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Q.(a) Show that the oscillations produced in a simple pendulum are simple harmonic.

(2)
(b) Write an expression for time period of oscillation.
(1)
(c) What is seconds pendulum ? Find the length of seconds pendulum. (2)
Kerala DhseKerala DHSE Plus One Board 2022Subjective· 5mImportance★★★★★
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A pendulum bob displaced from its equilibrium (lowest) position experiences a restoring force that, for small swings, is directly proportional to the displacement and always directed back toward equilibrium — which is precisely the defining condition for simple harmonic motion (SHM), F = −kx.

Setup

Consider a simple pendulum: a bob of mass m attached to a light, inextensible string of length L, free to swing about a fixed support. Let θ be the angular displacement of the string from the vertical at some instant, and let x be the corresponding linear (arc) displacement of the bob along its (nearly straight, for small θ) path, so x = Lθ.

Forces on the bob

Two forces act on the bob: gravity mg (vertically down) and the tension T along the string. Resolve gravity into two components:

  • Along the string (toward the support): mg cosθ, balanced by the tension T (this component does not affect the swinging motion).
  • Perpendicular to the string, i.e. tangential to the bob's arc: mg sinθ, directed back toward the equilibrium (mean) position — this is the restoring force.

So the net tangential (restoring) force is

F=−mgsin⁡θF = -mg\sin\theta

(the negative sign shows it always opposes the displacement, pulling the bob back toward θ = 0).

Small-angle approximation

For small angular displacements (θ typically less than about 10°, in radians),

sin⁡θ≈θ\sin\theta \approx \theta

so

F≈−mgθF \approx -mg\theta

Since x = Lθ, i.e. θ = x/L,

F=−mg(xL)=−(mgL)xF = -mg\left(\frac{x}{L}\right) = -\left(\frac{mg}{L}\right)x

Identifying SHM

This is exactly of the form

F=−kx,k=mgLF = -kx, \qquad k = \frac{mg}{L}

A force directly proportional to displacement and always directed toward the equilibrium position (mean position) is, by definition, the condition for simple harmonic motion. Hence the small-angle oscillations of a simple pendulum are simple harmonic.

Angular frequency and period

Comparing with the standard SHM equation of motion ma=−kxma = -kx, i.e. md2xdt2=−(mgL)xm\dfrac{d^2x}{dt^2} = -\left(\dfrac{mg}{L}\right)x, the angular frequency is

ω=km=gL\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{g}{L}}

and the time period is

T=2πω=2πLgT = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{g}}

Notably, T (and ω) is independent of the mass of the bob and of the amplitude (for small oscillations) — it depends only on the pendulum's length L and the local value of g.

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