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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/s 2.

  • A 6.3 m/s<sup>2</sup>
  • B 25.3 m/s<sup>2</sup>
  • C 3.2 m/s<sup>2</sup>
  • D 12.6 m/s<sup>2</sup>

Correct answer: D. 12.6 m/s<sup>2</sup>

Explanation: &omega; = 2&pi;/T = 4&pi; rad/s. a<sub>max</sub> = &omega;<sup>2</sup>A = (4&pi;)<sup>2</sup> &times; 0.08 &asymp; 12.6 m/s<sup>2</sup>.

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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