If a particle of mass m slides back and forth between two smooth planes as shown  in figure, the correct statement is

The particle of mass m is at rest on top of the inclined plane. Let it start sliding  down and attain a speed of V at the bottom of the inclined plane.According to the law of conservation of energy, mgh=12mv 2 .On simplifying this we get  v=2gh……………………..(1)The acceleration of the particle down the inclined plane is obtained by resolving g  in the direction of the plane of inclination.We get the acceleration of the particle as  gsinθ . 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=gsinθ,We get  2gh=u+gsinθ  t,Therefore,  2ghgsinθ  =t.= time of descent The total time taken=time period of motion will be four times the time taken to  reach the ground,Hence, time period of oscillation =42ghgsinθ   .and frequency of oscillation =  gsin2θ32h