A simple pendulum of length 0.8 m is displaced from its equilibrium position by an angle of 5° and released. Using the small angle approximation, find the period of its oscillation.

easy 1 mark Class 11 · Physics · Oscillations

Answer

1.80 s (approximately)

The correct answer is 1.80 s (approximately) — Round the result to two significant figures.

Step-by-step solution

Compute the period using the formula $T = 2 π \frac{l}{g}$ .
Insert the given values $l = 0.8 m$ and $g = 9.8 m/s^{2}$ to get $T = 2 π \frac{0.8}{9.8}$ .
Evaluate the square root $0.08163 = 0.2857$ .
Multiply by $2 π$ to obtain $T = 2 π \times 0.2857 \approx 1.80 s$ .
Round the result to two significant figures.

Common mistake: Using the full pendulum formula with

sin θ

instead of the small-angle approximation, leading to an overestimate of the period.

Concept tested: Period of simple pendulum (small-angle approximation)