Solution:
$k = F/x = 10 / 0.05 = 200 \text{ N/m}$. Time period $T = 2\pi \sqrt{m/k} = 2\pi \sqrt{2/200} = \frac{2\pi}{10} = 0.628 \text{ s}$.
A spring is stretched by $5 \text{ cm}$ by a force $10 \text{ N}$. The time period of the oscillations when a mass of $2 \text{ kg}$ is suspended by it is: (2021)
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