The time period of oscillation of a mass M attached to a spring of force constant k is:
A. 2π√(k/M)
B. 2π√(g/l)
C. 2π√(M/k)
D. π√(M/k)
Answer: Option C
Solution (By JKSSB Mock Tests)
For spring-mass system undergoing SHM, time period T = 2π√(M/k), derived from F = -kx = Ma ⇒ a = -(k/M)x, so ω² = k/M, T=2π/ω=2π√(M/k). Option B has inverse ratio; C misses factor 2; D is for pendulum. Memory tip: 'Spring: T ∝ √(M/k); Pendulum: T ∝ √(l/g)'. This standard formula is frequently tested in oscillations sections of competitive exams. Always verify dimensions: √(kg / (N/m)) = √(kg·m/N) = √(s²) = s, correct for time period.
Explanation:
Work = force × displacement in direction of force. Displacement zero, so work zero. Physiology work is done internally, but physics work zero.
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