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Definition/Summary
Displacement is change in position (of the same thing).
For example, in the physics of vibrations and stress, the displacement of each point in the material is the vector from its "natural" position to its present position.
And in mechanics, work done is force "dot" displacement, where displacement means the change in position of the point of application of that force.
Displacement is a vector, but the word also sometimes means the magnitude of that vector (a non-negative scalar): in examination questions, the context should make it clear which is intended.
Scalar displacement is different from distance: for example, if you go half-way round a square of side r, then you have gone a distance (or arc-length) of 2r, but your displacement is only r\sqrt{2}: and if you go all the way round, you have gone a distance of 4r, but your displacement is 0.
Displacement has dimensions of length (L).
Equations
Vector displacement:
\mathbf{r}(b)\ -\ \mathbf{r}(a)
Scalar displacement:
|\mathbf{r}(b)\ -\ \mathbf{r}(a)|
Distance (arc-length) along curve defined by \mathbf{r}(t)\ =\ (x(t),y(t),z(t)):
ds^2\ =\ (d\mathbf{r})^2
s \ =\ \int\ ds \ =\ \int_a^b\ \sqrt{\mathbf{r}'\,^2}\ dt\ =\ \int_a^b\ \sqrt{x'\,^2 + y'\,^2 + z'\,^2}\ \ dt
Extended explanation
In a one-dimensional situation, vector and scalar displacement are the same (arc-length from the "origin"), except that vector displacement can be "negative".
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
Displacement is change in position (of the same thing).
For example, in the physics of vibrations and stress, the displacement of each point in the material is the vector from its "natural" position to its present position.
And in mechanics, work done is force "dot" displacement, where displacement means the change in position of the point of application of that force.
Displacement is a vector, but the word also sometimes means the magnitude of that vector (a non-negative scalar): in examination questions, the context should make it clear which is intended.
Scalar displacement is different from distance: for example, if you go half-way round a square of side r, then you have gone a distance (or arc-length) of 2r, but your displacement is only r\sqrt{2}: and if you go all the way round, you have gone a distance of 4r, but your displacement is 0.
Displacement has dimensions of length (L).
Equations
Vector displacement:
\mathbf{r}(b)\ -\ \mathbf{r}(a)
Scalar displacement:
|\mathbf{r}(b)\ -\ \mathbf{r}(a)|
Distance (arc-length) along curve defined by \mathbf{r}(t)\ =\ (x(t),y(t),z(t)):
ds^2\ =\ (d\mathbf{r})^2
s \ =\ \int\ ds \ =\ \int_a^b\ \sqrt{\mathbf{r}'\,^2}\ dt\ =\ \int_a^b\ \sqrt{x'\,^2 + y'\,^2 + z'\,^2}\ \ dt
Extended explanation
In a one-dimensional situation, vector and scalar displacement are the same (arc-length from the "origin"), except that vector displacement can be "negative".
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
