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 small body attached to one end of a vertically hanging spring is performing SHM about its mean position with angular frequency $$\omega$$ and amplitude $$a$$. If at a height $$y^{\prime}$$ from the mean position, the body gets detached from the spring, calculate the value of $$y^{\prime}$$ so that the height $$\mathrm{H}$$ attained by the mass is maximum. The body does not interact with the spring during its subsequent motion after detachment $$\left(a \omega^{2}>g\right)$$

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