Sign of force for assumed $mgy$ gravity [closed]

Question

If gravity force of earth is $mg$:

1. if positive y is pointing upwards, then: $m\vec a = -mg\hat y$ and $\ddot y = -g$
2. if positive y is pointing down, then: $m\vec a = mg\hat y$ and $\ddot y = g$

If you assume force is $mgy$:

3. if positive y is pointing upwards, then: $m\ddot y = -mgy$ and $\ddot y = -gy$

4. if positive y is pointing down, then: $m\ddot y = mgy$ and $\ddot y = gy$

Solving the (3) and (4) differential eqs end up having completely different trajectories((3) with $c_1cos(\sqrt{2g}t)$ while (4) with $Ae^{\sqrt{g}t} + Be^{-\sqrt{g}t}$ which shouldn't be the case. There seems to be a flaw in (3) or (4), but no idea what.

Answer 1

If you assume force is $mgy$:

After reading OP's comments, it seems that OP just wants to understand how to solve a problem where the force is proportional to the displacement.

The main problem the commenters (myself included) seem to have is one of units. In particular, OP already said that $m$ and $g$ are the usual mass and gravitational constant. Therefore, $mgy$ can not have units of force.

So, we can introduce another parameter with units of length $\ell$ and write, more appropriately: $$ F = \frac{mg}{\ell}y\;. $$

In fact, to save us some writing, let's define: $$ k\equiv -\frac{mg}{\ell} $$ and thus we can write: $$ F = -ky\;. $$

Such a force has been studied since at least the 17th century. In particular, by Robert Hooke, who famously quipped "ut tensio sic vis." Such a force is often called a spring force.

Certainly one can solve the equation: $$ m\frac{d^2 y}{dt^2} = -k y\;, $$ (given appropriate boundary conditions).

The solutions for position $y(t)$ are well-known to be sinusoidal. For example $$ y(t) = A\sin(\sqrt{\frac{k}{m}} t)\; $$ is one such solution.