Derive the formula for length of a simple pendulum

Consider a pendulum of mass $m$ and length $L$, displaced by an angle $\theta$ from the vertical. The restoring force acting tangentially to the arc is the component of gravity, $F_t = -mg \sin \theta$. According to Newton's second law, $F_t = ma_t$, where $a_t = L \frac{d^2\theta}{dt^2}$ is the tangential acceleration.

$$-mg \sin \theta = mL \frac{d^2\theta}{dt^2}$$

Divide both sides by $m$ and rearrange the terms.

$$\frac{d^2\theta}{dt^2} + \frac{g}{L} \sin \theta = 0$$

For small angles of oscillation (typically $\theta < 10^\circ$), we can approximate $\sin \theta \approx \theta$ (where $\theta$ is in radians). This simplifies the differential equation.

$$\frac{d^2\theta}{dt^2} + \frac{g}{L} \theta = 0$$

This equation is in the standard form for simple harmonic motion, $\frac{d^2x}{dt^2} + \omega^2 x = 0$, where $\omega$ is the angular frequency. By comparing, we find the angular frequency squared.

$$\omega^2 = \frac{g}{L}$$ $$\omega = \sqrt{\frac{g}{L}}$$

The period $T$ of simple harmonic motion is related to the angular frequency $\omega$ by the formula $T = \frac{2\pi}{\omega}$. Substitute the expression for $\omega$.

$$T = \frac{2\pi}{\sqrt{\frac{g}{L}}}$$ $$T = \boxed{2\pi \sqrt{\frac{L}{g}}}$$