Question

A simple pendulum performing small oscillations at a height R above Earth's surface has a time period of \(T_1 = 4\) s. What would be its time period at a point which is at a height \(2R\) from Earth's surface?

• A. \(T_1 = T_2\)
• B. \(2T_1 = 3T_2\)
• C. \(3T_1 = 2T_2\)
• D. \(2T_1 = T_2\)

Hint:

The time period of a simple pendulum is independent of the height above Earth's surface for small oscillations.

Answer 1

The Correct Option is A

The time period of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] Where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity. At height \(R\) above Earth's surface, \(g\) is reduced slightly, but since we're dealing with small changes in height, the time period will remain approximately the same at a height \(2R\). Thus, \(T_1 = T_2\).