The vertical extension in a light spring by a weight of 1 kg suspended from the lower end is 9.8 cm. The period of oscillation of the spring is:

A
$20 π \sec$
B
$(2 π / 10) \sec$
C
$2 π / 100 \sec$
D
$(200 π) \sec$

Solution:

Spring constant $k = \frac{F}{x} = \frac{mg}{x} = \frac{1 \times 9.8}{9.8 \times 10^{- 2}} = 100 N / m T = 2 π \sqrt{\frac{M}{k}} \Rightarrow 2 π \sqrt{\frac{1}{100}} = \frac{2 π}{10} \sec$

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