Time period of a spring-mass system (spring constant $k$, mass $m$) is:
- $T = 2\pi\sqrt{m/k}$
- $T = 2\pi\sqrt{k/m}$
- $T = 2\pi mk$
- $T = 2\pi/(km)$
Show answer and 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).*