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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 $mgy$:

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

  2. 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.

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    $\begingroup$ "If you assume force is mgy..." The force is not $mgy$. The quantity $mgy$ is the gravitational potential energy (or negative of the gravitational potential energy depending on your conventions). The force is $mg$ in magnitude and its direction is towards the earth, regardless of your conventions for which direction of y is positive. $\endgroup$
    – hft
    Nov 12, 2023 at 20:48
  • $\begingroup$ @user1079505 correct and i know potential is $mgy$, but i am assuming that force is $mgy$. $\endgroup$
    – Dimitri
    Nov 12, 2023 at 21:08
  • $\begingroup$ In all 4 equations units on the left and right sides disagree. $\endgroup$ Nov 12, 2023 at 21:22
  • $\begingroup$ Otherwise, there are systems where the force grows linearly with the distance: harmonic oscillators. You may want to learn about them. $\endgroup$ Nov 12, 2023 at 21:25
  • $\begingroup$ i know potential is $mgy$, but i am assuming that force is $mgy$ That makes no sense at all. Force and potential don’t have the same dimensions, so mgy can’t be a force. It’s like saying “distance is $vt$ but I am assuming mass is $vt$.” You can’t make arbitrary assumptions. Physics has rules, and dimensional consistency is one of them. $\endgroup$
    – Ghoster
    Nov 12, 2023 at 23:20

1 Answer 1

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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.

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