Question

Derive an expression for time period of simple pendulum ( for 5marks question)

Answer 1

We know that acceleration in SHM is given by
a=W²y .....(1)
and we know that W is 2π/T
putting this value in (1) we get a=(2π/T)²y
Therefore, T=2π√(y/a).....(2)
Now for a simple pendulum,we consider that it's radius is equal to length "l" and the displacement from the mean position is "y".There are few forces that act on the pendulum:- tension(T),and weight of the pendulum bob (mg) (we consider no weight of the string).
when the pendulum moves by an angle θ.
therefore, tension is cancelled by mgcosθ and the net restoring force is mgsinθ.
thus, F=mgsinθ
so ma=mgsinθ
​thus , a=gsinθ
Now, if θ is very small,then sinθ≃θ
Hence, a=gθ.we know θ=length of arc/radius
hence,θ=y/l.
therefore, a=gy/l
and,y/a=l/g
now just put this value in equation (2) above and we get T=2π√(l/g) .
This is the expression for time period of a simple pendulum .