Simple Harmonic Motion — Maximum Speed For a particle executing SHM with angular frequency ω and amplitude r (displacement amplitude), what is the maximum speed attained during the motion?

Introduction / Context: In SHM, displacement x, velocity v, and acceleration a vary sinusoidally and are 90 degrees out of phase. Knowing maximum speed helps in sizing components, estimating kinetic energy, and checking resonance risks in machines.

Given Data / Assumptions:

• Displacement: x(t) = r sin(ωt + φ).
• Angular frequency: ω; amplitude: r.

Concept / Approach: Differentiate displacement to get velocity and evaluate its maximum magnitude over one cycle. Alternatively, use the energy relation v^2 = ω^2 (r^2 − x^2).

Step-by-Step Solution:

Verification / Alternative check: From v^2 = ω^2 (r^2 − x^2), at x = 0 we get v_max = ω r, and at x = ±r we get v = 0, matching physical expectations.

Why Other Options Are Wrong: 'ω' ignores amplitude units; 'ω^2 r' and 'ω/r' have incorrect dimensions for speed.

Common Pitfalls: Confusing amplitude r with radius of circular motion or mixing up where v and a are maximum (v is max at x = 0; a is max at |x| = r).

Final Answer: ωr.