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⚛️ Physics  ·  Oscillations  ·  NEET & JEE

For a seconds pendulum on Earth (T = 2 s, L = 1 m), what must be the length on a planet where g = 4 m/s² for the same period?

Answer: 0.4 m.

  • A 0.4 m
  • B 1 m
  • C 4 m
  • D 2.5 m

Correct answer: A. 0.4 m

Explanation: T = 2π√(L/g). For T = 2 s and g = 4 m/s²: 2 = 2π√(L/4) → 1/π = √(L/4) → L/4 = 1/π² ≈ 0.101 → L ≈ 0.405 m.

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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