Frequency–period relationship in simple harmonic motion (SHM) For a particle executing SHM with angular frequency ω and period T, which statement correctly gives the relationship involving the frequency f?

Introduction / Context:
Simple harmonic motion (SHM) appears in oscillating systems such as springs, pendulums (for small angles), and electrical LC circuits. Understanding how frequency, period, and angular frequency interrelate is fundamental for designing timing systems and analyzing vibrations.

Given Data / Assumptions:

• SHM defined by sinusoidal time dependence.
• Frequency f (cycles per second), period T (time per cycle), and angular frequency ω (radians per second).
• Small-angle approximation may apply for pendulums, but the relationships f = 1/T and ω = 2π f always hold.

Concept / Approach:
By definition, period T is the time for one complete cycle, and frequency f is the number of cycles per unit time. Therefore, f = 1/T. Angular frequency relates to frequency by ω = 2π f, so f = ω / (2π), linking all three quantities.

Step-by-Step Solution:

Verification / Alternative check:
Dimensional consistency: [f] = s^-1, [T] = s, [ω] = s^-1. f = 1/T and ω = 2π f satisfy these units exactly.

Why Other Options Are Wrong:

• Direct proportionality to T is incorrect; as T increases, f decreases.
• Inversely proportional to ω is incorrect; f increases with ω.
• Independence from T and ω contradicts definitions.
• Proportional to T^2 is not physically meaningful for SHM.

Common Pitfalls:
Confusing linear frequency f with angular frequency ω; forgetting the factor 2π; assuming frequency changes with amplitude (it does not for ideal SHM).

Final Answer:
Frequency is inversely proportional to the period (f ∝ 1/T)