Period of simple harmonic motion (relationship with angular frequency): How does the time period of an SHM vary with its angular velocity ω?
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Adirectly proportional to its angular velocity
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Bdirectly proportional to the square of its angular velocity
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Cinversely proportional to the square of its angular velocity
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Dinversely proportional to its angular velocity
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E—
Answer
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:
- Pure SHM with angular frequency ω.
- Linear restoring law; no damping or forcing.
- Standard kinematic relations of SHM hold.
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:
- Direct or square relations contradict T = 2π/ω.
- Inverse square would imply T ∝ 1/ω^2, which is incorrect.
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