Q.Show that for a particle executing simple harmonic motion a. the average value of kinetic energy is equal to the average value of potential energy. b. average potential energy = average kinetic energy = $\dfrac12$(total energy). Hint: average kinetic energy = $\langle$Kinetic energy$\rangle$ = $\dfrac{1}{T}\displaystyle\int_0^T(\text{Kinetic energy})\,dt$ and average Potential energy = $\langle$Potential energy$\rangle$ = $\dfrac{1}{T}\displaystyle\int_0^T(\text{Potential energy})\,dt$

Simple Harmonic Motion Energy: From Intuition to Precision

Imagine a pendulum swinging, or a mass bouncing on a spring. You push it once, and it keeps moving back and forth. Where does that energy go? It doesn't vanish — it just changes form. That's the core idea.

The Intuition: A Trade Between Two Forms

Think of a child on a swing. At the highest point, the swing is momentarily still — all the energy is stored as potential energy (the height you could fall from). At the lowest point, the swing is moving fastest — all that stored energy has turned into kinetic energy (the energy of motion). In between, it's a mix of both.

For a spring-mass system (the simplest SHM), the same trade happens:

• When the mass is at the extreme position (maximum displacement), it's momentarily at rest — all energy is potential.
• When the mass passes through the equilibrium position (the centre), it's moving fastest — all energy is kinetic.
• Everywhere else, it's a blend.

The key insight: total mechanical energy stays constant (if no friction). Energy is never created or destroyed — it just shifts between potential and kinetic.

The Precise Statement

For a particle of mass $m$ executing SHM with angular frequency $\omega$ and amplitude $A$:

This is a constant. At any displacement $x$ from equilibrium:

• Kinetic energy: $K = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 (A^2 - x^2)$
• Potential energy: $U = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2$ (since $k = m\omega^2$)
• Total energy: $E = K + U = \frac{1}{2} m \omega^2 A^2$

Notice: when $x = \pm A$, $K = 0$ and $U = E$. When $x = 0$, $K = E$ and $U = 0$.

Why This Matters for Exams

Three things to remember:

1. Total energy depends only on amplitude and frequency — not on the mass's position or speed at any instant. It's a fixed number for a given oscillation.

2. Energy is proportional to the square of amplitude: double the amplitude, quadruple the energy. This is a common exam trap — students think doubling amplitude doubles energy. It doesn't.

3. The potential energy curve is a parabola: $U = \frac{1}{2} k x^2$. This is why SHM is called "harmonic" — the restoring force ($F = -kx$) comes from this parabolic potential well.

A Quick Check

A mass of 0.5 kg oscillates on a spring with $k = 8$ N/m and amplitude 0.1 m. Find total energy. …