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 Î¸ 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 Î¸ is balanced by the tension in the string which is clear from fig. 2 below.

Hence the equation becomes: T âˆ’ mgcosÎ¸ = 0

The force that tends to move it towards the centre O, in the direction opposite to increasing Î¸ = -mg sinÎ¸

i.e, F = -mgsinÎ¸

and F = ma

âˆ´ -mg sinÎ¸ = 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, Î¸ is small, such that the arc OA forms an approximately straight line if OA = x.

When Î¸ is small, sin Î¸ = Î¸ in radians.

Also from fig. 1

Î¸ = \( \frac{x}{l}\)

âˆ´ from equation **

-mg sinÎ¸ = 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 = -Ï‰^{2}x

where Ï‰^{2} = \( \frac{g}{l} \) or Ï‰ = \(\sqrt { \frac{g}{l}}\)

angular velocity (Ï‰) 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 = \(\frac{2 \pi}{Ï‰} \)

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/s^{2}

**Solution:**

Use the formula;

l = 1.10m

g = 9.8 m/s^{2}

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 20Âº, 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

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

open the brackets

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

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