==>$\frac{{{d^2}y}}{{d{t^2}}} = {a_y} = 2K= 2 \times 1 = 2\,m/s^2 (K= 1\,m/s^2)$
Now, ${T_1} = 2\pi \sqrt {\frac{l}{g}} $ and ${T_2} = 2\pi \sqrt {\frac{l}{{(g + {a_y})}}} $
Dividing, $\frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{g + {a_y}}}{g}}= \sqrt {\frac{6}{5}} $
==> $\frac{{T_1^2}}{{T_2^2}} = \frac{6}{5}$



