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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 = 1 $m / s^{2}$ . If new time period is $T_{2}$ then find ratio of $T_{1}^{2}$ and $T_{2}^{2}$

\frac{2}{3}

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\frac{5}{6}

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\frac{6}{5}

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\frac{4}{5}

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Solution

The correct option is D $\frac{6}{5}$
$y = K t^{2} \Rightarrow \frac{d^{2}}{d t^{2}} = a_{y} = 2 K = 2 \times 1 = 2 m / s (∵ K = 1 m / s)$

Now, $T_{1} = 2 π \sqrt{\frac{l}{g}}$

and $T_{2} = 2 π \sqrt{\frac{1}{(g + a_{y})}}$

Dividing $\frac{T_{1}}{T_{2}} = \frac{g + a_{y}}{g}$

$\Rightarrow \sqrt{\frac{6}{5}}$

$\Rightarrow \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{6}{5}$

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