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