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 $\frac{T_1^2}{T_2^2}$ is (Take $g=10m{s}^{-2}$)

• A. $\frac{6}{5}$
• B. $\frac{5}{6}$
• C. $\frac{1}{2}$
• D. $\frac{4}{5}$

Answer 1

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

Upward displacement $y=k{t}^2$

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

or $a_y=2m/s^2[ask=1m/s^2]$

$T_1=2\pi \sqrt{\frac{l}{g}}$

and $T_2=2\pi \sqrt{\frac{l}{g+a_y}}$

$\therefore \frac{T{}_1^2}{T{}_2^2}=\frac{g+a_y}{g}$

$=\frac{10+2}{10}=\frac{6}{5}$