Time period of a long Pendulum

( 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 = − mgL2 θ(1/L + 1/R)

Iα = − mgL2 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 ?

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