To solve this problem, we need to understand how the upward motion of the suspension point affects the time period of the simple pendulum. The basic time period \(T_1\) of a simple pendulum, when the point of suspension is stationary, is given by the formula:
\(T_1 = 2\pi \sqrt{\frac{L}{g}}\)
where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.
When the point of suspension is accelerating vertically upwards with acceleration \(a = \frac{d^2y}{dt^2} = 2k = 2\text{ m/s}^2\), the effective gravitational acceleration becomes \(g' = g + a\).
So, the modified time period \(T_2\) is given by:
\(T_2 = 2\pi \sqrt{\frac{L}{g + a}}\)
Substituting the given values, \(g = 10\text{ m/s}^2\) and \(a = 2\text{ m/s}^2\), we find:
\(T_2 = 2\pi \sqrt{\frac{L}{10 + 2}} = 2\pi \sqrt{\frac{L}{12}}\)
Thus, the ratio of the squares of the time periods is:
\(\frac{T_1^2}{T_2^2} = \frac{\left(2\pi \sqrt{\frac{L}{10}}\right)^2}{\left(2\pi \sqrt{\frac{L}{12}}\right)^2} = \frac{L/10}{L/12} = \frac{12}{10} = \frac{6}{5}\)
Therefore, the correct option is \(\frac{6}{5}\).