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 ?

Also Read :

→ Stable , Unstable & Neutral Equilibrium
→ S.H.M :Linear SHM & Angular SHM
→ Analytical Treatment to SHM
→ Kinetic Energy & Potential Energy & Total Energy in SHM
→ Average Value of P.E. & K.E. of Harmonic Oscillator
→ SHM as a projection of circular motion
→ Simple Pendulum in Inertial & Non Inertial Frame
→ SHM of Spring Mass System
→ Physical Pendulum & Torsional Pendulum
→ Undamped & Damped simple harmonic oscillations

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