Simple harmonic motion (SHM) energy at the mean position: A particle executes simple harmonic motion. While passing through the mean (equilibrium) position, which form(s) of mechanical energy does the particle possess at a maximum or minimum level?

Introduction / Context:
In simple harmonic motion (SHM), such as a mass–spring system or a small-angle pendulum, energy oscillates between kinetic and potential forms. Understanding how kinetic energy (KE) and potential energy (PE) vary with displacement and velocity is fundamental in vibration analysis and engineering dynamics.

Given Data / Assumptions:

• Motion is ideal SHM with no damping.
• Mean position is the equilibrium point where displacement is zero.
• Total mechanical energy remains constant (no losses).

Concept / Approach:

For SHM with displacement x(t) and amplitude A, potential energy is proportional to displacement squared (for a spring, PE = 1/2 k x^2). Kinetic energy depends on velocity squared (KE = 1/2 m v^2). At the mean position, x = 0 → PE is minimum (zero for a spring), and velocity is maximum → KE is maximum. At extreme positions, velocity is zero → KE is zero and PE is maximum.

Step-by-Step Solution:

Verification / Alternative check:

Energy conservation: total energy E = 1/2 k A^2 is constant. At x = 0, all of E is kinetic; at x = ±A, all of E is potential—consistent with the conclusion.

Why Other Options Are Wrong:

(b) PE cannot be maximum at the mean position; (c) KE is not minimum at x = 0; (d) both minimum is impossible because total energy is constant and nonzero for A ≠ 0; (e) is incorrect because one option exactly describes the SHM energy distribution.

Common Pitfalls:

Confusing displacement and velocity extrema; assuming KE and PE are equal at the mean position (they are equal only at intermediate displacements, not at x = 0).

Final Answer:

maximum kinetic energy and minimum potential energy