Time Period of a Simple Pendulum (Small Angle Approximation)

Problem Statement

A pendulum of length 0.9 m oscillates with small amplitudes. Find the time period of oscillation. Also determine how the period changes if the pendulum is taken to a location where g = 9.75 m/s^2 instead of 9.8 m/s^2.

Given / Required

• Length l = 0.9 m.
• g_1 = 9.8 m/s^2, g_2 = 9.75 m/s^2.

We need T_1 and T_2.

Hint

Use the small-angle result \(T = 2\pi \sqrt{l / g}\). Note that mass and amplitude do not appear in the expression.

Step-by-Step Solution

Step 1 – Time Period at g = 9.8 m/s^2

\[ T_1 = 2\pi \sqrt{\frac{0.9}{9.8}} = 2\pi \sqrt{0.09184} \approx 2\pi (0.303) \approx 1.90,\text{s} \]

Step 2 – Time Period at g = 9.75 m/s^2

\[ T_2 = 2\pi \sqrt{\frac{0.9}{9.75}} = 2\pi \sqrt{0.09231} \approx 1.91,\text{s} \]

Final Answer

• Original location: T approx 1.90 s.
• Slightly weaker gravity: T approx 1.91 s (a tiny increase).

Extension / Variation

• Try changing the length to see how sensitive the period is.
• Explore large-angle oscillations with numerical methods.

Key Concept Recap

• For small angles, period depends only on length and gravity.
• Reducing gravity lengthens the period slightly.
• Mass and amplitude do not appear in the small-angle formula.