Q: A simple pendulum has time period T_{1}. The point of suspension is now moved upward according to the relation y = kt^{2},(k = 1m/s^{2}) where y is the vertical displacement. The time period now becomes T_{2}. The ratio of (T_{1}^{2})/(T_{2}^{2} ) is (Take g=10 m/s^{2})

(a)6/5

(b)5/6

(c)1

(d)4/5

Ans: (a)

Sol: $\large y = k t^2$

Differentiating w.r.t time

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

a_{y} = 2 m/s^{2} (as , k = 1)

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

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

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

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