Question

A simple pendulum has time period $T_1$. The point of suspension is now moved upward according to the relation $y = kt^2, (k = 1ms^{-2})$ where y is the vertical displacement. The time period now becomes $T_2$. The ratio of $\frac{T^2_1}{T^2_2}$ is
$(Take\,g = 10ms^{-2})$

• A. 6/5
• B. 5/6
• C. 1
• D. 4/5

Answer 1

Answer: (A)

$y=kt^2 \Rightarrow \frac{d^2y}{dt^2}=2k$
$a_y=2ms/s^2$ $(as k=1m/s^2)$
$T_1=2\pi\sqrt{\frac{l}{g}}$
and $T_2=2\pi\sqrt{\frac{l}{g+a_y}}$
$\therefore \frac{T^2_1}{T^2_2}=\frac{g+a_y}{g}=\frac{10+2}{10}=\frac{6}{5}$
$\therefore $ Correct answer is (a)