# Finding the total space that an oscillating body has gone through via complex analysis

I was solving my homework and I got to an exercise that stated:

An harmonic oscillating body has an equation of $$y(t) = A \sin(t)$$ Find the total space that the body has travelled during $t \in [t_0,t_1]$.

Using the pythagorean thm I have figured out that the solution is the following: $$s = \int^{t_1}_{t_0} \sqrt{1+[y'(t)]^2} dt = \int^{t_1}_{t_0} \sqrt{1+ v^2(t)} dt$$ (where $v(t)$ is the velocity function).

For personal research reasons (occurring through my leisure time), I have thought that this exercise can be solved using complex analysis. So:

Let $f:\mathbb{R} \to \mathbb{C}$ be a function to the complex numbers with $$f(t) =x(t) + iy(t) = A \cos(t) + iA \sin(t) = A e^{it}$$ or $f(z) = z$

so the total space travelled is equal to $s = \mathrm{Im}{(I)}$ where:

$$I = \int^{t_1}_{t_0} f(z) dz = \int^{t_1}_{t_0} z \cdot z'(t) dt = i \int^{t_1}_{t_0} A^2e^{2it}dt = i A^2 \bigg [ \frac{e^{2it}}{2i} \bigg | ^{t_1}_{t_0} \bigg ] \\ = \frac{A^2}{2}\bigg ( e^{2it_1} - e^{2it_0} \bigg )$$

Am I correct?

• Are you sure that $s$ is given by $s = \int^{t_1}_{t_0} \sqrt{1+[x'(t)]^2} dt = \int^{t_1}_{t_0} \sqrt{1+ v^2(t)} dt$ and not by $s = \int^{t_1}_{t_0} \sqrt{[x'(t)]^2} dt = \int^{t_1}_{t_0} \sqrt{v^2(t)} dt = \int^{t_1}_{t_0} | v(t) | dt$ – Lelesquiz Jun 30 '14 at 19:15
• What does "total space" mean? Is the motion one-dimensional? Are you looking for the vector displacement travelled by the particle? Are you just adding up the total scalar distance travelled? If the latter, it's easier to just answer $A\,{\rm trunc}\left(\frac{t_{1}-t_{0}}{\pi}\right) + A |\sin(t_{1}-t_{0})|$, which is obviously the answer by inspection. – Jerry Schirmer Jun 30 '14 at 19:36
• @Lelesquiz It is the length of the curve between two points. – bolzano Jun 30 '14 at 19:56
• @JerrySchirmer The motion is 1D – bolzano Jun 30 '14 at 19:56
• @MaR1oC if the motion is 1D then you are completely wrong – Lelesquiz Jul 1 '14 at 22:04