The correct question is;
An object of mass m attached to a spring of force constant K oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.
What is the system's potential energy when its kinetic energy is equal to ¾E?
Answer:
P.E = ⅛KA²
Step-by-step explanation:
From conservation of energy, the total energy in the system is given as the sum of potential and kinetic energy.
Thus,
Total Energy; E = K.E.+P.E.
In simple harmonic motion, the total energy is given by;
E = ½KA²
We are told that kinetic energy is ¾E.
Thus, ½KA² = ¾(½KA²) + P.E
P.E = ½KA² - ⅜KA²
P.E = ⅛KA²