Q.Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. The mass and radius of a planet are double that of the earth. If the time period of a simple pendulum on the earth is T, find the time period on the planet.
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Start your 14-day free trial to unlock the full solution →For small angular displacement, the restoring force on a simple pendulum's bob is proportional to its displacement and directed opposite to it — the defining condition for SHM — giving T = 2π√(l/g). Since the planet described has half Earth's surface gravity, the pendulum's period there becomes √2 times its period on Earth.
Part 1 — Showing the motion is SHM.
Consider a simple pendulum: a bob of mass m suspended by a light, inextensible string of length l from a fixed support, allowed to oscillate in a vertical plane. Let θ be the small angular displacement of the string from the vertical at any instant.
The forces on the bob are its weight mg (vertically down) and the tension T along the string. Resolving the weight along and perpendicular to the string, the component of weight along the string is balanced by tension; the component perpendicular to the string (tangential to the bob's circular arc) is mg sin θ, and this tangential component is what restores the bob toward the mean (vertical) position:
Restoring force, F = −mg sin θ
(the negative sign shows it acts opposite to the displacement, always trying to bring the bob back to θ = 0).
For small angular displacements (θ small, measured in radians), sin θ ≈ θ, so:
F ≈ −mg θ
Since the arc length (linear displacement along the arc) is x = lθ, we have θ = x/l, so:
F = −mg(x/l) = −(mg/l) x
This is of the form F = −kx with k = mg/l — the force is directly proportional to the displacement x and directed opposite to it. This is exactly the defining condition for simple harmonic motion (SHM). Hence, for small oscillations, the simple pendulum executes SHM.
Part 2 — Deriving the time period.
For SHM under a force F = −kx, comparing with Newton's second law F = ma = m(d²x/dt²):
m(d²x/dt²) = −kx, so d²x/dt² = −(k/m)x = −ω²x, where ω² = k/m
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