A particle executes simple harmonic motion between x = - A and x = + A . If it takes a time T 1 to g 0 from…

Equation of SHM $x=a\sin \left(\omega t\right)$

at $x=0 ; t_o=0$

at 

$$\begin{gathered} x=\frac{A}{2} \\ from equation \\ \frac{A}{2}=a\sin \left(\frac{2\pi }{T}{t}_1\right) \\ \Rightarrow t_1=\frac{T}{12} \end{gathered}$$

at $x=A$

from equation 

$$\begin{gathered} A=a\sin \left(\frac{2\pi }{T}{t}_1\right) \\ \Rightarrow t_1=\frac{T}{4} \end{gathered}$$

so time from $x=0 to x=\frac{A}{2}$

$T_1=t_1-t_o=\frac{T}{12}$

and time from $x=\frac{A}{2} to x=A$

$T_2=t_2-t_1=\frac{T}{4}-\frac{T}{12}=\frac{T}{6}$

so $T_1<T_2$