alleryThe time period of a spring-mass system with mass $m$ and spring constant $k$ is:
A$T = 2\pi \sqrt{\dfrac{k}{m}}$
B$T = 2\pi \sqrt{\dfrac{m}{k}}$
C$T = 2\pi k m$
D$T = 2\pi (k + m)$
Answer & Solution
Correct answer: B. $T = 2\pi \sqrt{\dfrac{m}{k}}$
Spring-mass system: Newton's second law gives $m\ddot x = -kx$, hence $\omega = \sqrt{k/m}$.
Period: $T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{m/k}$.
A stiffer spring (larger $k$) gives a shorter period. A heavier mass gives a longer period. Option A inverts the ratio, the classic trap.
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