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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – The Photon
    Sep 15, 2020 at 3:28

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.