a SHO has a maximum speed of 5m/s. How fast will it be going when the potential energy is seven-ninths of the kinetic energy.
when U, the potential energy is 0, the thing goes 5 m/s
Its total energy is kinetic
E = (1/2) m v^2 = 25 m/2
so at our condition
25 m/2 = U + Ke = (7/9)(1/2)mv^2 + (1/2)m v^2
so
25 = (7/9)v^2 + v^2 = (16/9) v^2 (nice square numbers)
v^2 = 25*9/16
v = 5*3/4 = 15/4 = 3.75
check
ke = (1/2)mv^2 = 7.03 m
u = (7/9) ke= = 5.57 m
ke+u = 12.6 m
total should be 25/2 m = 12.5 m
close enough