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PhysicsSHMCalculation of time periodMedium3 minQB

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A simple pendulum has time period $T_{1}$ . The point of suspension is now moved upward according to the relation $y = t^{2}$ where y is the vertical displacement and t is time. The time period now becomes $T_{2}$ . The ratio of $\frac{T _{1} ^{2}}{T _{2} ^{2}}$ is :-

Options:

Answer:

Solution:

Suspension point accelerates upward at $a = \frac{d ^{2} y}{d t ^{2}} = 2 m / s^{2}$ , so $g_{eff} = g + a = 12$ .

$\frac{T _{1}^{2}}{T _{2}^{2}} = \frac{g _{eff}}{g} = \frac{12}{10} = \frac{6}{5}$

Answer: A

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Calculation of time period

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