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⚛️ Physics  ·  Oscillations  ·  NEET & JEE

The velocity of a particle in SHM at displacement x from equilibrium (amplitude A) is:

Answer: ω√(A²-x²).

  • A ω√(A²-x²)
  • B ωx
  • C ωA
  • D ω√(A²+x²)

Correct answer: A. ω√(A²-x²)

Explanation: v = ω√(A² - x²). At x=0: v = ωA (maximum). At x=A: v = 0 (at extremes).

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.

Read the full Oscillations notes →