The 'natural frequency' ω of a single-degree-of-freedom (SDOF) spring-mass system is:
Choose the correct answer
ω = √(m/k) (inverted)
ω_n = √(k/m) — square root of stiffness to mass ratio
ω = 2πk/m
ω = k×m
Correct Answer
B. ω_n = √(k/m) — square root of stiffness to mass ratio
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SDOF: ω_n = √(k/m) (rad/s) where k = stiffness (N/m), m = mass (kg). Cyclic frequency f_n = ω_n/(2π) Hz; period T_n = 1/f_n. For a simply supported beam with mass M at centre: equivalent stiffness k = 48EI/L³ → ω_n = √(48EI/mL³). Damped natural frequency ω_d = ω_n√(1-ζ²) where ζ = damping ratio. Buildings: T_n ≈ 0.1N (N = storeys) empirical.
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