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Standard XII

Physics

Pendulum in Non Inertial Frame

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 becomes $T_{2}$ . The ratio of

$\frac{T_{1}^{2}}{T_{2}^{2}}$ is : $(g = 10 m / s^{2})$

6/5

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Solution

The correct option is A
6/5

$y = K t^{2}$

$\frac{d^{2} y}{d t^{2}} = 2 K$

$o r a_{y} = 2 m / s^{2}$ $(a s k = 1 m / s^{2})$

$T_{1} = 2 π \sqrt{\frac{l}{g}}$

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

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

$= \frac{10 + 2}{10} = \frac{6}{5}$

$∴ c o r r e c t a n s w e r i s (a)$

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