Oscillations — Simple Pendulum Period What is the periodic time T (time for one oscillation) of a simple pendulum of length l (small-angle assumption)?
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AT = 2π * sqrt(l / g)
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BT = 2π * sqrt(g / l)
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CT = π * sqrt(l / g)
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DT = 4π * sqrt(g / l)
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Answer
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).