Consider a simple pendulum of length l = 0.9 m which is properly placed on a trolley rolling down on a inclined plane which is at `0 = 45°` with the horizontal. Assuming that the inclined plane is frictionless. Assuming that the time period of oscillation of the simple pendulum is T. Find the value of T.

Explanation

The effective value of acceleration due to gravity (g) will be equal to the component of g normal to the inclined plane which is
`g' = g cos alpha`
`T= 2pi sqrt((l)/(g))= 2pi sqrt((l)/(g cos theta))`
Length of the pendulum `l=0.9m`
Angle of inclination `theta=45^(@)`
`:.` Time period `2pi sqrt((l)/(g))`
`g' =g cos theta= g cos 45^(@)`
`= gxx(1)/(sqrt(2))`
`=(g)/(1.414)`
`=(9.8)/(1.414)`
`=6.9306 m//s^(2)`
Time period `T= 2pi sqrt((l)/( g cos theta))`
`=2xx3.14xxsqrt((0.9)/(6.9306))`
`=6.28xx sqrt(0.129858)`
`=6.28xx 0.3602`
`=2.2628` second