At what distance from equilibrium is the energy of a sho half kinetic and half potential?

Distance from equilibrium x = 0.707A

Where A is the amplitude of the SHO

Explanation:

all the energy of a SHO is a potential energy at

PE = 0.5kA^2

Where A is the amplitude of the SHO

Half that energy means

PE = 0.25kA^2

The energy of a SHO is half KE and half PE at a displacement of

PE = 0.5kx^2

Where x is the displacement the displacement from equilibrium when the energy is half potential energy and half kinetic energy.

Equating both equations and solving, we have,

0.25kA^2 = 0.5kx^2

We solve for x

x^2 = (0.25kA^2)/(0.5kx^2)

x^2 = 0.5A^2

x = 0.707A.

answer: its c lololo

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.