Animated Solution for Physics - Oscillations: For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

The phenomenon of Simple Harmonic Motion (SHM) is one of the most elegant and ubiquitous concepts in physics. Whether it's a pendulum swinging, a mass bobbing on a spring, or atoms vibrating in a crystal lattice, the underlying principles remain the same. At the heart of this motion is a continuous, mesmerizing dance between two forms of energy: Kinetic Energy and Potential Energy.

In this article, we will dive deep into the mechanics of SHM to find the exact point where these two energies are perfectly balanced.

To truly understand what's happening, we must first visualize the energy landscape of a particle executing SHM. Imagine a block attached to a spring, oscillating back and forth on a frictionless surface.

Let the displacement of the block from its mean (equilibrium) position be $x$, and let the maximum displacement—the amplitude—be $A$.

Potential Energy (PE): When the block is at the mean position ($x=0$), the spring is neither compressed nor stretched. The potential energy is zero. As the block moves towards the extremes ($x=\pm A$), the spring stretches or compresses, storing energy. The potential energy is given by the formula: Graphically, this represents a beautiful parabola opening upwards.

Kinetic Energy (KE): Conversely, when the block passes through the mean position, it is moving at its maximum speed. Here, the kinetic energy is at its peak. As it approaches the extreme positions, it slows down, momentarily stopping before reversing direction. The kinetic energy is given by: This forms an inverted parabola on our energy graph.

The question asks us a very specific and intriguing question: At what displacement $x$ do these two energies become exactly equal?

If we look at our energy graph, we are searching for the exact points where the upward-opening parabola of Potential Energy intersects the downward-opening parabola of Kinetic Energy.

To find this mathematically, we simply set the two energy equations equal to each other:

Substituting our formulas into this equation, we get:

At first glance, this equation might look a bit cluttered, but physics often rewards us with elegant simplifications. Notice that the term $\frac{1}{2}m{\omega }^2$ is present on both sides of the equation. This term represents the physical constants of our specific system (mass and angular frequency).

Because it is a non-zero common factor, we can divide both sides by it, effectively canceling it out. This leaves us with a beautifully simple geometric relationship:

Now, we need to isolate $x$. We can do this by adding $x^2$ to both sides of the equation:

Next, we divide by 2:

Finally, to find the displacement $x$, we take the square root of both sides. Remember that taking the square root yields both a positive and a negative result, representing the two sides of the mean position:

We have found our answer, but what does it actually mean?

The value $\frac{1}{\sqrt{2}}$ is approximately $0.707$. This tells us that the kinetic and potential energies are perfectly equal when the particle has covered about $70.7\%$ of the distance from the mean position to the extreme position.

At this exact spot, the total mechanical energy of the system is split exactly 50/50 between motion (kinetic) and position (potential). It is a point of perfect symmetry in the dynamic world of Simple Harmonic Motion!