Question

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

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

Answer 1

The correct option is A

$\frac{6}{5}$

Point of suspension is moving as

$y=k{t}^2(k=1\frac{m}{s^2})$

$y=t^2$

$\frac{dy}{dt}=2t$

$\frac{d^{2}y}{d{t}^2}=2$

So accleration of suspension point =2$\frac{m}{s^2}$

So from non-intertial frame the bob will feelg and a${}_{\circ }$ (pseudo scceleration) downward

$T=2\pi \sqrt{\frac{l}{g}}$

$g_{effective}$ will be $(g+a_{\circ })=10+2=12\frac{m}{s^2}$

$T_2=2\pi \sqrt{\frac{l}{g_{eff}}}$

$\frac{T_1^2}{T_2^2}=\frac{12}{10}=\frac{6}{5}$