From the book entitled Classical Mechanics written by John R Taylor, chapter no 5, Simple Harmonic Motion. I'm just citing the lines.

$$x(t)=\text{Re }Ce^{i\omega t}=\text{Re }A e^{i(\omega t-\delta)}$$ Where $C$ is the constant, $A$ is the amplitude. the complex number $e^{i(\omega t-\delta)}$ moves counter clock wise with angular velocity $\omega$ around a circle of radius $A$. Its real part is the projection of the complex number on to the real axis. While the complex number goes around the circle this projection oscillates back and forth on the $x$ axis with angular velocity $\omega$ and amplitude $A$.

It's basically the solution of the simple harmonic motion. I just don't understand a bit of those words. How they decide only the complex number moves on? Why here comes the perspective of a circle? Why it does have to move counter clock wise? I know i lack some basic physics. But i really need help please. Thank you

  • $\begingroup$ Is this your first encounter with complex numbers in polar form ($re^{i\theta}$)? $\endgroup$ – jacob1729 May 23 at 16:02
  • $\begingroup$ no.Its not. I had done some previous math but those were shallow. $\endgroup$ – F.sharmin May 23 at 16:04
  • $\begingroup$ Try to plot $e^{it}$ for a few values of $t$ in an Argand diagram. Then try plotting $\mathfrak{Re}(e^{it})$ against $t$. $\endgroup$ – jacob1729 May 23 at 16:08

Consider the differential equation for a SHM:

$\frac{d^2x}{dt^2}+\omega x = 0$.

The solution to this actually lies on the complex plane, just like the solution to equation $x^2+1=0$. The solution being:

$x(t)=C_1e^{-i\omega t}+C_2e^{i\omega t}$.

Here both the constants $C_1, C_2$ are complex constants. Let the initial condition be $x(0)=A, \dot{x}(0)=B$. Then if you plug these into the solution, we will get:

$A=C_1+C_2$ and $B=-i\omega C_1+i\omega C_2$


$C_1+C_2=A$ and $-C_1+ C_2=\frac{B}{i\omega}$.

See what happened! The complex constants are actually conjugates to each others. Plugging these in the solution and using Euler's formula for $e^{i\theta}$, we get:

$x(t) = A cos(\omega t) + \frac{B}{\omega} sin(\omega t)$

The solution is totally real! Further using cosine and $sine$ formulas, you can reduce the solution to what you have cited.

The recourse to the complex number space while analyzing the solution is purely a matter of convenience. The complex phasor $x(t)=A e^{i(wt-\delta)}$ is itself a solution: both real and imaginary parts are solutions. We can use properties of phasors to simplify our understanding. For example, it is not so obvious, but we can see that the velocity leads the position from the solution by angle $\pi/2$. This can easily be seen from the complex notation because $\dot{x}(t)= i \omega x$, a rotation by $\pi/2$ and scaling of amplitude.

The phasor rotates counter clock wise because the angle is continuously changing as $\theta=+\omega t$.

I would like you to remember that complex notation greatly simplifies the mathematical analysis and that's why we use it. In and itself there is no connection between CN and reality. It only happens so that our ideas can better be coded into complex space.


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