A simple pendulum has a time period T_1. The | vedclass

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(C) The displacement of the point of suspension is given by $y = kt^2$.The acceleration of the point of suspension is $a_y = \frac{d^2y}{dt^2} = 2k$.G

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$6/5$

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(C) The displacement of the point of suspension is given by $y = kt^2$.The acceleration of the point of suspension is $a_y = \frac{d^2y}{dt^2} = 2k$.Given $k = 1\,m/s^2$,so $a_y = 2 	imes 1 = 2\,m/s^2$.The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.Initially,$T_1 = 2\pi \sqrt{\frac{l}{g}}$.When the suspension point moves upward with acceleration $a_y$,the effective acceleration due to gravity is $g_{eff} = g + a_y$.Taking $g = 10\,m/s^2$ (standard approximation),$g_{eff} = 10 + 2 = 12\,m/s^2$.Thus,$T_2 = 2\pi \sqrt{\frac{l}{g + a_y}} = 2\pi \sqrt{\frac{l}{12}}$.The ratio is $\frac{T_1^2}{T_2^2} = \frac{g + a_y}{g} = \frac{10 + 2}{10} = \frac{12}{10} = \frac{6}{5}$.

$A$ 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:

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