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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 the relation $y = k t^{2} (k = 1 m s^{2})$ where y is the vertical displacement. The time period now become $T_{2}$ . The ratio of $\frac{T_{1}^{2}}{T_{2}^{2}}$ is $g = 10 m / s^{2}$ .

A

6 / 5

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B

5 / 6

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C

1

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D

4 / 5

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Solution

The correct option is A $6 / 5$
$y = k t^{2}$
$= \frac{d^{2} y}{d t^{2}} = 2 k$
$a = 2 m / s^{2}$
$T_{1} = 2 π \sqrt{\frac{l}{g}}$
$T_{2} = 2 π \sqrt{\frac{l}{g + a_{y}}}$
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{g + a_{y}}{g}$
$= \frac{10 + 2}{10} = \frac{6}{5}$ .

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