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

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(c) $y = K{t^2}$

==>$\frac{{{d^2}y}}{{d{t^2}}} = {a_y} = 2K= 2 	imes 1 = 2\,m/s^2 (K= 1\,m/s^2)$

Now, ${T_1} = 2\pi \sqrt {\frac{l}{g}} $ an

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$6/5$

Vedclass · Answer

(c) $y = K{t^2}$

==>$\frac{{{d^2}y}}{{d{t^2}}} = {a_y} = 2K= 2 	imes 1 = 2\,m/s^2 (K= 1\,m/s^2)$

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}} $

==> $\frac{{T_1^2}}{{T_2^2}} = \frac{6}{5}$

A simple pendulum has time period $T_1$. The point of suspension is now moved upward according to equation $y = k{t^2}$ where $k = 1\,m/se{c^2}$. If new time period is $T_2$ then ratio $\frac{{T_1^2}}{{T_2^2}}$ will be

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