Question

A simple pendulum has time period $2\, s$. The point of suspension is now moved upward according to relation $y=\left(6 t-3.75 t^{2}\right) m$ where $t$ is in second and $y$ is the vertical displacement in upward direction. The new time period (in s) of simple pendulum will be $\left(g=10\, ms ^{-2}\right)$

Answer 1

$T_{1}=2 s =2 \pi \sqrt{\frac{l}{g}}, T_{2}=2 \pi \sqrt{\frac{l}{g^{\prime}}}$
where $g'=g+\frac{d^{2} y}{d t^{2}}=10-7.5=2.5=\frac{g}{4}$
$\Rightarrow T_{2}=2 \pi \sqrt{\frac{l}{g / 4}}=2 \times 2=4\, s$