Q193 / 200 · Manipal (MET) 2014 · Physics
$T_1$ is the time period of simple pendulum. The point of suspension moves vertically upwards according to $y=k t^2$, where $k=1 \mathrm{~m} / \mathrm{s}^2$. New time period is $T_2$, then $\frac{T_1^2}{T_2^2}=?\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
- A$4 / 5$
- Correct answerB$6 / 5$
- C$5 / 6$
- D$1$
Solution
Acceleration of the point of suspension
$\begin{aligned}
a & =\frac{d^2 y}{d t^2}=2 k=2 \mathrm{~m} / \mathrm{s}^2 \\
T & =2 \pi \sqrt{\frac{L}{g_{\mathrm{eff}}}} \Rightarrow T_1=2 \pi \sqrt{\frac{L}{10}} \text { and } T_2=2 \pi \sqrt{\frac{L}{12}} \\
\frac{T_1^2}{T_2^2} & =\frac{6}{5}
\end{aligned}$
Hence, the correct answer is option: 'B'.