Simple harmonic motion (SHM) — Maximum acceleration:\nA particle performs SHM of amplitude r and angular frequency ω. What is the maximum magnitude of its acceleration?

Introduction / Context:
In simple harmonic motion, acceleration is proportional to displacement and directed toward the equilibrium (restoring) position. Being able to relate maximum acceleration to amplitude and angular frequency is essential for vibration analysis and design (e.g., springs, pendulums, machinery).

Given Data / Assumptions:

• SHM with displacement x(t) = r * sin(ωt + φ) or r * cos(ωt + φ).
• Amplitude = r (maximum displacement from equilibrium).
• Angular frequency = ω (radians per second).

Concept / Approach:
For SHM, the defining relation is a = −ω^2 * x. The maximum magnitude of a occurs when |x| is maximum, i.e., at the extremes where |x| = r. Therefore, |a|max depends on ω^2 times r.

Step-by-Step Solution:
Start from acceleration–displacement relation: a = −ω^2 * x.Maximum |x| = r at the ends of motion.Thus maximum |a| = ω^2 * r.Therefore the correct expression is ω^2r.

Verification / Alternative check:
Differentiation method: x = r cos(ωt) ⇒ v = dx/dt = −rω sin(ωt); a = dv/dt = −rω^2 cos(ωt) = −ω^2 x. Maximum |a| occurs at cos(ωt) = ±1 → |a| = rω^2.

Why Other Options Are Wrong:

• ω: Has units of 1/s, not acceleration.
• ωr: Has units of speed, not acceleration.
• ω / r: Units inconsistent; not from SHM relations.

Common Pitfalls:

• Confusing maximum speed (ωr) with maximum acceleration (ω^2r).
• Dropping the square on ω in a = −ω^2 x.

Final Answer:
ω^2r