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?
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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 = 2π * (l / g)
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ET = (1 / 2π) * sqrt(l / g)
Answer
Correct Answer: T = 2π * sqrt(l / g)
Explanation
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:
Set up equation: angular acceleration = −(g / l) * θ.Compare with SHM form: θ¨ + ω^2 * θ = 0, so ω = sqrt(g / l).Compute period: T = 2π / ω = 2π * sqrt(l / g).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)