A mass $m$ on a vertical spring stretches it by $x_0$ when in equilibrium. The time period of small oscillations about this equilibrium is:
A$2\pi\sqrt{m/(mg/x_0)}$
B$2\pi\sqrt{x_0/g}$ (since $kx_0 = mg$ ⇒ $k/m = g/x_0$)
CBoth A and B (same expression)
D$2\pi\sqrt{g/x_0}$
Answer & Solution
Correct answer: C. Both A and B (same expression)
Equilibrium: $k x_0 = mg$ ⇒ $k = mg/x_0$. Then $\omega = \sqrt{k/m} = \sqrt{g/x_0}$ ⇒ $T = 2\pi\sqrt{x_0/g}$. Note this looks like a pendulum formula with x_0 playing the role of length! Often called 'spring pendulum analogue'.
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