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:

(1) 0.0628 s

(2) 6.28 s

(3) 3.14 s

(4) 0.628 s

*Solution*

We have, $T = 2\pi \sqrt {\frac{m}{k}} $

$F=kx$

$\Rightarrow k=\frac {10N}{5cm}=200 N/m$

$\therefore T=2\pi \sqrt {\frac {2}{200}}=\frac {\pi}{5}=0.628 s$

Answer: Option (4)