Difficulty: Easy
Correct Answer: inversely proportional to its angular velocity
Explanation:
Introduction / Context:The time period T of a simple harmonic oscillator specifies the time for one complete oscillation. Connecting T with the angular frequency ω is central to vibration analysis, pendulum approximations, and spring–mass systems.
Given Data / Assumptions:
Concept / Approach:By definition, angular frequency ω relates to the time period by ω = 2π / T, or equivalently T = 2π / ω. Thus, T varies inversely with ω: doubling ω halves T, etc.
Step-by-Step Solution:
Start from x = A sin(ωt + φ); one full cycle when argument increases by 2π.Hence ωT = 2π ⇒ T = 2π/ω.Therefore T ∝ 1/ω (inverse proportionality).Verification / Alternative check:For a spring–mass system, ω = sqrt(k/m) ⇒ T = 2π sqrt(m/k), again showing inverse relation to ω.
Why Other Options Are Wrong:
Common Pitfalls:Confusing angular frequency ω (rad/s) with cyclic frequency f (Hz); note that T = 1/f and ω = 2πf.
Final Answer:inversely proportional to its angular velocity
Discussion & Comments