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A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is :

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0.628 s

0.0628 s

6.28 s

3.14 s

answer is A.

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Detailed Solution

$\begin{array}{l} K = \frac{10}{5 \times 10^{- 2}} N / m = 200 N / m \\ ∴ T = 2 π \sqrt{\frac{m}{K}} = 2 π \sqrt{\frac{2}{200}} s = \frac{π}{5} s \end{array}$
T = 0.628 s

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