Question

show that for small oscillations the motion of the simple pendulum is simple harmonic.derive an expression for its time period.

Answer 1

A motion be Simple harmonic motion only when ,
1. Acceleration of particle is just opposite to motion of body
2. Acceleration is directly proportional to displacement e.g., a = -ω²x

A pendulum moves in such a way that angle is formed by it is θ , as you know , along tangent , motion of pendulum is just opposite to force. As shown in figure.
now, restoring force , F = -mgsinθ ,
When displacement of pendulum is very small , then sinθ ≈ θ
so, F = -mgθ, also here it is clear , θ = x/L
∴F = -mgx/L
Now use F = ma { Newton's second law }
ma = -mgx/L ⇒ a = -gx/L
Here what you see both the above conditions are satisfied when displacement of pendulum is small.
Now, comapre both the expressions ,
∴ω² = g/L
we know, ω = 2π/T , here T is time period .
so, {2π/T}² = g/L
⇒ T = 2π√{L/g}

Hence, for pendulum time period is T = 2π√{L/g}