# Direction of displacement

Displacement is defined as the vector obtained by joining the final position to the initial position (head towards the final position). Well,i know this is silly but what are these final and initial positions?Common sense says that initial position must be the one from which we started observing the motion and the final position should be the one where we are asked to end our observation as per the problem and it works almost everywhere except in SHM. Consider a body performing SHM about it's mean position O. Say, one of it's extreme positions is E (somewhere on positive x axis) and M be a point somewhere in between O and E. Now what is the direction of displacement as the body moves from E to M? Using my 'common sense defintions of final and initial position' displacement must be in negative x direction but i was told that this is not the case (it is towards positive x) because the mean position is taken to be the initial position no matter where the body starts from. So my question is how do we know what are the initial and final positions? Are there any conventions?

• It depends on what you're asked. If you're asked for the displacement from the mean position, then the mean position is the starting point. If you're asked for the displacement from its position at some time t, then whatever position it was at at time t is the starting point. Etc. Commented Sep 15, 2020 at 3:28

Instead of initial and final position, I think it's better to think of reference position and current position. You're free to choose any reference position that suits your taste, and then you measure the current position relative to that. Essentially, you're choosing the origin of your coordinate system.

For a harmonic oscillator, choosing its resting or mean position is sensible, because that point is meaningful to the problem at hand. But we could have some fun and decide to use a different reference point, say, one that is $$1$$ unit to the left and $$1$$ unit above the resting position. Then the displacement of an oscillator oscillating vertically would be

$$\vec r=\begin{pmatrix}1\\0\\-1+A\cos(\omega t+\varphi_0).\end{pmatrix}$$

But it's a lot easier on the eyes to choose the resting point as a reference, because then it's just $$(0,0,A\cos(\omega t+\varphi_0))$$. But displacement doesn't have to be relative to some physically meaningful point. They're just more convenient.

I think you need to define first what exactly does it mean to measure the position of a point (or a particle).

You have some reference points (sometimes called the origin) and some unit directions (sometimes called the basis vectors) along which you can measure distances.

You define the position of a particle as from the three distances along the three direction vectors relative to the origin.

$$\vec{\rm pos} = x \hat{i} + y \hat{j} + z \hat{k}$$

or with the shorthand notation

$$\vec{\rm pos} = \pmatrix{x\\y\\z}$$

Now a displacement can be described as the relative position of a point as measured at different times

$$\vec{\rm displacement} = \vec{\rm pos}(t_{\rm final}) - \vec{\rm pos}(t_{\rm initial})$$

but you are free to call the time frames anything you want.

Similarly a distance can be described as the relative position of a point, to some other point at the same time.

$$\vec{\rm distance} = \vec{\rm pos}_2(t) - \vec{\rm pos}_1(t)$$

Displacement is relative. You cannot just be asked to provide a displacement. You must be asked to provide a displacement from somewhere.

In a case like your SHM example, it would be reasonable for many purposes to keep the starting point the same, so that you always are considering displacement from the point O, for example. But that is just my assumption. It must be clear from your question - otherwise you cannot answer it. Just like how you can't answer what the length is to some point without knowing where to measure from.