Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the o

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Q. Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.

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Concept understanding — Simple Harmonic Motion Energy

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 $A$ :

E_{\text{total}} = \frac{1}{2} m \omega^2 A^2

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:

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.
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.
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.

Watch out

A frequent mistake: writing $U = \frac{1}{2} m \omega^2 x^2$ but forgetting that $\omega^2 = k/m$ . Both forms are equivalent — use whichever is given in the problem.

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. …

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The potential energy of a simple harmonic oscillator is half its maximum total energy when its displacement is $\pm \frac{A}{\sqrt{2}}$ , where $A$ is the amplitude.

A simple harmonic oscillator (SHO) is a system that, when displaced from its equilibrium position, experiences a restoring force proportional to the displacement. This leads to oscillatory motion. Key to understanding its energy is recognizing that the total mechanical energy (sum of kinetic and potential energy) remains constant throughout the motion, assuming no damping.

The potential energy (PE) of an SHO is stored due to its displacement from the equilibrium position. It's maximum at the extreme points of oscillation (amplitude $A$ ) and zero at the equilibrium position. The kinetic energy (KE) is maximum at the equilibrium position (where velocity is maximum) and zero at the extreme points (where velocity momentarily becomes zero before reversing direction).

The "maximum energy of the oscillator" refers to this constant total mechanical energy. We can find its value by considering the point where all the energy is potential (at maximum displacement, $x = \pm A$ , where KE is zero) or where all the energy is kinetic (at equilibrium, $x = 0$ , where PE is zero). Since the problem involves potential energy, it's most convenient to use the total energy expressed in terms of the amplitude.

Here's how to find the displacement:

Define the potential energy of a simple harmonic oscillator. The potential energy $PE$ of an SHO at a displacement $x$ from its equilibrium position is given by:

PE = \frac{1}{2}kx^2

where $k$ is the spring constant (or force constant) of the oscillator.

2. Determine the maximum energy of the oscillator.

The total mechanical energy $E$ of an SHO is conserved. At the extreme positions of oscillation, $x = \pm A$ (where $A$ is the amplitude), the oscillator momentarily comes to rest, meaning its kinetic energy is zero. At these points, all the total energy is in the form of potential energy.

Therefore, the maximum energy of the oscillator, which is its total energy, is:

E_{max} = \frac{1}{2}kA^2

This is the constant total energy of the system.

> [!FORMULA]
> The total energy of a simple harmonic oscillator is $E = \frac{1}{2}kA^2$.

3. Set up the condition given in the problem.

The problem states that the potential energy ( $PE$ ) is half of the maximum energy ( $E_{max}$ ).

$PE = \frac{1}{2} E_{max}$ …

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Q.Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.

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