What happens to the period of a simple pendulum if its length is doubled?

What happens to the period of a simple pendulum if its length is doubled?

• A. Increases by a factor of 2
• B. Increases by a factor of √2
• C. Remains the same
• D. Decreases by a factor of 2

Answer: (B)

The period of a simple pendulum is determined by the formula \(T = 2\pi\sqrt{\frac{L}{g}}\), where \(T\) is the period, \(L\) is the length of the pendulum, and \(g\) is the acceleration due to gravity. When the length \(L\) of the pendulum is doubled, we can express the new period as follows: 1. Substituting \(2L\) in place of \(L\) in the original formula gives us \(T' = 2\pi\sqrt{\frac{2L}{g}}\). 2. This can be simplified to \(T' = 2\pi\sqrt{2} \times \sqrt{\frac{L}{g}}\). 3. Recognizing that the original period \(T\) is \(2\pi\sqrt{\frac{L}{g}}\), we find that the new period \(T'\) equals \( \sqrt{2} \times T\). Therefore, the period increases by a factor of \(\sqrt{2}\). This demonstrates that when the length of a simple pendulum is doubled, the period does not simply increase linearly but grows

The period of a simple pendulum is determined by the formula (T = 2\pi\sqrt{\frac{L}{g}}), where (T) is the period, (L) is the length of the pendulum, and (g) is the acceleration due to gravity.

When the length (L) of the pendulum is doubled, we can express the new period as follows:

1. Substituting (2L) in place of (L) in the original formula gives us (T' = 2\pi\sqrt{\frac{2L}{g}}).

2. This can be simplified to (T' = 2\pi\sqrt{2} \times \sqrt{\frac{L}{g}}).

3. Recognizing that the original period (T) is (2\pi\sqrt{\frac{L}{g}}), we find that the new period (T') equals ( \sqrt{2} \times T).

Therefore, the period increases by a factor of (\sqrt{2}). This demonstrates that when the length of a simple pendulum is doubled, the period does not simply increase linearly but grows