# 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)

$\displaystyle \tau = -mgL\theta (1 + \frac{\phi}{\theta})$

But θ ≈ sin θ = y/L

and φ ≈ tan φ = y/R

$\displaystyle \tau = -mgL\theta (1+\frac{L}{R})$

$\displaystyle \tau = -mgL^2\theta (\frac{1}{L} +\frac{1}{R})$

$\displaystyle I\alpha = -mgL^2\theta (\frac{1}{L} +\frac{1}{R})$

$\displaystyle \alpha = -g\theta (\frac{1}{L} + \frac{1}{R})$

α ∝ − θ

i.e. Oscillations is simple harmonic in nature

$\displaystyle \omega^2 = g(\frac{1}{L}+\frac{1}{R})$

Special cases :

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

$\displaystyle T = 2\pi \sqrt{\frac{L}{g}}$

(b) If L → ∞ then

$\displaystyle T = 2\pi \sqrt{\frac{R}{g}}$

≈ 84.6 minutes.

(c) If L = R then

$\displaystyle T = 2\pi \sqrt{\frac{R}{2g}}$

≈ 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 ?