Simple pendulum is a simple mechanical system which executes the periodic motion. A simple pendulum consists of a bob attached to a string that is fastened such that the pendulum assembly can swing or oscillate freely in a plane. For a simple or ideal pendulum, all the mass is considered to be concentrated at a point at the center of the bob. It follows simple harmonic motion in smaller angles. |

A simple pendulum consists of a concentrated weight, such as a metal ball, suspended from a fixed point by a string so that it can swing back and forth under the influence of gravity. The length (l) of the thread of the pendulum, from the point of suspension to the center of the metal ball, is called the length of the pendulum. A single swing of the pendulum from one extreme limit to the other and back is known as an oscillation.

**The period of a can be defined as the time required to complete one oscillation by a metal bob.**The mathematical expression of period is given as

T = 2$\pi$$\sqrt{\frac{l}{g}}$

g = gravitational accelerationl = pendulum length

**Simple pendulum frequency can be explained as how many number of oscillations can complete in one second.**Frequency of a simple pendulum is given as

*f = $\frac{1}{T}$*

f = $\frac{1}{2\pi}\sqrt{\frac{g}{l}}$

T is the period
Simple harmonic motion occurs in any system where the tendency to return to equilibrium increases in direct proportion to the displacement from equilibrium.

**A simple pendulum follows a simple harmonic motion at a smaller angle (less than 10°). All higher angles it is not follows the simple harmonic motion.***A condition for simple harmonic motion is that the restoring force is proportional to the displacement from the equilibrium position.***Some of the problems related to the topic is discussed here.**

### Solved Examples

**Question 1:**Determine the frequency of a simple pendulum of length 2m? Take g = 9.81m/s

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**Solution:**

Given that,

l = 2m, g = 9.81m/s

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Frequency, f = $\frac{1}{2\pi}\sqrt{\frac{g}{l}}$

f = $\frac{1}{2\times3.14}\sqrt{\frac{9.81}{2}}$

f = 0.352Hz

**Question 2:**If the pendulum of length 3m and its frequency of oscillation is 0.2875Hz, determine the value of g?

**Solution:**

Given that,

f = 0.2875Hz, l = 3m

Frequency, f = $\frac{1}{2\pi}\sqrt{\frac{g}{l}}$

0.2875 = $\frac{1}{2\times3.14}\sqrt{\frac{g}{3}}$

$(0.2875)^2 \times (2\pi)^2$ = $\frac{g}{3}$

g = 3$\times$3.263 = 9.789m/s

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