Mass Spring System: PE & KE in SHM Motion

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In a simple harmonic motion (SHM) system, potential energy (PE) and kinetic energy (KE) exhibit a complementary relationship. In a vertical mass-spring system, both elastic potential energy and gravitational potential energy must be considered, leading to a combined potential energy expression. The total potential energy can be represented as U(x,h) = -mgh + 1/2kx^2, where h and x must be defined consistently. Generally, both types of potential energy can be added since they arise from conservative forces. Understanding the reference position is crucial for accurately defining potential energy in these systems.
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In a SHM motion, the PE and KE of the system is of a complement relationship. However, in a vertical mass spring system, there seem to be two kinds of PE, elastic PE in the mass spring system as well as the gravitational PE in the Earth mass system. So, what should be the PE of the SHM, and what objects should be considered as a whole system? (As far as I know, we have to define a system when we want to talk about PE, right? ) And is the sum of the elastic pe and the gravitational pe equals the pe in a horizontal mass spring system, and why? I am rather confused with the concepts of pe. Can pe be simply added?
 
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The PE has to be defined with respect to some "reference position" of the system, usually the "rest position" of the mass-spring system.

If you orient the spring vertically, then the mass will naturally dangle a bit lower due to its weight, i.e., the spring will be slightly stretched at "rest."

What happens when you add the PE due to gravity and the PE due to the spring deformation?
 
The system concerning elastic pe is the mass spring and that concerning gravitational pe is mass Earth system. Can they be added?
 
In general they can be added. Each "type" of potential energy arises from a conservative force. Gravitational potential energy follows from -mgh and a elastic potential energy follows from 1/2kx^2. So in the spring mass system, the potential energy becomes U(x,h) = -mgh + 1/2kx^2. In general for a vertical spring system, we define the potential so that h and x are consistent, that is U(x) = -mgx + 1/2kx^2. Likewise, the concept is analogous in a horizontal spring system where we take mgh = 0 for ease of calculation since gravitational potential does not change.
 
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