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 : $(g=10m/s^2)$

• A. 6/5
• B. 5/6
• C. 1
• D. 4/5

Answer 1

The correct option is A

6/5

$y=K{t}^2$

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

$or{a}_y=2m/s^2$ $(ask=1m/s^2)$

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

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

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

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

$\therefore correctansweris\left(a\right)$