T = 2π√(L/g). So L = g(T/2π)² = 10 × (2/2π)² ≈ 1 m. Classic SHM calculation.

A Simple Pendulum Has Period 2s — What is its Length? (g = 10)

Question

A simple pendulum has a time period of 2 seconds. Find its length. (Take g = 10 m/s²)

Solution — Step by Step

The time period of a simple pendulum is:

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

Here, $T$ is the time period, $L$ is the length of the pendulum, and $g$ is acceleration due to gravity.

We need $L$ , so rearrange. Square both sides:

T^2 = 4\pi^2 \cdot \frac{L}{g}

Now multiply both sides by $g$ and divide by $4\pi^2$ :

L = \frac{g T^2}{4\pi^2}

Plug in $T = 2$ s and $g = 10$ m/s²:

L = \frac{10 \times (2)^2}{4\pi^2} = \frac{10 \times 4}{4\pi^2} = \frac{40}{4\pi^2} = \frac{10}{\pi^2}

We know $\pi^2 \approx 9.87$ , so:

L = \frac{10}{9.87} \approx 1.01 \text{ m} \approx \mathbf{1 \text{ m}}

The length of the pendulum is approximately 1 metre.

Why This Works

The formula $T = 2\pi\sqrt{L/g}$ comes from the SHM analysis of a pendulum. When a pendulum is displaced by a small angle, the restoring force is $mg\sin\theta \approx mg\theta$ (for small $\theta$ ). This gives simple harmonic motion with angular frequency $\omega = \sqrt{g/L}$ , and since $T = 2\pi/\omega$ , we get the formula directly.

The key physical insight: a longer pendulum swings slower (larger $L$ → larger $T$ ), and a stronger gravitational field speeds it up (larger $g$ → smaller $T$ ). A 1 m pendulum taking exactly 2 seconds to complete one oscillation is actually the definition of the “seconds pendulum” — a concept that comes up in PYQs quite often.

Notice that mass does not appear anywhere in the formula. This is not a coincidence — heavier bobs don’t swing faster. Galileo’s observation about pendulums is baked right into the math.

Alternative Method

If you remember that a seconds pendulum (T = 2 s) has length approximately 1 m on Earth, you can directly state the answer in MCQs without calculation. This is a high-value shortcut for JEE Main.

We can also verify using dimensional analysis. The only combination of $g$ (m/s²) and $T$ (s) that gives length (m) is:

L \sim g T^2

Up to the constant $4\pi^2 \approx 39.5$ , the answer is $L = \frac{10 \times 4}{39.5} \approx 1$ m. Dimensional reasoning confirms we haven’t made a structural error.

Common Mistake

Forgetting to square T before substituting. A very common slip is writing $L = gT / 2\pi$ instead of $L = gT^2 / 4\pi^2$ . This happens when students try to “shortcut” the rearrangement step. Always square the entire equation first, then isolate $L$ — do not try to take shortcuts with square roots algebraically.

Another trap: using $\pi \approx 3$ instead of $\pi^2 \approx 9.87$ . When the question gives $g = 10$ m/s² and $T = 2$ s, the numbers are chosen to give a clean answer of ~1 m — but only if you use $\pi^2 \approx 10$ as an approximation (which gives exactly $L = 1$ m). Many CBSE board marking schemes accept this approximation. JEE questions will usually specify whether to use $\pi^2 = 10$ or the exact value, so read carefully.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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