# What's the relation between acceleration, position and angular velocity?

I just encountered a problem involving lift and oscillations where I found the following differential equation: $$\ddot y = -\frac{ \rho gA}{m}y = -\omega^2 y$$

What's the relation between $$\ddot y$$, $$y$$ and $$\omega$$? Does $$\ddot y = -\omega^2y$$ or $$\ddot y =\omega^2y$$ apply in all situations? If yes, what's the reasoning behind it? I know it makes sense in terms of the units but what's the physical reason behind it?

## 2 Answers

The equation for a simple harmonic oscilator is $$\ddot{x}=-\omega^2x$$, so by comparison in your equation you can state that in your system there is oscilatory motion with $$\omega^2=\rho g A /m$$.

Oscillations such as simple harmonic motion works well (the equation you wrote in the question) for small amplitudes. Consider a simple pendulum oscillating about a fixed point joined by a string. If you gently push the bob sideways, you would generate oscillations. The restoring force acting here is the component of the force acting along the trajectory:

$$F=-mgsin(\theta)$$But for the motion to be simple harmonic motion requires $$F$$ to be direcly proportional to $$-\theta$$ here. That works only if $$sin(\theta)$$ is approximately $$\theta$$ (which is true for very small angles of $$\theta$$ like 15 degrees or smaller.

For larger angles, the motion is more complicated and your expression will not be accurate.