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

Analyze the question: The length of a simple pendulum is multiplied by 4. We need to find how many times the period increases.

Concept: The period of a simple pendulum is

T = 2π√(L/g)

So T is proportional to √L (g and 2π stay constant).

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)

Therefore:

T₂/T₁ = √4 = 2

The period increases by a factor of 2.

Checking the options:

• A (0.5π) and C (2π) are wrong because π is already part of the formula and cancels out in the ratio.
• D (4) would be correct only if T were directly proportional to L, but it depends on the square root of L.

Tip: For a pendulum, T ∝ √L. Whatever factor you multiply the length by, take its square root to get the factor for the period. For example, if L is multiplied by 9, T is multiplied by 3. If L is multiplied by 4, T is multiplied by 2. The constants such as 2π and g cancel out in ratio questions, so you never need to include them in the final factor.