A simple pendulum has length $L$ and small oscillations with period $T$. Assuming local gravitational acceleration is $g$, which expression correctly calculates $g$ from $L$ and $T$?

• A. $g=\frac{4{\pi }^{2}L}{T^2}$
• B. $g=\frac{2\pi L}{T}$
• C. $g=\frac{T^2}{4{\pi }^{2}L}$
• D. $g=\frac{L}{2T}$

Answer: (A)

Verified step by step guidance

1. Recall the formula for the period of a simple pendulum undergoing small oscillations: \(T = 2\pi \sqrt{\frac{L}{g}}\).
2. Square both sides of the equation to eliminate the square root: \(T^2 = (2\pi)^2 \frac{L}{g}\).
3. Rewrite the squared term: \(T^2 = 4\pi^2 \frac{L}{g}\).
4. Isolate \(g\) by multiplying both sides by \(g\) and then dividing both sides by \(T^2\): \(g = \frac{4\pi^2 L}{T^2}\).
5. This expression shows how to calculate the gravitational acceleration \(g\) from the pendulum length \(L\) and the period \(T\).