Q. A simple pendulum has time period $T_{1}$. The point of suspension is now moved upward according to the relation $y = k\ t^{2},\ \left( k = 1\ ms^{- 2} \right)$ where $y$ is the vertical displacement. The time period now becomes $T_{2}$. The ratio of $\frac{T_{1}^{2}}{T_{2}^{2}}\ is\ \left( g = 10\ ms^{- 2} \right)$
A) $\frac{6}{5}$
B) $\frac{5}{6}$
C) $1$
D) $\frac{4}{5}$
Explanation
(a) $y = kt^{2}$
$\frac{d^{2}y}{dt^{2}} = 2k$
or $a_{y} = 2\ ms^{- 2}$ ($as\ k = 1\ ms^{- 2}$)
$T_{1} = 2\pi\sqrt{\frac{l}{g}}$
and $T_{2} = 2\pi\sqrt{\frac{l}{g + a_{y}}}$
$\therefore$ $\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{g + a_{y}}{g} = \frac{10 + 2}{10} = \frac{6}{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.