A simple pendulum has time period T₁. The point of suspension is now moved upward according to the relation y = Kt², (K = 1 m/s²) where y is the vertical displacement. The time period now become T₂. The ratio of $${{T_1^2} \over {T_2^2}}$$ is (g = 10 m/s²)

A particle executes simple harmonic motion between x = - A to x = + A. The time taken for it to go from 0 to $${A \over 2}$$ is T₁ and to go from $${A \over 2}$$ to A is T₂. Then

The period of oscillation of a simple pendulum of length $$L$$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $$\alpha$$, is given by

A particle free to move along the x-axis has potential energy given by $$U\left( x \right) = k\left[ {1 - \exp \left( { - {x^2}} \right)} \right]$$ for $$ - \infty \le x \le - \infty $$, where k is a positive constant of appropriate dimensions. Then