The length of a second's pendulum on the surface of Earth is 1m. What will be the length of a second's pendulu

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Q. The length of a second's pendulum on the surface of Earth is 1m. What will be the length of a second's pendulum on the moon?

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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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A second's pendulum has a period of 2 s. Since the period depends on $\sqrt{\frac{L}{g}}$ , reducing gravity by a factor of 6 (Earth to Moon) requires reducing length by the same factor to maintain the period. The length on the Moon is $\frac{1}{6}$ m ≈ 0.167 m.

Why the period of a pendulum depends on gravity

A simple pendulum swings because gravity provides the restoring force. The period—the time for one complete oscillation—is given by

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

where $L$ is the length and $g$ is the acceleration due to gravity. This formula tells us that a longer pendulum swings more slowly (larger $T$ ), and stronger gravity makes it swing faster (smaller $T$ ).

A "second's pendulum" is defined as one whose half-period is exactly one second, so its full period is $T = 2$ s. On Earth, with $g_{\text{Earth}} \approx 9.8 \text{ m/s}^2$ , this requires $L = 1$ m.

When we move to the Moon, gravity is weaker—about $\frac{1}{6}$ of Earth's value. To keep the same 2 s period, we must adjust the length.

Finding the length on the Moon

Write the period equation for Earth.

On Earth, the second's pendulum has:

T = 2\pi\sqrt{\frac{L_{\text{Earth}}}{g_{\text{Earth}}}} = 2 \text{ s}

with $L_{\text{Earth}} = 1$ m.

Write the period equation for the Moon.

On the Moon, we want the same period $T = 2$ s, so:

T = 2\pi\sqrt{\frac{L_{\text{Moon}}}{g_{\text{Moon}}}} = 2 \text{ s}

Equate the two expressions.

Since both equal 2 s:

2\pi\sqrt{\frac{L_{\text{Earth}}}{g_{\text{Earth}}}} = 2\pi\sqrt{\frac{L_{\text{Moon}}}{g_{\text{Moon}}}}

Cancel $2\pi$ :

\sqrt{\frac{L_{\text{Earth}}}{g_{\text{Earth}}}} = \sqrt{\frac{L_{\text{Moon}}}{g_{\text{Moon}}}}

Square both sides and rearrange. …

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Q.The length of a second's pendulum on the surface of Earth is 1m. What will be the length of a second's pendulum on the moon?

Concept understanding — Simple Pendulum Period

Simple Pendulum Period: From Intuition to Formula

The Precise Statement

Why does this formula make sense?

Common exam pitfalls

Length of a Second's Pendulum on the Moon

Why the period of a pendulum depends on gravity

Finding the length on the Moon

Q.The length of a second's pendulum on the surface of Earth is 1m. What will be the length of a second's pendulum on the moon?

Why the period of a pendulum depends on gravity

Finding the length on the Moon

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