The time period of a 1 m long pendulum approximates to

[amp_mcq option1=”6 s” option2=”4 s” option3=”2 s” option4=”1 s” correct=”option3″]

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UPSC NDA-1 – 2022

The time period of a 1 m long pendulum approximates to 2 s.

The time period ($T$) of a simple pendulum for small oscillations is given by the formula $T = 2\pi \sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Given length $L = 1$ m. Taking the standard value of $g \approx 9.8$ m/s² and $\pi \approx 3.14$: $T = 2 \times 3.14 \times \sqrt{\frac{1}{9.8}} \approx 6.28 \times \sqrt{0.102} \approx 6.28 \times 0.319 \approx 2.005$ s. A common approximation for calculation purposes is sometimes $\pi^2 \approx g$, which gives $T = 2\pi \sqrt{\frac{1}{\pi^2}} = 2\pi \times \frac{1}{\pi} = 2$ s. Both calculations yield a value very close to 2 seconds.

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