Simple Harmonic Motion (SHM) The periodic time T of a particle in SHM is ________ proportional to the angular frequency ω.

Introduction / Context: In SHM, the time for one complete oscillation is called the period T, and the angular frequency is ω. Their relationship is a fundamental property that appears across mechanical and electrical oscillators.

Given Data / Assumptions:

• Standard SHM with displacement x(t) = A sin(ωt + φ).
• Angular frequency ω is constant for the system being considered.

Concept / Approach: By definition, the period is the time taken for the phase to increase by 2π. Thus T and ω are tied by the identity T = 2π / ω, regardless of amplitude (for linear SHM).

Step-by-Step Solution:

Verification / Alternative check: For a mass–spring oscillator, ω = sqrt(k / m), so T = 2π * sqrt(m / k); increasing stiffness increases ω and decreases T. For a simple pendulum (small angle), ω = sqrt(g / l), so T = 2π * sqrt(l / g), again showing inverse proportionality to ω.

Why Other Options Are Wrong: 'Directly' contradicts T = 2π / ω. 'Depends on amplitude' is false for ideal linear SHM (period is independent of amplitude). 'Neither' is incorrect because there is a clear inverse relationship.

Common Pitfalls: Confusing frequency f with angular frequency ω, and forgetting the factor 2π (i.e., ω = 2πf).

Final Answer: inversely.