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)