Q: A Simple pendulum has time period ‘T1’. The point of suspension is now moved upwards according to the relation y = kt², (k = 1 m/sec²) where y is the vertical displacement. The time period now becomes ‘ T2 ’, then find the ratio of $\frac{T_1^2}{T_2^2}$

Sol: y = k t^{2}

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

$\large \frac{d^2y}{dt^2} = 2 k = 2 $

a = 2 m/s^{2} (acceleration)

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

$\large T_2 = 2\pi \sqrt{\frac{l}{g+a}}$

$\large \frac{T_1^2}{T_2^2} = \frac{g+a}{g}$

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