Question

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

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

Answer 1

Answer: (B)

Given, $y = kt^2$
Differentiating both sides of the above equation w.r.t. time $'t'$, we get
$\therefore \frac{dy}{dt} = 2kt$
Similarly, $\frac{d^2y}{dt^2} = 2k$
$\Rightarrow a = \frac{d^2y}{dt^2} = 2\times 1 $
$ = 2\,m\,s^{-2}$
For a pendulum, $T = 2\pi \sqrt{\frac{1}{g_{eff}}}$ or
$ T \propto \sqrt{\frac{1}{g_{eff}}}$
$\therefore (\frac{T_1}{T_2})^2 = \frac{g +a}{g}$
$ = \frac{12}{10} = \frac{6}{5}$