Animated Solution for Physics - Oscillations: When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is

When we study Simple Harmonic Motion (SHM), we usually picture a block bouncing on a spring or a pendulum swinging back and forth. We visualize its position changing as time ticks forward. But what if we remove the clock entirely? What if we want to see how the particle's velocity behaves strictly based on where it is, rather than when it is there?

This leads us into a beautiful concept called a Phase Space Diagram, where we plot velocity ($v$) on the y-axis and displacement ($x$) on the x-axis. Let's uncover the geometry of this hidden space.

To find the relationship between velocity and displacement, we must start with the fundamental equations of motion that govern SHM. The displacement of a particle executing SHM is given by:

Here, $A$ is the maximum displacement (amplitude), and $\omega$ is the angular frequency.

Velocity is simply the rate of change of displacement. By differentiating our position equation with respect to time, we get:

We now have two equations, both tied to the variable of time, $t$. Our mission is to forge a direct link between $v$ and $x$, which means time must be eliminated.

To do this, we isolate the trigonometric functions from both equations:

Now, we call upon the most famous identity in trigonometry:

By substituting our isolated terms into this identity, we seamlessly erase time from our reality:

Take a step back and look at the structure of our final equation:

This perfectly mirrors the standard mathematical equation of an ellipse ($\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$).

This tells us a profound truth: if you track the state of an oscillating particle by plotting its velocity against its position, it doesn't move randomly. It traces out a perfect, continuous elliptical loop.

- The ellipse crosses the x-axis at $+A$ and $-A$, which are the extreme positions where velocity is zero. - It crosses the v-axis at $+A\omega$ and $-A\omega$, which is the mean position where the particle reaches its maximum speed.

Is it ever possible for this path to be a circle? Yes! If the angular frequency $\omega$ happens to be exactly $1\text{ rad/s}$, then the denominator for the velocity term becomes $A^2$. The equation simplifies to $x^2+v^2=A^2$, which is the equation of a circle. However, because $\omega$ can be any value depending on the mass and spring constant, the general, most accurate representation is always an elliptical path.