Question

An object executing SHM has an acceleration of $0.2$ m/$s^2$ at a distance of $0.05$ m from the mean position. The period of SHM is

• A. 2.12 s
• B. 2.72 s
• C. 3 s
• D. 3.142 s

Answer 1

The correct option is D 3.142 s

$a=-{\omega }^{2}x$
$\frac{0.2\frac{m}{s^2}}{0.05m}={\omega }^2=\frac{2}{1\text{/}0}\times \frac{10\text{/}0}{5}=\frac{20}{5}=4$
${\omega }^2=4;\omega =2$
$\frac{2\pi }{T}=2;T=\pi sec$