Why linear displacement is length of the arc?

Question

I mean isn't the word itself says "linear" meaning straight and "displacement" meaning shortest path between two points?
we even used equations of motion in circular motion like $v^2 = u^2 + 2a_t(l)$ where $a_t$ is tangential acceleration, $l$ is lenght of arc. But displacement here is not a straight line
but we did problems related to bob and pendulum like

A simple pendulum of length L is swingging from A to B, forms angle $2\theta$ find the distance travelled and displacement travelled

Since the motion of the pendulum is also circular why didn't we treat displacement as length of the arc?, we got displacement as chord ($2L\sin\theta)$ rather than an arc, and arc was treated as distance ($2L\theta)$.
I'm confused regarding this

Answer 1

First I think it is useful to give a brief definition of displacement and distance. Distance depends on the path which an object moves whereas displacement is the shortest distance between two points, usually initial and final positions. Displacement and Linear distance imply the same.

You have mentioned in your question that you have used equations of motion for solving questions regarding circular motion. But you CAN'T use equations of motions unless acceleration is not constant, therefore here you can clearly see $a_t$ is not constant as it changes its direction every second. You may wonder then how to solve such questions. Yes, you have to use this equation but with a slightly different meaning. What is this contradiction. Let me explain.

I wonder whether you have a clear idea about from where these equations of motion come. This is basically from the definition of acceleration. Acceleration is defined as rate of change of velocity. This can be shown as,$$\text{rate of change of velocity}=\frac{\text{final velocity} -\text{initial velocity}}{t}$$ $$a=\frac{v-u}{t}$$ where $a$ is constant. From this you can derive $$v=u+at\tag{1}$$ And from the defintion of average velocity, $$\text{average velocity}=\frac{\text{total displacement}}{\text{total time}}=\frac{\text{initial velocity} +\text{final velocity}}{2}$$ $$\frac St=\frac{u+v}{2}$$ $$S=\frac{(u+v)}{2} t\tag{2}$$ From these two equations you can derive other equations.

Then how do you get correct answer in your question regarding circular motion using that equation? It's simply because your $a$ and $S$ do not have usual meanings here. At this situation your $a$ represents rate of change of speed and $S$ represents distance. This comes from the definition like before: $$\text{rate of change of speed}=\frac{\text{final speed} -\text{initial speed}}{2}$$ So you may see $l$ in your equation is not for displacement, it is obviously for length. Also you can use this equation because the magnitude of $a_t$ is constant, eventhough its direction is not (because the equation we defined is not a vector equation). Thus this conclusion of yours is not correct:

we even used equations of motion in circular motion like $v^2=u^2+2a_t(l)$ where at is tangential acceleration, $l$ is lenght of arc. But displacement here is not a straight line

Hence, sorry to say, the title of the question is incorrect too.

Answer 2

I think you are misreading the page/slide that you show. I don't think it is saying that linear displacement is the length of the arc. It says linear displacement is $\Delta x$ (not $\Delta s$). Although $x$ is not shown in the diagram, I imagine it is referring to a one dimensional linear motion $x=f(t)$ where displacement between times $t_1$ and $t_2$ is $f(t_2) - f(t_1) = \Delta x$. It is contrasting this with circular motion $\theta = g(t)$ where the angular displacement between times $t_1$ and $t_2$ is $g(t_2) - g(t_1) = \Delta \theta$.

Although this is not explicitly stated, the linear displacement corresponding to an angular displacement of $\Delta \theta$ is the length of the chord between start and end positions i.e. $\Delta x = 2r \sin (\frac {\Delta \theta} 2)$

Answer 3

The swing of a pendulum is often approximated by Simple Harmonic Motion (SHM) for small swings, i.e. if $\theta$ is small. SHM takes place over a straight line.

Even though the pendulum really moves through a distance $2L\theta$, The displacement, the straight line distance, is $2L\sin\theta$. Distance and displacement are almost the same for a pendulum if the angle is small.