Lesson 8, Topic 3
In Progress

# For a Simple Pendulum

Lesson Progress
0% Complete

Below we show that the motion of the bob of a simple pendulum is simple harmonic.

If a bob of mass, m, of a simple pendulum is slightly displaced through a small angle and allowed to oscillate about O.

Suppose it is at A, at any instance in time during oscillations and θ is the angle POA, subtended by the string with the vertical.

mg is the force acting on the bob at point A in a vertically downward direction.

At A, the bob is at its extreme position. its velocity is zero, and its component mg cos θ is balanced by the tension in the string which is clear from fig. 2 below.

Hence the equation becomes: T − mgcosθ = 0

The force that tends to move it towards the centre O, in the direction opposite to increasing θ = -mg sinθ

i.e, F = -mgsinθ

and F = ma

∴ -mg sinθ = ma ………**

Period (T) is the time it takes the pendulum to complete one cycle. i.e to swing from A to O to B, and from B to O back to A.

For the period, T, of oscillation;

The angle of displacement, θ is small, such that the arc OA forms an approximately straight line if OA = x.

When θ is small, sin θ = θ in radians.

Also from fig. 1

θ = $$\frac{x}{l}$$

∴ from equation **

-mg sinθ = ma, becomes

$$\scriptsize -mg \normalsize \frac{x}{l}$$ = ma

let’s make a the subject of the formula

ma = $$\scriptsize -mg \normalsize \frac{x}{l}$$

a = $$\scriptsize -\normalsize \frac{\not{m}}{\not{m}} \scriptsize g \normalsize \frac{x}{l}$$

a = $$\frac{-g}{l}\scriptsize x$$

a = -ω2x

where ω2 = $$\frac{g}{l}$$ or ω = $$\sqrt { \frac{g}{l}}$$

angular velocity (ω) of oscillating pendulum

Since the acceleration a of the oscillating mass is proportional to the distance, x from O, we can say that the motion is simple harmonic.

Since T = $$\frac{2 \pi}{ω}$$

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

The formula only works for the oscillations through small angles, as it was something we assumed in the process of proof.

### Example 1:

What is the period and frequency of a simple pendulum that has a length of 1.10m?
g = 9.8 m/s2

Solution:

Use the formula;

l = 1.10m

g = 9.8 m/s2

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

T $$\scriptsize = 2 \pi \sqrt { \normalsize \frac{1.10m}{9.81}}$$

T $$\scriptsize = 2 \pi \sqrt { 0.112s}$$

T = 2.10s

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

f = $$\frac{1}{2.10s}$$

f = 0.48 Hz

### Example 2:

A pendulum with a mass of 0.25 kg has a period of 1.15 seconds. If the initial displacement of the pendulum is 20º, what is the length of the pendulum?

Solution:

To find the length of the pendulum use the formula;

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

male l the subject of the formula

square both sides

T2 $$\scriptsize = \left(2 \pi \sqrt { \normalsize \frac{l}{g}} \right)^2$$

open the brackets

T2 $$\scriptsize = 4 \pi^2 \normalsize \frac{l}{g}$$

$$\scriptsize T^2g = 4 \pi^2 l$$

$$\scriptsize l = \normalsize \frac{T^2g}{4 \pi^2}$$

$$\scriptsize l = \normalsize \frac{1.15^2 \: \times \: 9.81 }{4 \pi^2}$$

$$\scriptsize l = \scriptsize 0.33 \: m$$

error: