Simple harmonic motion (SHM): A particle executes simple harmonic motion. At which position in its path does it attain the maximum speed, and why?

Introduction / Context:
In simple harmonic motion (SHM), a particle oscillates about a mean (equilibrium) position under a restoring force proportional to displacement. Understanding where speed is maximum is fundamental to vibration theory and helps connect energy concepts with kinematics in oscillatory systems.

Given Data / Assumptions:

• Displacement x varies sinusoidally about the mean position.
• Amplitude A is the maximum displacement on either side.
• Angular frequency is ω (constant).
• No damping or external forcing; pure SHM.

Concept / Approach:
For SHM, velocity v is given by v = ω * sqrt(A^2 - x^2). Alternatively, v = dx/dt when x = A sin(ωt + φ). Maximum speed occurs when the square root term is largest, i.e., when x = 0. Energy view: at the mean position, potential energy (stored in the spring or restoring element) is minimum and kinetic energy is maximum, so speed peaks there.

Step-by-Step Solution:

Verification / Alternative check:
At extremes x = ±A, v = 0, matching the observation that the particle turns around at the endpoints. Between extreme and mean, v increases; it is largest exactly at the centre.

Why Other Options Are Wrong:

• Extreme point: speed is zero there (turning points).
• Half amplitude: speed is non-zero but not maximum.
• “None of these” contradicts the SHM velocity relation.

Common Pitfalls:
Confusing maximum displacement with maximum speed; in SHM they do not coincide. Also, mixing angular frequency ω with linear frequency f without conversion.

Final Answer:
through the mean position