Simple pendulum

Watch a swing in a park, a wrecking ball drifting to rest, or the pendulum bob behind the glass of an old clock, and you are watching the same fifteen-line piece of algebra play out in every case: a mass on a string, gravity pulling it back toward the bottom, over and over, in a rhythm that barely cares how far it swings or how heavy the bob is. That last part is the strange, useful fact at the heart of this article — a heavier bob does not swing slower, and a bigger swing (as long as it stays small) does not take longer. Galileo is said to have noticed exactly this while watching a lamp sway in Pisa cathedral, timing it against his own pulse, and the observation eventually became the mechanical heartbeat of clocks for three centuries. This article derives that heartbeat from scratch. Starting from nothing more than $F=ma$ , the geometry of a circular arc, and the small-angle approximation $\sin\theta\approx\theta$ , we will show that a swinging bob obeys the same restoring-force law as a mass on a spring, $F=-kx$ , with an effective spring constant $k=mg/L$ that depends only on the pendulum's length and gravity — never on mass. That single substitution hands us the famous result $T=2\pi\sqrt{L/g}$ , one of the two or three formulas every Class 11 physics student is expected to be able to derive on demand, in an exam, from the free-body diagram up. Beyond the derivation, this is also a genuinely useful piece of physics: it is how school laboratories measure $g$ , how a "seconds pendulum" is built to have a period of exactly two seconds, and how you can sanity-check a numeric physics problem in your head — if $L$ doubles, $T$ grows by only $\sqrt{2}$ , not by 2. We will work through both a straightforward board-style numeric problem and a slightly trickier competitive-exam question, and flag the small-angle assumption every single time it does the real work — because that one assumption is exactly where most marks are lost.

Key takeaways

A simple pendulum is an idealized point mass $m$ (the bob) on a light, inextensible string of length $L$ , swinging in a vertical plane under gravity.
For a small angular displacement $\theta$ , the restoring force along the arc is $F=-mg\sin\theta\approx-mg\theta=-\dfrac{mg}{L}x$ , exactly the SHM form $F=-kx$ with $k=mg/L$ .
The angular frequency is $\omega=\sqrt{g/L}$ , giving the time period $T=2\pi\sqrt{L/g}$ .
$T$ depends only on $L$ and $g$ — not on the mass of the bob and not on the amplitude, provided the swing stays small (roughly under $10°$ ).
A "seconds pendulum" (period exactly $2\ \text{s}$ ) on Earth ( $g=9.8\ \text{m/s}^2$ ) has a length of about $0.99\ \text{m}$ , close to $1\ \text{m}$ — a classic NCERT check number.
The same formula, run in reverse, is the standard school-lab method to measure $g$ : time several oscillations of a known-length pendulum and solve for $g=4\pi^2L/T^2$ .
Key idea: for small swings a pendulum's restoring force reduces to the SHM law $F=-kx$ with $k=mg/L$ , so $T=2\pi\sqrt{L/g}$ depends only on length and gravity — never on mass or amplitude.

The core insight

For small angular displacements a simple pendulum's tangential restoring force $-mg\sin\theta$ collapses to $-mg\theta=-(mg/L)x$ , the SHM law $F=-kx$ with $k=mg/L$ , giving the length-and-gravity-only period $T=2\pi\sqrt{L/g}$ .

Where it fits in the syllabus

The simple pendulum is the concluding worked application of the CBSE Class 11 Physics chapter "Oscillations," coming right after the loaded-spring case ( $T=2\pi\sqrt{m/k}$ ) and the energy analysis of SHM. You are expected to draw the free-body diagram, resolve the forces on the bob, apply the small-angle approximation, derive $T=2\pi\sqrt{L/g}$ , and use it numerically to solve for $T$ , $L$ , or $g$ . The 2025-26 CBSE prescription explicitly stops here: the physical (compound) pendulum, damped and forced oscillations, and resonance are all kept out of this chapter and are not developed below — they belong to a later stage of your physics education, if you choose to study more.

