How to double the time period of a simple pendulum For a simple pendulum operating at small amplitude, what change is required to double its period of oscillation?

Introduction / Context:
The period of a simple pendulum guides timing devices and dynamic isolation systems. Understanding parameter sensitivity helps engineers tune response without trial-and-error changes in mass or structure.

Given Data / Assumptions:

• Small-angle oscillations so that T = 2π * sqrt(l / g) applies.
• Mass of bob is concentrated; string is light and inextensible.
• Local gravity g is constant.

Concept / Approach:

Period T depends on length l and gravity g only. Mass does not appear in T. To change T by a factor, scale l accordingly using the square-root relationship.

Step-by-Step Solution:

Verification / Alternative check:

Plug into T = 2π * sqrt(l / g). If l → 4l, then T → 2π * sqrt(4l/g) = 2 * 2π * sqrt(l/g) = 2T, confirming the requirement.

Why Other Options Are Wrong:

(a) and (b) change mass, which does not affect T in the simple model. (d) doubling length increases T by sqrt(2), not 2. (e) while halving g would increase T by sqrt(2); it does not double T unless g is reduced by a factor of 4, which is unrealistic.

Common Pitfalls:

Believing mass influences period; forgetting the square-root dependence; applying the formula outside small-angle limits.

Final Answer:

quadruple the length of the pendulum