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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.Initially,the point of suspension is at rest,so $g_{eff} = g

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$\frac{6}{5}$

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.Initially,the point of suspension is at rest,so $g_{eff} = g = 10 \ m/s^2$. Thus,$T_1 = 2\pi \sqrt{\frac{L}{g}}$.When the point of suspension moves upward with an acceleration $a$,the effective acceleration is $g_{eff} = g + a$.Given the displacement $y = kt^2$,the acceleration $a$ is the second derivative of displacement with respect to time: $a = \frac{d^2y}{dt^2} = \frac{d^2}{dt^2}(kt^2) = 2k$.Given $k = 1 \ m/s^2$,we have $a = 2(1) = 2 \ m/s^2$.Therefore,$g_{eff} = g + a = 10 + 2 = 12 \ m/s^2$.The new time period is $T_2 = 2\pi \sqrt{\frac{L}{g_{eff}}} = 2\pi \sqrt{\frac{L}{12}}$.Now,the ratio $\frac{T_1^2}{T_2^2} = \frac{4\pi^2 (L/g)}{4\pi^2 (L/g_{eff})} = \frac{g_{eff}}{g} = \frac{12}{10} = \frac{6}{5}$.

$A$ simple pendulum has a time period $T_1$. The point of suspension is now moved upward according to the equation $y = kt^2$,where $k = 1 \ m/s^2$. If the new time period is $T_2$,then the ratio $\frac{T_1^2}{T_2^2}$ will be (take $g = 10 \ m/s^2$):

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