Q.a) Prove that the oscillations of a simple pendulum are simple harmonic and hence derive an expression for the time period of a simple pendulum.
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Start your 14-day free trial to unlock the full solution →A simple pendulum's restoring force is proportional to its (small) angular displacement, which is the defining condition of SHM; this gives T = 2π√(L/g). A pendulum that "ticks seconds" has T = 2 s (one tick per second, so 2 s per full oscillation), giving a length of about 0.993 m.
a) Showing the pendulum's oscillation is SHM, and finding its time period
Consider a simple pendulum: a bob of mass m attached to a string of length L (assumed massless and inextensible), free to swing about a fixed support. Let the bob be displaced through a small angle θ from its equilibrium (vertical) position and released.
Two forces act on the bob: gravity mg (vertically downward) and the tension T along the string. Resolve gravity into two components:
- mg cosθ, along the string (balanced by the tension, since the bob moves along a nearly straight arc for small θ)
- mg sinθ, perpendicular to the string, tangential to the arc — this is the RESTORING force, always directed back toward the equilibrium (mean) position
So the net tangential (restoring) force is:
F = −mg sinθ
(The negative sign shows the force opposes the direction of increasing θ, i.e., it always pushes the bob back toward the vertical.)
For SMALL angular displacements (θ in radians, θ small), sinθ ≈ θ. Also, the arc length displacement of the bob from equilibrium is x = Lθ, so θ = x/L. Substituting:
F ≈ −mgθ = −mg(x/L) = −(mg/L)x
This is exactly of the form F = −kx, with k = mg/L — a restoring force that is directly proportional to the displacement x and always directed opposite to it. This is precisely the defining condition for Simple Harmonic Motion. Hence, for small angular displacements, the oscillation of a simple pendulum is simple harmonic.
Finding the time period. For any SHM system, Newton's second law gives F = ma = −mω²x, where ω is the angular frequency of oscillation. Comparing this standard SHM form with F = −(mg/L)x obtained above:
mω² = mg/L
ω² = g/L
ω = √(g/L)
The time period T (time for one complete oscillation) is related to ω by T = 2π/ω, so:
T = 2π√(L/g)
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