Time period of a spring-mass system (spring constant $k$, mass $m$) is:
A$T = 2\pi\sqrt{m/k}$
B$T = 2\pi\sqrt{k/m}$
C$T = 2\pi mk$
D$T = 2\pi/(km)$
Answer & Solution
Correct answer: A. $T = 2\pi\sqrt{m/k}$
Newton's law + Hooke: $m\ddot x = -kx$ ⇒ $\omega = \sqrt{k/m}$ ⇒ $T = 2\pi/\omega = 2\pi\sqrt{m/k}$. Note: independent of gravity (works horizontally and vertically — gravity only shifts equilibrium).
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