I Should we consider GPE in the vertical case of SHM?

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In the vertical case of simple harmonic motion (SHM), gravitational potential energy (GPE) should be considered, but it does not affect the period of oscillation. The equilibrium length of a spring increases when a mass is suspended vertically, yet the oscillation period remains dependent solely on the mass and spring constant. Mathematical analysis shows that GPE cancels out in the equations of motion, confirming that the system behaves like a harmonic oscillator. The general solution for the motion includes a constant offset due to gravity, but the angular frequency remains unchanged. Thus, while GPE is relevant, it does not alter the fundamental characteristics of SHM.
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For the horizontal case of SHM, we only need to consider KE and EPE. But should we also take GPE into consideration when we are dealing with a vertical case?
 
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Yes, you should.
Gravity action is unidirectional.
 
Jason Ko said:
For the horizontal case of SHM, we only need to consider KE and EPE. But should we also take GPE into consideration when we are dealing with a vertical case?
Not necessarily. The equilibrium length of a spring will increase if a mass is hanging vertically. But, the period of oscillation is unaffected. It depends only on the mass and the spring constant.

If you do the maths, you'll see where the GPE cancels out.

Or, simply Google for SHM mass spring system. There's a good explanation on phys.libretexts.org.
 
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PeroK said:
Not necessarily. The equilibrium length of a spring will increase if a mass is hanging vertically. But, the period of oscillation is unaffected. It depends only on the mass and the spring constant.

If you do the maths, you'll see where the GPE cancels out.

Or, simply Google for SHM mass spring system. There's a good explanation on phys.libretexts.org.
Thks a lot
 
It's also seen easily with math. Let ##x=0## be the position, where the spring is relaxed and the ##x## axis pointing downward (in direction of ##\vec{g}##). Then the equation of motion reads
$$m \ddot{x}=-D x + m g$$
or
$$\ddot{x} + \omega^2 x = g, \quad \omega=\sqrt{D/m}.$$
The general solution of this linear differential equation is given as one special solution of the equation + the general solution of the homogeneous equation, i.e., with setting the right-hand side to 0.

A special solution of the inhomogeneous equation is obviously given for the case that the mass stays at rest, i.e., for ##x=x_0=\text{const}##. This leads to ##x_0=g/\omega^2 = m g/D##. That's easy to understand: It's the elongation of the spring such that the restoring force of the spring compensates the gravitational force, i.e., ##D x_0=mg##.

The general solution for the homogeneous equation is given by
$$x_{\text{hom}}(t)=C_1 \cos(\omega t) + C_2 \sin(\omega t).$$
So the general solution of full equation of motion is
$$x(t)=C_1 \cos(\omega t) + C_2 \sin(\omega t) + x_0,$$
i.e., you have a harmonic oscillator around the equilibrium position, ##x_0##, with the same angular frequency as if there were no gravitation, as already stated above.
 
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
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