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⚛️ Physics  ·  Oscillations  ·  NEET & JEE

A particle executes SHM of amplitude 8 cm with a period of 0.5 s. Its maximum acceleration is about:

Answer: 12.6 m/s2.

  • A 6.3 m/s2
  • B 25.3 m/s2
  • C 3.2 m/s2
  • D 12.6 m/s2

Correct answer: D. 12.6 m/s2

Explanation: ω = 2π/T = 4π rad/s. amax = ω2A = (4π)2 × 0.08 ≈ 12.6 m/s2.

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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