Question

A simple pendulum has time period $T_1$. The point of suspension is now moved upward according to the relation $y=k{t}^2\left(k=1m{s}^{-2}\right)$ where $y$ is the vertical displacement. The time period now becomes $T_2$. The ratio of $\frac{T_1^2}{T_2^2}$ is (take $g=10m{s}^{-2}$ )

• A. $\frac{6}{5}$
• B. $\frac{5}{6}$
• C. 1
• D. 4

Answer 1

Answer: (A)

According to the relation, $y=k{t}^2$
$\Rightarrow \frac{dy}{dt}=2kt$
$\therefore a=\frac{d^{2}y}{d{t}^2}=2k=2\times 1=2$
$[\because k=1]$
The point of suspension is moving upward with acceleration a, then effective acceleration due to gravity on pendulum
$g^{{}^{\prime }}=g+a=10+2=12m/s^2$
$\therefore T_1=2\pi \sqrt{\frac{l}{g}}$
And $T_2=2\pi \sqrt{\frac{l}{g+a}}$
$\therefore \frac{T_1^2}{T_2^2}=\frac{g+a}{g}$
$=\frac{12}{10}=\frac{6}{5}$