Relation between shm and circular motion

I recently read in a book that combination of two simple harmonic motions of equal amplitude in perpendicular directions differing in phase by pi/2 is circular motion. I don't seem to understand this because I am not able to figure which two forces in circular motion are acting to cause two different simple harmonic motions.

Any explanation would be thoroughly appreciated

We have two harmonic oscillators in perpendicular directions and phase difference $\pi/2$, therefore the initial conditions are $x(0) = 0$, $\dot{x}(0) = a$, $y(0) =a$, $\dot{y}(0) = 0$, $a>0$, and $$\begin{split}\ddot{x} &= -x,\\ \ddot{y}&= -y,\end{split}$$ with solutions $$\begin{split}x(t) &= a \sin t,\\ y(t) &= a\cos t,\end{split}$$ which satisfy $x^2 + y^2 = a^2$, the equation of a circle.

• Note also that $x=a \sin(t)$ $y=a \sin(t\pm \pi/2)$ satisfies $x^2+y^2=a^2$, hence the "differing in phase by $\pi/2$". – user12029 Apr 13 '14 at 9:14
• Yes, I tried to make things simple, so that the idea is easier to convey. This is the reason the equations above have $\omega =1$, so that we don't need to set initial conditions that might seem arbitrary at first sight. – auxsvr Apr 13 '14 at 9:20
• But what are the 2 forces? – user34304 Apr 13 '14 at 10:57
• They are $-a\cos t\hat{x}$, $-a\sin t\hat{y}$. – auxsvr Apr 13 '14 at 11:25

Consider a point moving in a circle with constant angular velocity $\omega$:

At a time $t$ the radius to the moving point $(x, y)$ makes an angle to the $x$ axis of $\omega t$, so the coordinates $x$ and $y$ are given by:

$$x = r \cos(\omega t)= r \sin(\tfrac{\pi}{2} - \omega t)$$

$$y = r \sin(\omega t)$$

So the points on the $x$ and $y$ axes move in simple harmonic motion with a phase difference of $\pi/2$.

Think about a point moving round a circle (centered at the origin) at constant speed. The point's x co-ordinate shows simple harmonic motion. Likewise the point's y co-ordinate shows shm. They are out of phase, since when for example the x co-ordinate is 0, the y co-ordinate will be +/- radius.

For a circular motion in the XY plane the two forces you are looking for are not physical (like the tension in the string the rotating object is attached to) but are the X and Y components of the centripetal force. Try to think about the position in polar coordinates and you will find that those two components are really easy to calculate.

The figures you get by combining two s.h.m. with different phase and frequency are called Lissajous figures. Basically the answer is in the Wikipedia article (and the other very good answers here), but maybe I'll expand a bit on this gif:

You would get the phase difference of $\pi/2$ for example by starting the first spring at half and the second spring at zero distance from the point the spring is mounted, assuming the maximum distance i.e. the amplitude for them is the same.
So in the end, the combination of two shm with phase difference $\pi/2$ and same amplitude gives you a circle. When you have zero phase difference - this means you start the two masses at the same point - you get a figure like this: