A particle of mass m slides back and forth between two inclined frictionless planes. If h is the initial height then period of the motion is

The mass begins at the top of incline at zero speed and begins to slide down until it reaches maximum speed at the bottom. The speed will be such that the kinetic energy equals the potential energy. Call this maximum v0:

(1/2)mv0^2 = mgh

v0 = sqrt(2gh)

The mass will begin the climb up at this speed against a gravity reduced by the incline angle:

a = -gsin(b)

Where b is the incline angle with respect to the horizontal.

Due to symmetry (see note at bottom), the time it takes to climb the incline is one-fourth of the period. The mass will continue to climb until its speed is zero.

v = v0 - gsin(b)*t

Which equals zero when:

t = v0/(gsin(b))

Substituting above result for v0:

t = sqrt(2h/g)/sin(b)

But this was one-fourth of the period:

T = sqrt(32h/g)/sin(b)

*Note that the time to climb the incline is equal to the time to fall back down. The mass must climb and fall twice for one full cycle. This means one climb is one-fourth of the cycle period.