If the length L of a simple pendulum is doubled, by what factor does its period increase?

If the length L of a simple pendulum is doubled, by what factor does its period increase?

• A. sqrt(2)
• B. 2
• C. 1
• D. sqrt(1/2)

Answer: (A)

The period of a simple pendulum (for small angles) depends on the length as T ∝ sqrt(L). More precisely, T = 2π sqrt(L/g). If you double the length, substitute 2L into the formula: T' = 2π sqrt(2L/g) = sqrt(2) × [2π sqrt(L/g)] = sqrt(2) × T. So the period increases by a factor of sqrt(2) (about 1.414, roughly 41% longer). This relationship relies on small-angle motion and neglects air resistance and other non-idealities.

The period of a simple pendulum (for small angles) depends on the length as T ∝ sqrt(L). More precisely, T = 2π sqrt(L/g). If you double the length, substitute 2L into the formula: T' = 2π sqrt(2L/g) = sqrt(2) × [2π sqrt(L/g)] = sqrt(2) × T. So the period increases by a factor of sqrt(2) (about 1.414, roughly 41% longer). This relationship relies on small-angle motion and neglects air resistance and other non-idealities.