Energy in Simple Harmonic Motion

In SHM, kinetic energy (KE) and potential energy (PE) oscillate out of phase with each other, both at twice the frequency of the position. But their sum is constant — that's energy conservation.

The Math

For $x(t) = A\cos(\omega t)$ and $v(t) = -A\omega\sin(\omega t)$:

$\text{PE}(t) = \tfrac{1}{2}k\,x^{2} = \tfrac{1}{2}k A^{2}\cos^{2}(\omega t)$

$\text{KE}(t) = \tfrac{1}{2}m\,v^{2} = \tfrac{1}{2}m A^{2}\omega^{2}\sin^{2}(\omega t) = \tfrac{1}{2}k A^{2}\sin^{2}(\omega t)$

(In the last step we used $m\omega^{2} = k$, the SHM definition.)

Adding them:

$E_{\text{total}} = \text{PE} + \text{KE} = \tfrac{1}{2}k A^{2}\bigl(\cos^{2}\omega t + \sin^{2}\omega t\bigr) = \tfrac{1}{2}k A^{2}$

The trig identity $\sin^{2} + \cos^{2} = 1$ does the magic. The total mechanical energy is $\tfrac{1}{2}kA^{2}$ at every instant — it depends only on the amplitude.

Step-by-Step Solution

Given: $m = 2\;\text{kg}$, $k = 50\;\text{N/m}$, $A = 0.2\;\text{m}$, $\omega = 5\;\text{rad/s}$.

Find: Total energy, max KE, max PE, and energy at $t = T/8$.

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Step 1 — Compute the total energy.

$E = \tfrac{1}{2}k A^{2} = \tfrac{1}{2}(50)(0.04) = 1.000\;\text{J}$

Step 2 — Maximum potential energy.

Reached at the extremes ($x = \pm A$, $v = 0$):

$\text{PE}_{\max} = \tfrac{1}{2}k A^{2} = 1.000\;\text{J}$

All the energy is potential.

Step 3 — Maximum kinetic energy.

Reached at the center ($x = 0$, $v = \pm A\omega$):

$\text{KE}_{\max} = \tfrac{1}{2}m(A\omega)^{2} = \tfrac{1}{2}(2)(1)^{2} = 1.000\;\text{J}$

Same as $\text{PE}_{\max}$ — they trade off perfectly.

Step 4 — Energy split at $t = T/8$.

At $t = T/8 = 2\pi/(8\omega)$, $\omega t = \pi/4$. So:

$x = A\cos(\pi/4) = A \cdot \dfrac{\sqrt 2}{2} \approx 0.1414\;\text{m}$

$v = -A\omega\sin(\pi/4) = -1 \cdot \dfrac{\sqrt 2}{2} \approx -0.7071\;\text{m/s}$

Energy components:

$\text{PE} = \tfrac{1}{2}(50)(0.0200) = 0.500\;\text{J}$

$\text{KE} = \tfrac{1}{2}(2)(0.5) = 0.500\;\text{J}$

At $T/8$, the energy is exactly half PE and half KE — the two are equal there.

Step 5 — Verify the sum.

$\text{PE} + \text{KE} = 0.500 + 0.500 = 1.000\;\text{J} = E_{\text{total}} \;\;\checkmark$

The total is unchanged.

Step 6 — Average values over one period.

A useful fact for many physics problems: averaged over one full cycle, $\langle\sin^{2}\rangle = \langle\cos^{2}\rangle = 1/2$. So:

$\langle \text{PE}\rangle = \langle \text{KE}\rangle = \dfrac{E_{\text{total}}}{2} = \tfrac{1}{4}kA^{2}$

On average, half the energy is potential and half is kinetic — even though at any instant they can be anything from 0 to $E_{\text{total}}$.

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Answer:

$\boxed{\;E_{\text{total}} = \tfrac{1}{2}kA^{2} = 1.000\;\text{J}\;\text{at every instant}\;}$

KE and PE oscillate at twice the frequency of $x(t)$, $\pi/2$ out of phase from each other. Their sum is exactly $\tfrac{1}{2}kA^{2}$ forever. At the extremes, all the energy is PE; at the center, all is KE; halfway between, it's split 50/50.

Try It

• Watch the bar graph beside the oscillator — KE (cyan) and PE (pink) trade off.
• The green total bar stays the same height — that's energy conservation.
• Adjust the amplitude: doubling it quadruples the total energy ($E \propto A^{2}$).