Q: A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y = kt2,(k = 1m/s2) where y is the vertical displacement. The time period now becomes T2. The ratio of (T12)/(T22 ) is (Take g=10 m/s2)
(a)6/5
(b)5/6
(c)1
(d)4/5
Ans: (a)
Sol: $\large y = k t^2$
Differentiating w.r.t time
$\large \frac{d^2 y}{dt^2} = 2 k$
ay = 2 m/s2 (as , k = 1)
$\large T_1 = 2\pi \sqrt{\frac{l}{g}}$
$\large T_2 = 2\pi \sqrt{\frac{l}{g + a_y}}$
$\large \frac{T_1^2}{T_2^2} = \frac{g + a_y}{g}$
$\large \frac{T_1^2}{T_2^2} = \frac{10 + 2}{10} = \frac{6}{5}$