Q.Given an one dimensional system with total energy $E = \dfrac{p_x^2}{2m} + V(x) = \text{constant}$, where $p_x$ is the x component of the linear momentum and $V(x)$ is the potential energy of the system. Show that total time derivative of energy gives us force $F_x = -\dfrac{d}{dx}V(x)$. Verify Hooke's law by choosing potential energy $V(x) = \dfrac{1}{2}kx^2$.

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.

$\omega = \sqrt{k/m} = \sqrt{8/0.5} = \sqrt{16} = 4$ rad/s

$E = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} (0.5)(4^2)(0.1^2) = \frac{1}{2} (0.5)(16)(0.01) = 0.04$ J

Or directly: $E = \frac{1}{2} k A^2 = \frac{1}{2} (8)(0.01) = 0.04$ J. Same result.

The Big Picture

SHM energy is a beautiful example of conservation of mechanical energy. The system constantly converts potential to kinetic and back, with the total never changing. This is why a pendulum (ideally) swings forever — and why real pendulums eventually stop (friction steals energy, converting it to heat).

For exams: if you know $A$ and either $k$ or $\omega$ and $m$, you can find total energy. If you know total energy and $x$, you can find $K$ and $U$ separately. It's all connected.

This is exactly the kind of concept that turns up under searches like "Simple Harmonic Motion Energy class 11 physics syllabus" or "Simple Harmonic Motion Energy solved examples" — and it belongs squarely in the Class 11 Physics NCERT/CBSE curriculum. Beyond board exams, it's a dependable scoring topic in JEE Main, NEET and state engineering/medical entrance exams once the core logic clicks.