A particle is executing simple harmonic motion. The equation of motion is given by   $\frac{d^{2}x}{d{t}^2}+4x=0$. Then their time period of oscillation is (in seconds)

A simple harmonic motion can be represented by the equation  $a=-{\omega }^{2}x$ . where  acceleration a  can be written as $\frac{d^{2}x}{d{t}^2}$ . 
Hence  $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$. 
Writing the given equation $\frac{d^{2}x}{d{t}^2}+4x=0$ in the above form, we get $\frac{d^{2}x}{d{t}^2}=-4x$ . 
Comparing this with the above equation, we get  ${\omega }^2=4$. 
Taking square root on both sides  $\omega =2$
Using the relation  $\omega =\frac{2\pi }{T}$, we get  $$

substitute   $\omega =2$

$$\begin{gathered} 2=\frac{2\mathrm{\pi }}{T} \\ \Rightarrow T=\mathrm{\pi } \mathrm{second}=time period of oscillation \end{gathered}$$