A simple pendulum of length 0.4m has a period of 2s. What is the period a similar pendulum of length 0.8m at the same place?

Analyze the question: The length of a simple pendulum is doubled (0.4 m to 0.8 m) at the same place, so g is unchanged. We need the new period.

Concept: The period of a simple pendulum is

T = 2π√(L/g)

So T is proportional to √L when g is constant. This gives:

T₂/T₁ = √(L₂/L₁)

Solution:

• T₁ = 2 s, L₁ = 0.4 m, L₂ = 0.8 m
• T₂ = T₁ × √(L₂/L₁)
• T₂ = 2 × √(0.8/0.4)
• T₂ = 2 × √2
• T₂ = 2√2 s ≈ 2.83 s

Check against the options:

• A (8 s) and B (4 s) are too large. Doubling the length does not double the period, it multiplies it by √2.
• D (√2 s) is too small, because the period must increase when the length increases.
• C (2√2 s) matches our result.

Exam tip: Remember that "T goes as the square root of L". If the length is multiplied by 2, the period is multiplied by √2. If the length is multiplied by 4, the period is multiplied by 2. This lets you skip the full formula and avoid computing π and g.