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Q.Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is seconds pendulum?

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2025Subjective· 8mImportance★★★★★
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A simple pendulum performs SHM for small angular amplitude because its restoring torque is proportional to (and opposite to) its angular displacement; this gives T = 2π√(L/g). A seconds pendulum has T = 2 s.

Setup: Consider a simple pendulum: a point mass mm (the bob) suspended from a fixed support by a light, inextensible string of length LL. When displaced from its equilibrium (vertical) position by a small angle θ\theta and released, it oscillates back and forth.

Showing the motion is SHM:

At angular displacement θ\theta, the forces on the bob are gravity mgmg (downward) and the string tension TT (along the string). Resolving gravity along and perpendicular to the string, the component of gravity that provides the restoring force (tangential to the arc, back towards equilibrium) is:

Frestoring=−mgsin⁡θF_{restoring} = -mg\sin\theta

(the negative sign shows it acts opposite to the displacement, trying to restore the bob to equilibrium)

For small angles (θ\theta typically less than about 10°), we use the small-angle approximation sin⁡θ≈θ\sin\theta \approx \theta (in radians):

Frestoring≈−mgθF_{restoring} \approx -mg\theta

Since the arc length displacement is x=Lθx = L\theta (so θ=x/L\theta = x/L):

Frestoring=−mgLxF_{restoring} = -\dfrac{mg}{L}x

This has exactly the form of the SHM restoring-force equation, F=−kxF = -k x, with an effective "spring constant" k=mgLk = \dfrac{mg}{L}. Since the restoring force is directly proportional to the displacement xx and directed opposite to it, the pendulum's motion is Simple Harmonic (for small θ\theta).

Deriving the time period:

For SHM, F=−mω2xF = -m\omega^2 x. Comparing with F=−mgLxF = -\dfrac{mg}{L}x:

mω2=mgL⇒ω2=gL⇒ω=gLm\omega^2 = \dfrac{mg}{L} \Rightarrow \omega^2 = \dfrac{g}{L} \Rightarrow \omega = \sqrt{\dfrac{g}{L}}

Since T=2πωT = \dfrac{2\pi}{\omega}:

T=2πLg\boxed{T = 2\pi\sqrt{\dfrac{L}{g}}}

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