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PhysicsSHMMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

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 becomes $T_{2}$ . The ratio of $\frac{T_{1}^{2}}{T_{2}^{2}}$ is (Take $g = 10 m s^{- 2}$ )

Options:

Answer:

Solution:

$y = k t^{2}$

$\frac{d^{2} y}{{dt}^{2}} = 2 k$

or $a_{y} = 2 m s^{- 2} (as k = 1 m s^{- 2})$

$T_{1} = 2 π \sqrt{\frac{l}{g}}$

and $T_{2} = 2 π \sqrt{\frac{l}{g + a_{y}}}$

$[∴ g_{eff} = g + a_{y}]$

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{g + a_{y}}{g} = \frac{10 + 2}{10} = \frac{6}{5}$

Hence, the correct answer is $\frac{6}{5}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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