Mistake Master
Mistake Master
Periodic motion is everywhere: heartbeats, planets, a kid kicking a swing. Most of it is not simple harmonic motion. SHM is the special case where the restoring force grows linearly with displacement from equilibrium. One equation does the work: $F = -kx$. This lesson is about that equation and the three places students reliably get it wrong.
Most periodic motion is not SHM. A planet on a circular orbit moves at constant speed; it traces the same loop forever, but the force on it is constant in magnitude. A ball bouncing elastically between two walls also repeats, but the force on it is zero almost everywhere except for sharp kicks at the walls. Periodic, yes. SHM, no.
Simple harmonic motion is what you get when the restoring force has one very specific form:
$$F = -kx$$Two ingredients matter. First, proportionality: the magnitude of the force grows linearly with the displacement $x$ from equilibrium. Double $x$, and you double the force. Second, the minus sign: the force always points back toward equilibrium, opposite to the displacement. That sign is what "restoring" means. A force that grew with $|x|$ but pointed away from equilibrium would push the block off to infinity, not oscillate it.
One more piece of vocabulary. Equilibrium is the place where the net force on the block is zero. For a horizontal spring on a frictionless surface, equilibrium is the natural-length position. For a mass hanging vertically from a spring, equilibrium is the stretched length where the spring force balances gravity. Whatever the setup, the net restoring force follows $F = -kx$ with $x$ measured from equilibrium, not from the wall or the ceiling. For a vertical spring, that net force is the spring force combined with gravity, not the raw spring force alone.
Pendulums at small angles obey the same kind of rule. The tangential force on the bob is $F_t = -mg\sin\theta$, which is not linear in $\theta$. But for small angles $\sin\theta \approx \theta$, so $F_t \approx -mg\theta$ becomes linear in the angular displacement, and the motion is SHM. Above roughly $15°$, the approximation breaks down and the motion drifts away from SHM, even though it is still periodic.
Once you take $F = -kx$ as the criterion, three consequences follow. Students get all three backward. Each one is a direct read off the equation.
The force is largest where the speed is zero. At a turning point, $|x| = A$, so $|F| = kA$ is at its peak. The block is momentarily stopped, but the spring is at maximum stretch and pulling its hardest. "At rest" describes the velocity, not the force. The force does not care that the block has paused.
The force changes during the motion. The spring constant $k$ is constant; the spring force $F = -kx$ is not. As the block moves, $x$ takes every value from $-A$ to $+A$ and back, and the force tracks $x$ at every instant. "The force on the block" is a moving target.
The sign-flip is the whole story. When the block is right of equilibrium ($x > 0$), the spring pulls it left ($F < 0$). When the block is left of equilibrium ($x < 0$), the spring pushes it right ($F > 0$). The negatives in $-kx$ multiply to a positive when $x$ itself is negative. That sign-flip is what makes the motion oscillate instead of running off in one direction.
A quick numerical example pins all of this down. Set $k = 4$ N/m, block mass $m = 1$ kg, amplitude $A = 1$ m. These are the defaults in the Restoring Lab. Apply $F = -kx$ and $a = F/m$ at four sample positions:
$F = -(4)(+1) = -4$ N (leftward). $a = F/m = -4/1 = -4$ m/s$^2$ (leftward).
$F = -(4)(+0.5) = -2$ N. $a = -2/1 = -2$ m/s$^2$. Both are half the turning-point value.
$F = -(4)(0) = 0$. $a = 0$. The force has dropped to zero; the block coasts through at maximum speed.
$F = -(4)(-1) = +4$ N (rightward). $a = +4$ m/s$^2$ (rightward). Same magnitude as the right turning point, opposite direction.
Trace those four lines. The force changes sign across equilibrium, peaks at the turning points, and zeroes at $x = 0$. Acceleration follows it exactly, because $a = F/m$ and $m$ is a constant. Velocity does the opposite: zero at the turning points, peak at equilibrium. Velocity and force are out of step. Most of §3 is about why.
One footnote. The motion repeats with a period that depends on $k$ and $m$, not on the amplitude. Double $A$, and the block travels twice as far, but it also feels a proportionally stronger force, so it moves proportionally faster. The round-trip time is the same. The formula for the period is Topic 7.2; for now it is enough to know the period exists and is set by the spring and block, not by how hard you pulled.
Three traps catch students reliably on this topic. Each one is a different way of treating the SHM force law as something it is not.
The intuition: when you pull the block to one side and let go, the force you feel against your hand is what makes the block bounce back. That force is a definite number. It is natural to imagine that same number is the force on the block for the whole cycle.
The reality: the spring force is different at every position. The spring constant $k$ is the constant; the spring force $F = -kx$ is not. As the block moves back toward equilibrium, the spring is less stretched, so the force drops. Halfway back, the force is half what it was at release. At equilibrium, the force is zero. Past equilibrium, it reverses.
Fix. Read $F = -kx$ literally: $F$ is a function of $x$. Every position gives its own force. The "force on the block" you feel at release is just one value of that function, the value at $x = A$.
The intuition: "at rest" sounds like nothing is happening. Zero velocity should mean zero everything. And in everyday speech, "stopped" usually implies a state of stillness, not a state of being whipped around in the opposite direction.
The reality: zero velocity and zero acceleration are different statements. Velocity is how fast the block is moving; acceleration is how fast the velocity is changing. The block has $v = 0$ at the turning point because its velocity has just been pulled to zero and is about to reverse, which is exactly the moment when velocity is changing fastest. At $|x| = A$, the force $|F| = kA$ is at its peak, so $|a| = kA/m$ is at its peak too.
Fix. The turning point is the place of maximum $|a|$, not zero $|a|$. $v = 0$ is a velocity statement; $a = F/m$ is set by the force, which is largest exactly here.
The intuition: maximum motion should mean maximum cause of motion. If the block is screaming through equilibrium at peak speed, surely the spring must be working hardest at that moment.
The reality: speed at equilibrium is the cumulative result of the force pulling on the block all the way from the turning point. By the time the block reaches $x = 0$, the spring is at natural length and the force is zero. The work has been done. The block is coasting. Velocity and force are out of step by a quarter cycle: velocity peaks where force is zero, force peaks where velocity is zero.
Fix. Force and acceleration follow position. Velocity is what the force has built up by pulling on the block through the path so far. They cannot peak together; one always lags the other by a quarter cycle.
Ten scenarios. Pick the best chip, then check. The traps explain why each wrong answer is tempting. Progress is saved as you go.