The physical picture

A simple pendulum is an idealization: a point mass $m$ (the bob) hangs from a fixed support on a string of length $L$ that is assumed massless and inextensible, and it swings without friction or air resistance in a single vertical plane. Two forces act on the bob at every instant — its weight $mg$ , acting straight down, and the tension $T_s$ along the string, acting toward the support. (We use $T_s$ for tension here to avoid clashing with the time period $T$ .) At any displaced angle $\theta$ from the vertical, it helps to resolve the weight into two perpendicular components: one along the string (balanced by tension, since the bob moves on a circular arc and does not fly outward), and one tangent to the arc, which is the only force with nothing to balance it and is therefore what accelerates the bob back toward the bottom. That tangential component has magnitude $mg\sin\theta$ , and it always points back toward $\theta=0$ , so:

F = -mg\sin\theta

where: $F$ is the net tangential (restoring) force along the arc, $m$ is the bob's mass, $g$ is the acceleration due to gravity, and $\theta$ is the angular displacement from the vertical, measured in radians. The minus sign is not decorative — it is the physical statement that whichever side the bob swings to, the force pushes it back the other way, exactly like the restoring force of a spring.

Figure

A pendulum bob displaced through a small angle from the vertical, showing the string tension along the string, the weight straight down, and the weight split into a component along the string balanced by tension and a tangential component along the arc that provides the restoring force

For small angular displacements, radian measure gives $\sin\theta\approx\theta$ to excellent accuracy (the error is under $0.5\%$ for $\theta$ up to about $10°$ ), so the restoring force becomes:

F \approx -mg\theta

If $x$ is the bob's displacement measured along the arc, then $x=L\theta$ , so $\theta=x/L$ , and:

F \approx -\frac{mg}{L}x

where: $x$ is the arc-length displacement from the mean position and $L$ is the string length. This is precisely the SHM restoring-force law $F=-kx$ from earlier in the chapter, with an effective force constant $k=mg/L$ . Because it has exactly this form, everything already known about SHM — sinusoidal motion, $\omega=\sqrt{k/m}$ , energy exchange between kinetic and potential — carries straight over, and substituting $k=mg/L$ gives:

\omega = \sqrt{\frac{g}{L}}, \qquad T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{g}}

where: $\omega$ is the angular frequency of the oscillation and $T$ is the time period, the time for one complete swing (out and back). Notice that $m$ never appears in the final formula — it cancelled out during the derivation of $k=mg/L$ because both the restoring force and the inertia it acts on scale with the same mass.

Derivation ADeriving the time period of a simple pendulum

The goal of this derivation is to start from nothing but Newton's second law applied to a pendulum bob on a circular arc, and to arrive at $T=2\pi\sqrt{L/g}$ using only the geometry of the arc and the small-angle approximation $\sin\theta\approx\theta$ — no calculus of variations, no torque about the pivot, just the same force-equals-restoring-constant-times-displacement logic already used for the spring. The one physical assumption that makes the whole thing work, and the one an examiner will specifically check for, is stating clearly why $\sin\theta\approx\theta$ is valid only for small swings.

Step 1. Draw the free-body diagram of the bob at a general angle $\theta$ from the vertical. Two forces act: tension $T_s$ along the string toward the support, and weight $mg$ vertically downward. Resolve the weight into a component $mg\cos\theta$ along the string (opposing tension) and a component $mg\sin\theta$ tangent to the circular arc, perpendicular to the string.

Step 2. Identify which component moves the bob. Because the bob is constrained to move on a circular arc of radius $L$ , motion along the string's direction is prevented by the string itself; tension adjusts to exactly cancel $mg\cos\theta$ in that direction. The only force left to accelerate the bob along its path is the tangential component, so the equation of motion along the arc is $ma = -mg\sin\theta$ , where the minus sign records that the force always points back toward $\theta=0$ .

Step 3. Convert the angle into a linear displacement. Let $x$ be the arc-length distance of the bob from its lowest point, so $x=L\theta$ (arc length equals radius times angle, in radians) and the tangential acceleration is $a=d^2x/dt^2$ . Substituting $\theta=x/L$ into $ma=-mg\sin\theta$ gives $ma=-mg\sin(x/L)$ .

Step 4. Apply the small-angle approximation. For $\theta$ small (in practice, under about $10°$ , so the string never swings far from vertical), $\sin\theta\approx\theta$ with the two differing by less than half a percent. This turns the equation into $ma\approx-mg\dfrac{x}{L}$ , or $m\dfrac{d^2x}{dt^2}=-\dfrac{mg}{L}x$ . This is exactly the SHM differential equation $m\ddot{x}=-kx$ with $k=mg/L$ .

