A body is dropped on to the floor from a height h. It makes elastic collision with the  floor bouncing back to the same height. The frequency of oscillation of its periodic  motion is

The body of mass m is at rest on top. It undergoes free fall and attains a speed of  v1 at the bottom.According to the law of conservation of energy, mgh=12mv12 .On simplifying this we get  v1=2gh…………..(1)Using this we find the time taken to go down the inclined plane using equations of  motion.v=u+at  , Substituting v=2gh ,   u=0 and a=g ,We get 2gh=u+g t ,Therefore, 2ghg =t .This is time of descent. The total time taken to complete one full oscillation will be  two times the time taken to reach the ground.Hence  22ghg =Ti.e.  T=22hg and  f=12g2h