At what displacement from the equilibrium is the energy of a Simple Harmonic Motion half kinetic energy and half the potential energy?
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Let A > 0 represent the amplitude of the motion (maximum distance from equilibrium).
If x represents the displacement from equilibrium, then the potential energy is (1/2)kx^2 where k is the spring constant.
Also, total energy is constant and is equal to potential energy + kinetic energy.
At the maximum displacement A from equilibrium, the potential energy is (1/2)kA^2, and the kinetic energy is zero since the velocity is zero at that moment. So the total energy is always (1/2)kA^2.
If we want the kinetic and potential energy each equal to half the total energy, it's enough to set just the potential energy equal to half the total energy (since then the other half would be kinetic energy).
So we want
(1/2)kx^2 = (1/2)(1/2)kA^2
x^2 = (1/2)A^2
x = +-A/sqrt(2).
So when the displacement from equilibrium is +-1/sqrt(2) or about +-0.707 times the maximum distance from equilibrium, the kinetic and potential energy each equal half the total energy.
Lord bless you today!
If x represents the displacement from equilibrium, then the potential energy is (1/2)kx^2 where k is the spring constant.
Also, total energy is constant and is equal to potential energy + kinetic energy.
At the maximum displacement A from equilibrium, the potential energy is (1/2)kA^2, and the kinetic energy is zero since the velocity is zero at that moment. So the total energy is always (1/2)kA^2.
If we want the kinetic and potential energy each equal to half the total energy, it's enough to set just the potential energy equal to half the total energy (since then the other half would be kinetic energy).
So we want
(1/2)kx^2 = (1/2)(1/2)kA^2
x^2 = (1/2)A^2
x = +-A/sqrt(2).
So when the displacement from equilibrium is +-1/sqrt(2) or about +-0.707 times the maximum distance from equilibrium, the kinetic and potential energy each equal half the total energy.
Lord bless you today!