# Acceleration due to gravity pendulum

1. Jul 14, 2011

### Alpharup

though my higher secondary book lays down procedures to find the acceleration due to gravity(g} and conclude that it there using a simple pendulum and gives the formula g= 4(pi)^2L/T^2 where L is the length of the string and T is the time period. the author has not given the derivations as my textbooks encourage only rote learning. could you please give me the derivation of this formula and also the proof that acceleration due to gravity exists?

2. Jul 14, 2011

### ZapperZ

Staff Emeritus
3. Jul 14, 2011

### Superstring

If a simple pendulum is displaced an angle $\theta$ from the vertical, then the force acting on the bob at any angle is given by:

$$F=-mgsin\theta$$

Where m is the mass of the bob and g is the acceleration due to gravity.

This means that the torque is given by:

$$\tau =-mgLsin\theta$$

Where L is the length of the string.

Since $\tau = I\alpha$ (where $I$ is the moment of inertia and $\alpha$ is the angular acceleration):

$$I\alpha =-mgLsin\theta$$

At small angles, $sin\theta\approx \theta$. This gives the approximate equation:

$$I\alpha =-mgL\theta$$

The moment of inertia for a point mass system is: $I=mr^2$. In this case, $r=L$. This simplifies the equation to:

$$\alpha +\frac{g}{L}\theta =0$$

Since the angular acceleration is the second derivative of the angle with respect to time, this yields the differential equation:

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

The solution to this differential equation is:

$$\theta=\theta_0cos(\omega t+\phi)$$

Where $\theta_0$ is the initial angular displacement (amplitude), $\phi$ is the phase shift, and $\omega = \sqrt{\frac{g}{L}}$

The period of this function is given by:

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

Solving for $g$, we obtain:

$$g=\frac{4\pi^2L}{T^2}$$