Step 5. Read off the angular frequency and period. Comparing with the standard SHM result $\omega=\sqrt{k/m}$ , substitute $k=mg/L$ : $\omega=\sqrt{\dfrac{mg/L}{m}}=\sqrt{\dfrac{g}{L}}$ — the mass cancels completely because it appears in both the inertia ( $m$ in $ma$ ) and the restoring-force constant ( $k=mg/L$ ). The time period follows immediately as $T=\dfrac{2\pi}{\omega}=2\pi\sqrt{\dfrac{L}{g}}$ , depending only on the length of the pendulum and the local acceleration due to gravity.

Worked examples

Board level - length of a seconds pendulum

Find the length of a simple pendulum whose time period is exactly $2\ \text{s}$ (a "seconds pendulum" — one that ticks once every second, i.e. beats the second on each swing). Use $g=9.8\ \text{m/s}^2$ .

Step 1. Write the period formula and square both sides to isolate $L$ : $T=2\pi\sqrt{L/g}$ gives $T^2=4\pi^2\dfrac{L}{g}$ , so $L=\dfrac{T^2g}{4\pi^2}$ .

Step 2. Substitute the given values, $T=2\ \text{s}$ and $g=9.8\ \text{m/s}^2$ , using $\pi^2\approx9.87$ : $L=\dfrac{(2)^2\times9.8}{4\times9.87}=\dfrac{4\times9.8}{39.48}=\dfrac{39.2}{39.48}$ .

Step 3. Compute the final value. $L\approx\mathbf{0.993\ m}$ , a length very close to $1\ \text{m}$ — this is the standard NCERT "seconds pendulum" result, and it is worth memorising as a sanity-check number: any pendulum problem giving $g\approx9.8\ \text{m/s}^2$ and $T=2\ \text{s}$ should always land near $L\approx1\ \text{m}$ .

Competitive level - finding g from a measured period

A simple pendulum of length $1.5\ \text{m}$ is taken to a location where its time period is carefully measured (using $50$ oscillations timed with a stopwatch) to be $2.53\ \text{s}$ . Find the acceleration due to gravity at that location, and comment on what the result implies about the location.

Step 1. Rearrange $T=2\pi\sqrt{L/g}$ to solve for $g$ . Squaring gives $T^2=4\pi^2L/g$ , so $g=\dfrac{4\pi^2L}{T^2}$ .

Step 2. Substitute $L=1.5\ \text{m}$ , $T=2.53\ \text{s}$ , and $\pi^2\approx9.87$ : $g=\dfrac{4\times9.87\times1.5}{(2.53)^2}=\dfrac{59.22}{6.4009}$ .

Step 3. Compute and interpret. $g\approx\mathbf{9.25\ m/s^2}$ , noticeably below the standard $9.8\ \text{m/s}^2$ . Since $g$ genuinely varies with altitude and latitude (it decreases going up, and decreases slightly going from the poles toward the equator due to Earth's rotation and equatorial bulge — both studied qualitatively elsewhere in mechanics), this measurement is consistent with the pendulum being at a high altitude or being affected by local variations, and is exactly the kind of experiment a school lab uses to measure $g$ rather than assume it. Timing note: in a real experiment you would never time a single swing with a stopwatch — reaction-time error of even $0.2\ \text{s}$ on a $2.53\ \text{s}$ period is an $8\%$ error. Instead time $n=50$ oscillations together and divide: $T=t_{50}/50$ , which shrinks the same $0.2\ \text{s}$ reaction-time error to under $0.2\%$ of $T$ . This "time many, divide" trick is a standard competitive-exam and practical-exam technique whenever a period is measured directly.

