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

[Jason Ko]

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?

 

[Lnewqban]

Yes, you should.
Gravity action is unidirectional.

 

[PeroK]

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.

 

[Jason Ko]

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

 

[vanhees71]

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.