If a simple pendulum's length is doubled, what happens to the period (small-angle approximation)?

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Multiple Choice

If a simple pendulum's length is doubled, what happens to the period (small-angle approximation)?

Explanation:
The period of a simple pendulum in small-angle motion scales with the square root of its length, T = 2π√(L/g). If the length is doubled, the period becomes T = 2π√(2L/g) = √2 times the original period. Since √2 ≈ 1.414, that’s about a 41% longer period. So the correct outcome is that the period increases by about 41%. Doubling doesn’t double the period, nor leave it unchanged; the 41% figure matches the √L dependence.

The period of a simple pendulum in small-angle motion scales with the square root of its length, T = 2π√(L/g). If the length is doubled, the period becomes T = 2π√(2L/g) = √2 times the original period. Since √2 ≈ 1.414, that’s about a 41% longer period. So the correct outcome is that the period increases by about 41%. Doubling doesn’t double the period, nor leave it unchanged; the 41% figure matches the √L dependence.

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