The acceleration due to gravity on the surface of moon is $1.7\ \text{m s}^{-2}$. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? ($g$ on the surface of earth is $9.8\ \text{m s}^{-2}$)

The acceleration due to gravity on the surface of moon is 1.7m s⁻². What is the time period of a simple pendul

Exercises · 13.15

1.7\ \text{m s}^{-2}

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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 period of a simple pendulum scales as $\sqrt{1/g}$ , so when gravity decreases by a factor of $9.8/1.7 \approx 5.76$ , the period increases by $\sqrt{5.76} = 2.4$ ; the moon period is 8.4 s.

Why the period depends on gravity

A simple pendulum's time period is determined by two things: the length of the string and the local acceleration due to gravity. The restoring force that pulls the bob back toward equilibrium is proportional to $g$ , so stronger gravity makes the pendulum swing faster (shorter period), while weaker gravity slows it down (longer period).

The exact relationship is

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

Notice that $T \propto \sqrt{1/g}$ . This square-root inverse dependence is the key: if you move the same pendulum (same length $L$ ) to a location with different gravity, the ratio of periods depends only on the ratio of gravitational accelerations.

Finding the moon period from the earth period

We don't need to find $L$ explicitly. Instead, we write the period on Earth and on the Moon, then take the ratio.

On Earth:

T_{\text{earth}} = 2\pi \sqrt{\frac{L}{g_{\text{earth}}}}

We're given $T_{\text{earth}} = 3.5\ \text{s}$ and $g_{\text{earth}} = 9.8\ \text{m s}^{-2}$ .

On the Moon:

T_{\text{moon}} = 2\pi \sqrt{\frac{L}{g_{\text{moon}}}}

with $g_{\text{moon}} = 1.7\ \text{m s}^{-2}$ .

Take the ratio to eliminate $L$ and $2\pi$ :

\frac{T_{\text{moon}}}{T_{\text{earth}}} = \sqrt{\frac{g_{\text{earth}}}{g_{\text{moon}}}}

Substitute the values: …

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Why the period depends on gravity

Finding the moon period from the earth period

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