__( when Length of the Pendulum is Comparable to Radius of the Earth )__

As shown in the figure, taking direction of g^{→} as directed towards the centre of the earth, we have

τ = − mg × OB

= mgL sin (θ + φ)

≈ − mgL (θ + φ)

τ = − mgL θ(1 + φ/θ)

But θ ≈ sin θ = y/L

and φ ≈ tan φ = y/R

τ = − mgL θ [1 + (L/R)]

τ = − mgL^{2} θ(1/L + 1/R)

Iα = − mgL^{2} θ(1/L + 1/R)

α = − g θ(1/L + 1/R)

α ∝ − θ

i.e. Oscillations is simple harmonic in nature

**Special cases :**

**(a)** If L < < R , 1/L > > 1/R and

**(b)** If L → ∞ then

≈ 84.6 minutes.

**(c)** If L = R then

≈ 1 hour.

**(d)** If L = 1 m then T = 2 seconds and it is called **Second pendulum.**

**Illustration : **Show that the period of oscillation of simple pendulum at depth h below earth’s surface is inversely proportional to √(R − h) where R is the radius of earth. Find out the time period of a second pendulum at a depth R/2 from the earth’s surface ?