Q 11-13-009NEETNEET 2021Top questionEasy
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 :
Answer: (A) $0.628$ s
$$k = \frac{F}{x} = \frac{10}{0.05} = 200\ \text{N/m}$$
$$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2}{200}} = \frac{2\pi}{10} = 0.628\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics