Question

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

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

Answer 1

Answer: (B)

Given, $y=k{t}^2$
Differentiating both sides of the above equation w.r.t. time ${}^{\prime }{t}^{\prime }$, we get
$\therefore \frac{dy}{dt}=2kt$
Similarly, $\frac{d^{2}y}{d{t}^2}=2k$
$\Rightarrow a=\frac{d^{2}y}{d{t}^2}=2\times 1$
$=2m{s}^{-2}$
For a pendulum, $T=2\pi \sqrt{\frac{1}{g_{eff}}}$ or
$T\propto \sqrt{\frac{1}{g_{eff}}}$
$\therefore (\frac{T_1}{T_2}{)}^2=\frac{g+a}{g}$
$=\frac{12}{10}=\frac{6}{5}$