Question-38 A simple pendulum has time period T_1. The point of suspension is now moved upward according to equation y=kt^2 where k=1\ m/s^2. If new time period is T_2 then \frac{T_1^2}{T_2^2} will be (g=10\ m/s^2)
(A) \frac{2}{3}
(B) \frac{5}{6}
(C) \frac{6}{5}
(D) \frac{3}{2}
Answer : (C)
Explanation:
T₁ = 2\pi \sqrt{\frac{l}{g}} ….(when stationary)
T₂ = 2\pi \sqrt{\frac{l}{g + a}}
….(When lift is accelerating upwards)
∴ y = kt²
v = \frac{dy}{dt} = 2kt
a = \frac{dv}{dt} = 2k = 2 m/s^2
T = 2\pi \sqrt{\frac{l}{10}} , T’ = 2\pi \sqrt{\frac{l}{12}}
⇒ \frac{T_1^2}{T_2^2} = \frac{12}{10} = \frac{6}{5}