A particle is executing simple harmonic motion along a straight line with time period T and amplitude A. The particle takes time ${\mathrm{t}}_1$ to move from $\mathrm{x}=+\mathrm{A} \mathrm{to} \mathrm{x}=+\mathrm{A}/2 \mathrm{and} {\mathrm{t}}_2$ to move from $\mathrm{x}=+\mathrm{A}/2 \mathrm{to} \mathrm{x}=0$. Then $\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}$ is equal to:

Let the equation of SHM be  x = cos $\omega$t

So the particle starts (t=0) its motion at the extreme position x = +A

When 

$$\begin{gathered} \mathrm{x}=+\frac{\mathrm{A}}{2}, \frac{\mathrm{A}}{2}=\mathrm{A} \cos \omega t_1 \\ \Rightarrow \omega t_1=\frac{\mathrm{\pi }}{3} \\ \Rightarrow {\mathrm{t}}_1=\frac{\mathrm{\pi }}{3\mathrm{\omega }}=\frac{\mathrm{T}}{6} \end{gathered}$$

When x = 0, 

$$\begin{gathered} 0=\mathrm{A} \cos \omega t_3 \\ \Rightarrow {\mathrm{t}}_3=\frac{\mathrm{\pi }}{2\mathrm{\omega }} \end{gathered}$$

$\therefore {\mathrm{t}}_2={\mathrm{t}}_3-{\mathrm{t}}_1=\frac{\mathrm{\pi }}{2\mathrm{\omega }}-\frac{\mathrm{\pi }}{3\mathrm{\omega }}=\frac{\mathrm{\pi }}{6\mathrm{\omega }}=\frac{\mathrm{\pi }}{6 \mathrm{x} {\displaystyle \frac{2\mathrm{\pi }}{\mathrm{T}}}}=\frac{\mathrm{T}}{12}$

$\therefore \frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}=\frac{\mathrm{\pi }/3\mathrm{\omega }}{\mathrm{\pi }/6\mathrm{\omega }}=\frac{\mathrm{T}/6}{\mathrm{T}/12}=2$