A simple pendulum of time period $1\ \text{s}$ and length $l$ is hung from a fixed support at O. The pendulum swings in a vertical plane; when the bob is at its lowest (mean) position it is at a point directly above a point A on the horizontal ground, at a height $H$ vertically above A (so $OA$ is vertical and $OP=l$, where P is the lowest position of the bob). The angular amplitude of the swing is $\theta_0$. While the bob is moving, the string suddenly snaps at the instant the angular displacement is $\theta=\theta_0/2$ (with the bob then moving outward, away from the mean position). Find the time taken by the bob to hit the ground and the horizontal distance from A at which it lands. Assume $\theta_0$ is small, so that $\sin\theta_0\approx\theta_0$ and $\cos\theta_0\approx1$.

A simple pendulum of time period 1s and length l is hung from a fixed support at O. The pendulum swings in a v

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Concept understanding — Simple Pendulum Period

Simple Pendulum Period: From Intuition to Formula

Imagine tying a small weight to a string, holding the other end fixed, and giving it a gentle push. It swings back and forth — that’s a simple pendulum. The question is: what determines how fast it swings? Does a heavier bob swing faster? Does a longer string make it slower?

Let’s start with what you already feel. If you hold a short string (say 20 cm) and swing it, the bob zips back and forth quickly. If you use a long string (say 1 m), the swing is noticeably slower. So length matters — longer means slower.

What about the weight? Try a light plastic bob and a heavy metal one of the same size, on the same string. You’ll find they swing at the same speed. That’s surprising — heavier things don’t fall faster, and here they don’t swing faster either. So mass does not affect the period (the time for one complete back-and-forth swing).

What about how hard you push? If you give a big push, the bob swings wider, but does it take more time? For small swings (small angles, say less than about 15°), the period is almost the same regardless of amplitude. That’s the key: for small oscillations, the pendulum is isochronous — its period is independent of amplitude.

Note

This is only true for small angles. If you pull the bob to 60° and let go, the period becomes noticeably longer. In most exam problems, you assume “small oscillations” (usually < 10°).

The Precise Statement

For a simple pendulum of length $L$ (measured from pivot to centre of bob), swinging with small amplitude in a uniform gravitational field $g$ , the time period $T$ (time for one complete oscillation) is:

T = 2\pi \sqrt{\frac{L}{g}}

That’s it. No mass term. No amplitude term (for small angles).

T = 2\pi \sqrt{\frac{L}{g}}

Why does this formula make sense?

$L$ in numerator: longer string → larger $T$ (slower swing). Doubling $L$ multiplies $T$ by $\sqrt{2} \approx 1.4$ .
$g$ in denominator: stronger gravity (larger $g$ ) → smaller $T$ (faster swing). On the Moon ( $g \approx 1.6$ m/s²), the same pendulum swings much slower.
$2\pi$ : comes from the mathematics of simple harmonic motion — the pendulum’s motion is approximately sinusoidal for small angles.

Tip

To remember: the formula is identical to that of a mass on a spring ( $T = 2\pi\sqrt{m/k}$ ), but here the “restoring force per unit displacement” is $mg/L$ , so the effective “ $k$ ” is $mg/L$ , giving $T = 2\pi\sqrt{L/g}$ .

Common exam pitfalls

Don’t confuse $L$ with amplitude. $L$ is the string length, not how far you pull it. …

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The bob's speed when the string snaps is found from SHM, $v=\tfrac{\sqrt3}{2}\omega l\theta_0$ . With $\theta_0$ small this velocity is horizontal and the bob is at height $\approx H$ , so it becomes a projectile: it falls for $t=\sqrt{2H/g}$ and moves horizontally $vt$ . Adding its initial horizontal offset $\approx l\theta_0/2$ from A gives the landing distance.

Step 1 — angular frequency

\omega=\frac{2\pi}{T}=\frac{2\pi}{1}=2\pi\ \text{rad s}^{-1}.

Step 2 — speed at $\theta=\theta_0/2$

Treating the swing as SHM of the arc, with angular amplitude $\theta_0$ , the speed at angular displacement $\theta$ is $v=\omega l\sqrt{\theta_0^2-\theta^2}$ . At $\theta=\theta_0/2$ :

v=\omega l\sqrt{\theta_0^2-\tfrac{\theta_0^2}{4}}=\omega l\,\theta_0\frac{\sqrt3}{2}=\frac{\sqrt3}{2}(2\pi)l\theta_0=\sqrt3\,\pi l\theta_0.

Step 3 — projectile motion after the snap

Because $\theta_0$ is small, the string is nearly vertical, so the velocity (perpendicular to the string) is nearly horizontal, and the bob's height is $\approx H$ (the small rise $l(1-\cos\tfrac{\theta_0}{2})$ is second order and negligible). Taking the initial vertical velocity as $\approx 0$ , the vertical fall gives

H=\frac12 g t^2\ \Rightarrow\ t=\sqrt{\frac{2H}{g}}.

Step 4 — horizontal distance from A

During the fall the bob moves horizontally by …

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Step 1 — angular frequency

Step 2 — speed at θ=θ0/2\theta=\theta_0/2θ=θ0​/2

Step 3 — projectile motion after the snap

Step 4 — horizontal distance from A

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Step 2 — speed at $\theta=\theta_0/2$

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