Common traps

Using $T=2\pi\sqrt{L/g}$ at large swing angles. The formula only exists because $\sin\theta\approx\theta$ was substituted in Step 4 of the derivation; once $\theta$ is no longer small (roughly past $10°$ ), the true motion is still periodic but is no longer simple harmonic, and the real period is slightly longer than this formula predicts. Examiners sometimes test this conceptually by asking what happens to $T$ for a large-amplitude swing — the answer is "it increases beyond $2\pi\sqrt{L/g}$ ," not "the formula still applies."
Believing the bob's mass affects $T$ . It is a very common wrong intuition that a heavier bob should swing slower (more inertia) or faster (more restoring force); the derivation shows both effects are proportional to $m$ and cancel exactly, so $T$ is mass-independent. This is precisely the observation credited to Galileo.
Confusing string length with pivot-to-center-of-mass distance. For the idealized point-mass bob of this chapter these coincide, and $L$ is simply the string length; a real bob of finite size (not required here) would need the distance to its center of mass instead, which belongs to the physical-pendulum treatment kept outside this chapter.
Mixing up degrees and radians when checking $\sin\theta\approx\theta$ . The approximation is a radian-measure statement; plugging a degree value directly into $\theta$ (instead of converting to radians first) gives nonsense, even though $\sin\theta$ itself can be evaluated from a degree value directly on a calculator.
Arithmetic slips squaring or unsquaring $T$ . Since $T$ appears squared in $T^2=4\pi^2L/g$ , a very common error is forgetting to square $T$ when solving for $L$ or $g$ , or forgetting to take the square root at the end when solving for $T$ itself.
Forgetting the idealizations. "Simple" pendulum means a massless, inextensible string, a point-mass bob, and no air resistance; real pendulums (thick string, sizeable bob, air drag slowly shrinking the amplitude) need corrections that are outside this chapter's scope, though the small-amplitude formula remains an excellent approximation for most lab-scale pendulums.

You should be able to…

Draw the free-body diagram of a pendulum bob at a general angle $\theta$ and identify the tangential restoring force $mg\sin\theta$ .
Explain, using $\sin\theta\approx\theta$ , why the pendulum's motion is only approximately SHM, and only for small angles.
Derive the SHM form $F=-kx$ for the pendulum with $k=mg/L$ , and from it derive $T=2\pi\sqrt{L/g}$ .
State clearly that $T$ is independent of both the bob's mass and the amplitude of swing (for small angles).
Solve numerically for any one of $T$ , $L$ , or $g$ given the other two, including in the "measuring $g$ " direction.
Recognise when a stated large angle means the small-angle period formula no longer applies.

Ask the tutor: Want to work through the free-body diagram and the small-angle step by step with your own numbers for $L$ , $T$ , or $g$ ? Ask the AI tutor to walk through it with you.

Why does the mass of the bob not matter at all? That feels wrong.

It feels wrong because in everyday life heavier things often behave differently under the same force — but here both the driving effect and the resistance to it scale with mass in exactly the same way. The restoring force is $mg\sin\theta$ , proportional to $m$ ; and by Newton's second law, the acceleration produced by any force is also inversely proportional to $m$ . When you write $ma=-mg\sin\theta$ and both sides carry an $m$ , it cancels, leaving an equation with no mass in it at all. A bowling ball and a marble on strings of the same length, released from the same small angle, swing with the same period — this is the pendulum version of "all objects fall at the same rate in a vacuum."

Does the amplitude of the swing really not matter, even a little?

For small amplitudes (under roughly $10°$ ), the effect of amplitude on the period is genuinely negligible — that is exactly what the small-angle approximation buys you, and it is why pendulum clocks stay accurate even as their swing very slowly dies down from air resistance. But this independence is not exact and not universal: for larger swings, the true period does grow with amplitude, because $\sin\theta$ falls increasingly short of $\theta$ as $\theta$ grows. The exact large-amplitude period involves an integral beyond this chapter's scope; the syllabus only requires you to know that the small-angle formula becomes inaccurate for large swings, not to compute the correction.

How is this actually used to measure g in a physics lab?

You set up a pendulum of carefully measured length $L$ , displace it through a small angle, and time a number of complete oscillations — typically $20$ to $50$ — with a stopwatch, then divide by that number to get an accurate single-swing period $T$ (timing many swings and dividing spreads out your reaction-time error, as in the competitive-level example above). Rearranging $T=2\pi\sqrt{L/g}$ to $g=4\pi^2L/T^2$ then gives $g$ directly from two measured quantities, $L$ and $T$ . This is one of the standard practical experiments in the Class 11 physics laboratory course, and it usually returns a value close to, but not exactly, $9.8\ \text{m/s}^2$ due to small experimental errors.

What about pendulums that swing damped, or friction, or resonance — are those in this chapter?

No — free, forced and damped oscillations, and resonance, are explicitly outside the 2025-26 CBSE Class 11 prescription for this chapter, and the compound (physical) pendulum with a finite-size bob is likewise kept out. Real pendulums do slowly lose amplitude to air resistance and pivot friction (that motion is called damped oscillation), and pushing a pendulum periodically can build up large swings (resonance) — both are genuinely important ideas in physics, studied in more advanced treatments later on, but this chapter's scope stops at the idealized, undamped, small-angle simple pendulum.