Q.The length of a second's pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on surface of planet X such that the acceleration of the planet X is $n$ times greater than the Earth is a) $0.9n$ m b) $0.9/n$ m c) $0.9n^2$ m d) $0.9/n^2$ m

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.

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:

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

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.

Common exam pitfalls

• Don’t confuse $L$ with amplitude. $L$ is the string length, not how far you pull it.
• Don’t include mass. The period does not depend on the bob’s mass — a common trick question.
• Small-angle assumption. If the problem says “small oscillations” or gives an angle < 10°, use this formula. If the angle is large, the formula changes (and is rarely asked in basic exams).

Quick check

A pendulum of length 1 m on Earth ($g = 9.8$ m/s²) has period:

That’s why a “seconds pendulum” (period exactly 2 s) has length about 1 m — a classic exam fact.

Final answer: For a simple pendulum with small amplitude, the period is $T = 2\pi \sqrt{L/g}$, independent of mass and amplitude.

Many students find this page while searching "Simple Pendulum Period formula physics" or "Simple Pendulum Period important questions and answers"; the concept sits firmly within the Class 11 Physics NCERT/CBSE syllabus. It's also a frequent building block for numericals in JEE Main, NEET and state engineering/medical entrance exams, so treating it as a one-time memorisation task rather than an understood idea tends to backfire later.