NECO 2005 Physics · Question 28

The length of a simple pendulum is increased by a factor of four. By what factor is its period Increased?

Options

A.0.25
B.0.5
C.1.66
D.2
E.4

Answer: D. 2

Explanation written with AI and checked against the answer key

Analyze the question: The length of a simple pendulum becomes 4 times its original length. We need to find how its period changes. Concept: The period of a simple pendulum is T = 2π√(L/g) So the period is proportional to the square root of the length (T ∝ √L), as long as g stays the same. Solution: Original period: T₁ = 2π√(L/g) New length: L₂ = 4L New period: T₂ = 2π√(4L/g) = 2π × √4 × √(L/g) = 2 × 2π√(L/g) = 2T₁ So T₂/T₁ = √4 = 2 The period is multiplied by a factor of 2, which is option D. Why not E (4)? A common mistake is to think that the period increases by the same factor as the length. But the period depends on the square root of the length, so the factor is √4 = 2, not 4. Tip: For a pendulum, remember "period goes as the square root of length." If length is multiplied by 4, 9 or 16, the period is multiplied by 2, 3 or 4. Look for perfect squares in such questions to save time in the exam.

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