Show that the motion of a simple pendulum is simple harmonic and hence derivean equation for its time period. Find the length of a simple pendulum which ticks seconds. `(g=9.8ms^(-2))`

Explanation

Simple pendulum:In a laboratory a heavy metallic bob is suspended from a rigid support with the help of a supunless thread. This arrangement is known as "simple pendulum".
Let the length of simple pendulum is 'l' and the point of suspension is 'S'. Let the pendulum is drawn to a side by a small angle `'theta'` and allowed free to oscillatioate in the veritcle plane. Then it will oscillate between the extreme position A and B with a displacement say 'X' at any given time.
Let the bob is at onc extreme position say B. The force vertically acting downwards is weight `W=mg.`

By resolving the weight into two perpendicular components:
1) The component `mg sin theta` is responsible for the to and fro motion of the bob.
2) The component `mg cos theta` will balance the tension in the string.
Force useful for motion `F=mgsin theta=ma`(From Newton's 2nd Law)
`thereforea=gsin theta`
When `theta` is small `sin theta~~theta=("arc")/("radius")=(x)/(l)`
`therefore a=g(x)/(l)...............(1)` or a `prop x`
Since acceleration is proportional to displacement and acceleration is always directed towards a fixed point the motion of simple pendulum is "simple harmonic"
Time period of simple pendulum:
Time period of a body in S.H.M`=2pisqrt((Y)/(a))=2pi(("displacement")/("acceleration"))`
From equation(1) `a=(g)/(l)x or (x)/(a)=("displacement"(y))/("acceleration"(a))=(l)/(g)`
`therefore` Time period of simple pendulum. `T=2pisqrt((Y)/(a))=2pisqrt(l/g)`
Problem:
In simple pendulum `T=2pisqrt(l/g) or t=(gt^(2))/(4pi^(2))`
For seconds pendulum `T=2srArrt^(2)=4`
`therefore l=(9.8xx3)/(4pi^(2))=1m(therefore pi^(2) "nearly" 9.8)`.