Difficulty: Easy
Correct Answer: T = 2π * sqrt(l / g)
Explanation:
Introduction / Context:
The simple pendulum is a classic oscillator used to illustrate periodic motion. Under small angular displacements, its motion approximates simple harmonic motion (SHM) with a well-known expression for the period T in terms of length l and gravitational acceleration g.
Given Data / Assumptions:
Concept / Approach:
Linearizing the pendulum’s equation leads to angular frequency ω = sqrt(g / l). Using the identity T = 2π / ω gives T = 2π * sqrt(l / g). The period is independent of amplitude for small angles and independent of mass.
Step-by-Step Solution:
Verification / Alternative check:
Dimensional analysis: sqrt(l / g) has units sqrt(m / (m/s^2)) = sqrt(s^2) = s; multiplied by 2π gives time (seconds).
Why Other Options Are Wrong:
Any expression with sqrt(g / l) in the numerator or missing the factor 2π does not match SHM theory. Options (b), (c), and (d) are dimensionally or numerically incorrect for a simple pendulum under small angles.
Common Pitfalls:
Forgetting the small-angle condition; mixing up g and l; dropping the 2π factor when converting from angular frequency to period.
Final Answer:
T = 2π * sqrt(l / g).
Discussion & Comments