Simple pendulum – determine the correct time period expression For a simple pendulum of length l oscillating with small angular displacement under gravity g, what is the correct expression for the period of one complete oscillation?

Introduction / Context:
The simple pendulum is a foundational vibration system used to approximate clocks, seismic instruments, and sway of slender structures. Its time period depends on geometric length and gravitational acceleration, assuming small-angle motion and a massless, inextensible string.

Given Data / Assumptions:

• Point mass bob, string length l, gravity g.
• Small angular amplitude so that sin θ ≈ θ (in radians).
• No air resistance or pivot friction; string mass negligible.

Concept / Approach:

For small oscillations, the linearized equation reduces to simple harmonic motion with angular frequency ω = sqrt(g / l). The period T equals 2π divided by ω, giving the classic square–root relationship between T, l, and g.

Step-by-Step Solution:

Verification / Alternative check:

Dimensional check: sqrt(l / g) has units of time^(1), ensuring T is in seconds. Longer pendulums have larger periods, consistent with observation.

Why Other Options Are Wrong:

(b) inverts the ratio. (c) multiplies l and g incorrectly. (d) lacks the square root. (e) incorrect constant factor.

Common Pitfalls:

Using large angles where the small-angle approximation fails; forgetting that mass cancels and does not affect T.

Final Answer:

T = 2π * sqrt(l / g)