Question

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

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

Answer 1

The correct option is A $\frac{6}{5}$

The point of suspension is moved up according to the relation :
$y=k{t}^2$
This implies that the simple pendulum is in an accelerated frame moving up with acceleration.
$a_y=\frac{d^{2}y}{d{t}^2}=2k$
$a_y=2{\text{ms}}^{-2}$ $(\because k=1{\text{ms}}^{-2})$
We know that,
$T_1=2\pi \sqrt{\frac{l}{g}}$
and
$T_2=2\pi \sqrt{\frac{l}{g_{eff}}}=2\pi \sqrt{\frac{l}{g+a_y}}$

$\therefore \frac{T_1^2}{T_2^2}=\frac{g+a_y}{g}=\frac{10+2}{10}=\frac{6}{5}$