theta$? Look at the figure: the tangential direction is perpendicular to the string. When you drop gravity's arrow onto that perpendicular direction, the geometry (a right triangle whose small angle equals $\theta$) gives the opposite side, and opposite/hypotenuse $=\sin \theta$. The radial slice lands on the adjacent side $=\cos \theta$.

PICTURE. The red arrow is the tangential slice $mg\sin \theta$ — the only piece that matters. The faint black arrows are the full gravity and its radial slice.

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Step 4 — Check every direction the bob can be (all cases)

WHAT. We must make sure that minus sign is right no matter where the bob is.

WHY. A formula is only trustworthy if it survives every case — left, right, and dead-centre.

PICTURE + cases:

• Right of centre ($\theta >0$): $\sin \theta >0$, so $F=-mg\sin \theta <0$ → force points left (home). ✓
• Left of centre ($\theta <0$): $\sin \theta <0$, so $F=-mg\sin \theta >0$ → force points right (home). ✓
• Exactly at centre ($\theta =0$): $\sin 0=0$, so $F=0$ → no push. This is equilibrium: the bob would hang here forever if placed at rest.

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Step 5 — Turn the force into an equation of motion

WHAT. The bob rides a circular arc of radius $L$. Its distance travelled along the arc is $s=L\theta$ (arc length = radius × angle — this is why we insisted on radians in Step 1). Since $L$ is fixed, the acceleration along the arc is

$a=L\frac{d^2\theta }{d{t}^2}.$

Now apply Newton's second law "$\text{mass}\times \text{acceleration}=\text{force}$" along the arc:

$\underset{\text{mass}}{\underbrace{m}}\underset{\text{tangential accel.}}{\underbrace{L\frac{d^2\theta }{d{t}^2}}}=\underset{\text{tangential force}}{\underbrace{-mg\sin \theta }}$

WHY. Newton's law is the bridge from "what force acts" to "how it moves." The symbol $\frac{d^2\theta }{d{t}^2}$ (read "d-two-theta-d-t-squared") just means how fast the angle's speed is changing — the angular acceleration.

PICTURE. The bob is drawn part-way along its arc; the arc length $s=L\theta$ is marked in red, the acceleration arrow tangent to the arc.

Now cancel the mass. The $m$ sits on both sides:

$\frac{d^2\theta }{d{t}^2}=-\frac{g}{L}\sin \theta .$

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Step 6 — Why this is NOT yet the "nice" equation

WHAT. The equation $\ddot{\theta }=-\frac{g}{L}\sin \theta$ has a $\sin \theta$ in it. Simple Harmonic Motion — the clean, solvable, sine-wave motion — needs the restoring term to be straight-line proportional to $\theta$, i.e. $\ddot{\theta }=-(\text{const})\cdot \theta$. Our $\sin \theta$ curves, so we're not there yet.

WHY care. Only when the equation is linear in $\theta$ can we say "the period is the same for every swing" and read off a clean formula. See Simple Harmonic Motion.

PICTURE. Plot $y=\sin \theta$ (red curve) against $y=\theta$ (black straight line). Near the origin they hug each other; far out they peel apart. That hugging near zero is the loophole we exploit next.

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Step 7 — The small-angle approximation, drawn

WHAT. For small $\theta$, the curve and the line are almost identical, so we replace one with the other:

$\sin \theta \approx \theta (\theta \text{ small, in radians}).$

This comes from the Taylor series:

$\sin \theta =\underset{\text{keep this}}{\underbrace{\theta }}-\underset{\text{tiny when }\theta \text{ small}}{\underbrace{\frac{{\theta }^3}{6}}}+\frac{{\theta }^5}{120}-\cdots$

WHY it's allowed. At $\theta =0.1$ rad ($\approx 5.7^{\circ }$), the dropped ${\theta }^3/6$ term is only $0.000167$ — a $0.17\%$ correction. The picture shows the gap is invisibly small there.

WHY radians (not degrees). The series above is only true in radians. In degrees, $\sin (10^{\circ })=0.174$ but "$\theta$" would be $10$ — off by a factor of 57. Radians make the arc length $s=L\theta$ honest, and that honesty is what makes $\sin \theta \approx \theta$ work.

PICTURE. Zoom into the region near $\theta =0$: the red $\sin \theta$ and black $\theta$ overlap; the shaded gap between them is the error, and it shrinks toward zero.

Substituting gives the clean equation:

$\frac{d^2\theta }{d{t}^2}=-\frac{g}{L}\theta .$

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Step 8 — Recognise SHM and read off the period

WHAT. Compare our clean equation with the master SHM equation:

$\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x⟺\frac{d^2\theta }{d{t}^2}=-\frac{g}{L}\theta .$

Matching term-by-term, the constant ${\omega }^2$ (the squared angular frequency) must equal $g/L$:

$\omega =\sqrt{\frac{g}{L}}.$

Then one full swing is $2\pi$ of phase, and $\omega$ is phase-per-second, so the time for one cycle is

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{L}{g}}.$

WHY $2\pi /\omega$. $\omega$ tells you radians of phase gained per second; a whole cycle is $2\pi$ radians of phase; dividing gives seconds per cycle. See Angular Frequency and Period.

PICTURE. The angle $\theta (t)$ traced against time is a pure cosine wave. The period $T$ is marked as the time between two crests; $\omega$ sets how tight the wave is.

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The one-picture summary

PICTURE. One figure, five panels left-to-right: (1) the swinging bob → (2) split gravity, keep the red tangential slice $-mg\sin \theta$ → (3) Newton's law $mL\ddot{\theta }=-mg\sin \theta$, mass crossed out → (4) $\sin \theta \to \theta$ straightening the curve into a line → (5) the cosine wave with period $T=2\pi \sqrt{L/g}$.

Recall Feynman retelling — the whole walkthrough in plain words

Hang a ball on a string. Gravity pulls it straight down, but the string forces it onto a curved path, so only the sideways slice of gravity actually pushes it along that path — and that slice always points back toward the bottom. Write Newton's "force makes motion" rule for that sideways push, and a small miracle happens: the ball's weight appears on both sides and cancels, so a heavy ball and a light ball swing in perfect time. The push depends on $\sin \theta$, which is a curve — but for small swings that curve is practically a straight line, so we straighten it. Once it's straight, the motion is the clean back-and-forth sine wave called SHM, and its swing-time is $T=2\pi \sqrt{L/g}$: longer string, slower swing; stronger gravity, quicker swing; and the ball's weight never enters at all.

Recall Checkpoints

Which slice of gravity drives the swing? ::: The tangential slice $mg\sin \theta$, perpendicular to the string. At which step does mass cancel? ::: Step 5, Newton's law $mL\ddot{\theta }=-mg\sin \theta$ — $m$ on both sides. Why must $\theta$ be in radians? ::: So $s=L\theta$ and $\sin \theta \approx \theta$ hold; the Taylor series is radians-only. Where does the $2\pi$ come from? ::: One full cycle is $2\pi$ radians of phase, and $T=2\pi /\omega$.

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Connections

• Simple Harmonic Motion — the clean equation we matched in Step 8
• Restoring Force and Equilibrium — the minus sign in Step 3/4
• Taylor Series and Small-Angle Approximations — the straightening in Step 7
• Angular Frequency and Period — $T=2\pi /\omega$
• Energy in Oscillations
• Mass-Spring System — same $\omega =\sqrt{k/m}$ pattern
• Measuring g with a Pendulum

Concept Map