The time period of a simple pendulum of length $L$ in a gravitational field $g$ is:
A$T = 2\pi \sqrt{\dfrac{L}{g}}$
B$T = 2\pi \sqrt{\dfrac{g}{L}}$
C$T = 2\pi L g$
D$T = \sqrt{\dfrac{L}{g}}$
Answer & Solution
Correct answer: A. $T = 2\pi \sqrt{\dfrac{L}{g}}$
For small oscillations the pendulum's equation of motion reduces to SHM with $\omega = \sqrt{g/L}$. So period $T = 2\pi/\omega = 2\pi\sqrt{L/g}$.
Independent of the mass of the bob (a key contrast with the spring-mass system, where $T \propto \sqrt{m}$).
Note: only valid for small-amplitude swings. At larger amplitudes the small-angle approximation fails and the period depends slightly on amplitude.
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