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=12mv 2.On simplifying this we get v=2gh ……………………..(1)Using this we find the time taken to go down 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