Oscillations — Simple Pendulum Period What is the periodic time T (time for one oscillation) of a simple pendulum of length l (small-angle assumption)?

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:

  • Point mass at the end of a massless string of length l.
  • Small-angle approximation so that sinθ ≈ θ (in radians).
  • Local gravitational acceleration g is constant.


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:

Start: ω = sqrt(g / l) for small angles. Use T = 2π / ω. Compute: T = 2π / sqrt(g / l) = 2π * sqrt(l / g).


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).

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