In the figure shown the time period and the amplitude respectively, when `m` is left from rest when spring is relaxed are (the inclined plane is smooth)

Options

A. `2pisqrt((m)/(k)),(mg sin theta)/(k)`
B. `2pi sqrt ((m sin theta)/(k))`, `(2 mg sin theta)/(k)`
C. `2pi sqrt ((m)/(k))`, `(mg cos theta)/(k)`
D. None of these

Answer

A

Explanation

In equilibrium, let `x_(0)` is extention in spring then
`kx_(0) = mg sin theta`
`:. x_(0) = (mg sin theta)/(k)` = amplitude of oscillations.