A simple pendulum has time period T₁, 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 as a function of time ‘t’. The time period now become $T_{2},$ the ratio of $\frac{{T_{1}}^{2}}{{T_{2}}^{2}}$ is $\frac{6}{N}$ where ‘N’ is equal to _______ $(g = 10 m / s^{2})$

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answer is 5.

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Detailed Solution

y=4t² ; $\frac{d y}{d x} = 2 k t$

$\frac{d^{2} y}{d t^{2}} = m / s^{2}$

$T = 2 π \sqrt{\frac{l}{g}}$

$\frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{12}{10} = \frac{6}{5}$

N =5

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