Answer:
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.
