# Simple Pendulum

A simple pendulum is a small mass suspended from a string that oscillates under the force of gravity. Here, the gravitational force provides the restoring torque required for oscillation. In a simple pendulum, the string or connecting rod is assumed to be either massless or small compared to the suspended mass that it can be neglected.

We will use the concept that total energy is constant for a body that is executing Simple Harmonic Motion to derive the expression for the period. Also we will neglect any air resistance and assume that the pendulum can oscillate forever after being disturbed.

Consider a pendulum with mass, m suspended from a length l from a pivot as shown below.

The force of gravity on the bob of mass m is mg. When the pendulum is at an angle θ to the vertical as shown, the gravitational force can be split into two components parallel and perpendicular to the string’s direction.

We can see that the restoring torque is $$mg~sin~θ$$ .

If the pendulum is travelling at a speed $$v$$ and has an acceleration $$a$$ ,

The Kinetic energy is

$$KE$$ = $$\frac{1}{2} mv^2$$

Let us consider the lowest point of the pendulum to have zero potential energy. When the pendulum travels upward, the potential energy increases. It can be seen from the diagram that the bob moves vertically a distance $$l~-~lcosϑ$$ as the bob swings by an angle $$θ$$ .

Thus the potential energy is:

$$PE$$ = $$mg(l~-~l ~cos~θ)$$

The total energy is a constant. Thus

$$Total~ Energy$$ = $$E$$ = $$KE + PE$$

$$E$$ = $$\frac{1}{2}~ mv^2 + mg(l~-~l ~cos~θ)$$

Let us differentiate with respect to time (the left hand side is zero since differential of a constant is zero)

$$\frac{dE}{dt}$$ = $$0$$ = $$\frac{d}{dt} ~\left(\frac{1}{2}~ mv^2 + mg(l~-~l ~cos~θ)\right)$$

$$0$$ = $$\frac{d}{dt}~ (\frac{1}{2}~ mv^2) + \frac{d}{dt} (mgl – mgl~ cos~θ)$$

Thus we have,

$$0$$ = $$mv. {dv}{dt} + 0~ -~ mgl(-sin~θ).\frac{dθ}{dt}$$

$$τ$$ = $$\frac{dθ}{dt}$$

And

$$v$$ = $$lω$$

(Linear velocity is the angular velocity (τ) times radius for motion along a circular arc)

$$0$$ = $$ml^2 τ.{dτ}{dt} + mgl sin~θ.τ$$

$$0$$ = $$ml ~\frac{d^2~ θ}{dt^2} + mg sin~θ$$

For small oscillations i.e. $$θ$$ being small, we can approximate

$$sin~θ ≈ θ$$

$$-mg~θ$$ = $$ml ~\frac{d^2 θ}{dt^2}$$

$$\frac{d^2 θ}{dt^2}$$ = $$-\frac{g}{l}~ θ$$

This is the equation of Simple Harmonic Motion whose general form is:

$$\frac{d^2 x}{dt^2}$$ = $$-ω^2 x$$

Where ω is the circular frequency for an oscillation time period T,

$$ω$$ = $$\frac{2π}{T}$$

Thus the angular frequency for a simple pendulum by comparing the two differential equations is:

$$ω$$ = $$\frac{2π}{T}$$ = $$√{\frac{g}{l}}$$

In other words,

$$Time~ period$$ = $$T$$ = $$2π√{\frac{l}{g}}$$