Mistake Master
Mistake Master
In SHM, energy bounces between two boxes: kinetic and spring potential. As the block swings back and forth, $K$ and $U_s$ trade places while their sum stays locked at $E = \dfrac{1}{2}kA^{2}$. Here's the catch: doubling the amplitude doesn't double the energy. It quadruples it.
The energy stored in a stretched or compressed spring is $U_s = \dfrac{1}{2}kx^{2}$, where $x$ is the displacement from the spring's natural length. Stretch the spring twice as far and the stored energy quadruples. That quadratic is the whole topic.
Why the square? Because the spring force isn't constant. At zero stretch the spring pulls with zero force; at displacement $x$ it pulls with $kx$. So the average force over the stretch is $\dfrac{1}{2}kx$, and the work you do against the spring is the area of a triangle with base $x$ and height $kx$:
$$U_s = \text{(area of triangle)} = \dfrac{1}{2} \cdot x \cdot kx = \dfrac{1}{2}kx^{2}$$That factor of $\dfrac{1}{2}$ is the correction for the growing force. If you used $F \cdot x$ with $F = kx$ (the final force), you'd get $kx^{2}$, which is twice the right answer. You'd be assuming the spring fought you with full force from the start.
So the bigger the amplitude $A$, the more the spring is stretched at the turn. More stretch means more stored energy. At the turning points, where $x = \pm A$, that stored energy hits its maximum: $U_{s,\max} = \dfrac{1}{2}kA^{2}$. As §2 shows, this is the system's total mechanical energy.
In a frictionless spring oscillator, the only thing doing work on the block is the spring. No friction, no heat, no losses, so the system's total mechanical energy is locked:
$$E = K + U_s = \dfrac{1}{2}kA^{2}$$At any moment in the cycle, $K$ and $U_s$ can be anywhere between $0$ and $E$, but their sum is always the same. The energy isn't going anywhere; it's trading sides.
At the turning points ($x = \pm A$), the block is momentarily at rest. $K = \dfrac{1}{2}m(0)^{2} = 0$ and the spring is at its maximum stretch, so $U_s$ is at its maximum: $\dfrac{1}{2}kA^{2}$. All the energy sits in the spring.
At equilibrium ($x = 0$), the spring is at its natural length, so $U_s = \dfrac{1}{2}k(0)^{2} = 0$. The block has its maximum speed, so $K$ takes the full energy: $K_{\max} = \dfrac{1}{2}mv_{\max}^{2}$. Set this equal to the total:
$$\dfrac{1}{2}mv_{\max}^{2} = \dfrac{1}{2}kA^{2} \quad\Longrightarrow\quad v_{\max} = A\sqrt{\dfrac{k}{m}} = A\omega$$That's the same $v_{\max} = A\omega$ you saw in 7.3, this time from energy conservation instead of from $x(t) = A\cos(\omega t)$. Two paths, same answer.
Two scaling laws fall out. First, $E \propto A^{2}$. Double the amplitude and the total energy quadruples, because $A$ shows up squared in $E = \dfrac{1}{2}kA^{2}$. Second, $E$ doesn't depend on $m$ at all. Heavier mass means slower oscillation (longer $T$), but for the same spring stretched to the same amplitude, you've stored the same energy.
If you spent Unit 3 thinking of $K = \dfrac{1}{2}mv^{2}$, the no-mass result is a surprise. Mass does appear inside $K$, but it shows up again (the other way) inside $v_{\max}^{2} = (k/m)A^{2}$. The two factors of $m$ cancel, and $E$ comes out clean.
Three pitfalls cover most of the wrong answers students give about energy in SHM. The first two are about scaling and where energy goes; the third is about the formula for spring PE itself.
You've seen this pattern before. In 7.3, doubling $A$ doubled $v_{\max}$ and doubled $a_{\max}$; both are linear in $A$. It's tempting to drop $E$ into the same bucket. Energy feels like another "size of the swing" quantity, so if the swing doubles, the energy doubles.
Fix. $E = \dfrac{1}{2}kA^{2}$ is quadratic in $A$. Doubling $A$ multiplies $E$ by $2^{2} = 4$. Tripling it multiplies by 9. The reason: when you grow $A$, two things grow at once. The spring force at the turn grows with $A$, and so does the distance the block travels against it. Force times distance gives $A \cdot A = A^{2}$.
This one comes from treating energy as a substance the block carries around with it. If the block has $K$ when it's moving, and $K = 0$ when it's at rest, then a block at rest looks like it has no energy. So if the block stopped at the turn, its energy must have leaked out somewhere.
Fix. The system's energy didn't go anywhere; it just moved from the block to the spring. At the turn, $K = 0$ but $U_s = \dfrac{1}{2}kA^{2}$, and the sum is unchanged. The spring is "holding" the energy and is about to give it back. Energy in an ideal SHM system is conserved at every instant of every cycle, never lost, never used up.
Work = force × distance is a clean formula when force is constant, and it's probably the first one you learned. But the spring force isn't constant. At zero stretch it's zero, at displacement $x$ it's $kx$, and it grows linearly the whole way.
Fix. Use the average force when the force is changing linearly. The spring force averages $\dfrac{1}{2}kx$ over a stretch from $0$ to $x$, so the work you do against it is $\dfrac{1}{2}kx \cdot x = \dfrac{1}{2}kx^{2}$. That factor of $\dfrac{1}{2}$ is the correction for the changing force. Without it, you'd be assuming the spring fought you with its maximum force all the way through.
Ten quick checks across the lesson's three pitfalls. Each one is a single-select question with a wrong-answer message that names what you were probably thinking when you picked it. Progress saves as you go.