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).
Related questions
A particle of mass 0.2 kg performs S.H.M. of amplitude 0.05 m with period π s. Its maximumIn damped harmonic motion, the amplitude of oscillationTwo springs of force constants k₁ and k₂ are connected in series and stretched by an exterA particle in S.H.M. of period 2 s starts from rest at the extreme position. Its displacemThe kinetic energy of a particle of mass m in linear S.H.M. of amplitude A at displacementA simple pendulum of length 1 m on the earth's surface (g = π² m/s²) has periodA spring-mass system of mass m and spring constant k oscillates with periodA particle executes S.H.M. with amplitude A and angular frequency ω. Its maximum accelerat