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⚛️ Physics  ·  Oscillations  ·  NEET & JEE

A particle executes SHM with T = 2 s and amplitude 5 cm. The distance covered by the particle in one full time period starting from extreme is:

Answer: 20 cm.

  • A 20 cm
  • B 5 cm
  • C 10 cm
  • D 40 cm

Correct answer: A. 20 cm

Explanation: In one full period, the particle goes from one extreme to the other and back: 4A = 4 × 5 = 20 cm total distance (not displacement).

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 →