Simple Harmonic Motion — Velocity at Mean Position For a particle in SHM, what is the velocity at the mean (equilibrium) position compared with elsewhere in the cycle?

Introduction / Context: SHM features an exchange between potential and kinetic energy. The mean position is where the restoring force (and hence potential energy for linear SHM) is zero, so kinetic energy — and velocity magnitude — are largest there.

Given Data / Assumptions:

• Displacement x, velocity v, acceleration a in standard SHM.
• Total energy E is constant: E = (1/2) k A^2 = (1/2) k x^2 + (1/2) m v^2.

Concept / Approach: At x = 0 (mean position), the potential energy part (1/2) k x^2 is zero, so kinetic energy is maximum; hence speed is maximum. Conversely, at extremes x = ±A, speed is zero and potential energy is maximum.

Step-by-Step Solution:

Verification / Alternative check: From sinusoidal form x = A sin(ω t + φ), v = dx/dt = A ω cos(ω t + φ). Cosine reaches magnitude 1 at the mean, giving maximum |v| = A ω.

Why Other Options Are Wrong: 'Zero' and 'minimum' occur at the extreme positions, not at the mean.

Common Pitfalls: Assuming acceleration and velocity peak at the same place; in SHM, they are out of phase by 90 degrees in time (π/2 radians).

Final Answer: maximum.