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⚛️ Physics  ·  Oscillations  ·  NEET & JEE

The restoring torque on a simple pendulum of mass m and length L displaced by small angle θ is:

Answer: -mgL sinθ ≈ -mgLθ.

  • A -mgL sinθ ≈ -mgLθ
  • B mgLθ in standard practice
  • C -mLθ under most conditions encountered
  • D -mgL cosθ as frequently observed in practice

Correct answer: A. -mgL sinθ ≈ -mgLθ

Explanation: Torque = -mg sinθ × L ≈ -mgLθ (for small θ). This gives α = -(g/L)θ, so ω² = g/L and T = 2π√(L/g).

Displacement, Velocity, Acceleration in SHMx(t)v(t)a(t)v leads x by 90°; a is exactly out of phase with x (a = −ω²x)

In SHM, displacement and velocity are 90° out of phase (v is maximum when x=0, and zero when x is extreme), while acceleration is always exactly opposite in sign to displacement, consistent with a = −ω²x.

Concept context

Simple harmonic motion, spring-mass, pendulum, energy in SHM, and resonance.

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