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Simple Pendulum

A simple pend...

Question

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

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Solution

The correct option is C $6 / 5$
$\begin{matrix} Initial Time period = T_{1} Final Time period = T_{2} Point of Suspension moved upwards with y = k \cdot t^{2} \end{matrix}$
$\begin{matrix} we have position of point of suspension y = k \cdot t^{2} \Rightarrow acceleration = \frac{d^{2} y}{d t^{2}} = 2 k = 2 m / s^{2} \end{matrix}$
$\begin{matrix} \Rightarrow effective acceleration = g + 2 m / s^{2} = 10 + 2 m / s^{2} g_{eff} = 12 m / s^{2} \end{matrix}$
$T_{2} = 2 π \sqrt{\frac{ℓ}{12}}$
$and we have T_{1} = 2 π \sqrt{\frac{ℓ}{10}}$
$\begin{matrix} \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{12}{10} = \frac{6}{5} \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{6}{5} \end{matrix}$

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