Simple harmonic motion

Lots of things oscillate — a mass bobbing on a spring, a child on a swing, a guitar string, atoms in a solid. Simple harmonic motion (SHM) is the cleanest, most important kind of oscillation, and it appears whenever a system obeys two conditions:

1. There is a restoring force that always acts to push (or pull) the object back towards a fixed equilibrium position.
2. That force — and therefore the acceleration — is directly proportional to the displacement from equilibrium.

Put those together in words: the further you pull the object from the middle, the harder it is pushed back, and the push is always aimed at the middle.

A mass on a spring is the model example. At equilibrium the spring is neither stretched nor compressed and the resultant force is zero. Pull the mass down and the stretched spring pulls it back up; push it up and the compressed spring pushes it back down. The force (via Hooke's law) grows in proportion to how far you have displaced it — exactly the SHM condition.

A simple pendulum (a small heavy bob on a light string) also performs SHM, but only for small swings (small angles, roughly under about 10°). For small angles the sideways restoring force is very nearly proportional to the displacement along the arc, so the motion is simple harmonic. For large swings this proportionality breaks down and the motion is only approximately SHM.

The quantities you will use throughout are the amplitude $A$ (the maximum displacement from equilibrium), the period $T$ (the time for one complete oscillation) and the frequency $f$ (oscillations per second, $f = 1/T$).

The two conditions of the previous section are captured in one compact equation, provided in the data booklet:

$a = -\omega^2 x$

Here $a$ is the acceleration, $x$ is the displacement from equilibrium, and $\omega$ (omega) is a positive constant called the angular frequency.

Read the equation carefully — every part earns marks:

• $\omega^2$ is always positive, so the size of the acceleration is proportional to the size of the displacement: $|a| = \omega^2 |x|$. Twice as far from the middle → twice the acceleration.
• The minus sign tells you the acceleration is always in the opposite direction to the displacement. When the object is displaced to the right (positive $x$), the acceleration is to the left (negative). When it is at the extreme and momentarily at rest, the acceleration is at its maximum and points straight back towards equilibrium. This minus sign is the "restoring" condition written mathematically.
• At $x = 0$ (equilibrium) the acceleration is zero — but the object is moving fastest here, so it overshoots and carries on to the other side.

Because $|a| = \omega^2|x|$, the acceleration is greatest at the amplitude ($x = \pm A$), where $a_{\text{max}} = \omega^2 A$, and zero at equilibrium.

Angular frequency connects the "how fast it oscillates" ideas together:

$\omega = \frac{2\pi}{T} = 2\pi f$

It is measured in radians per second (rad s⁻¹). One complete oscillation corresponds to $2\pi$ radians, so a shorter period (faster oscillation) means a larger $\omega$.

If you are ever asked to prove something undergoes SHM, the task is to show its acceleration can be written in the form $a = -(\text{positive constant})\times x$. Whatever that positive constant turns out to be, it equals $\omega^2$.

You are not asked to reproduce a full calculus derivation at SL, but seeing where the period formulae come from makes them impossible to misremember — and it shows the deep idea that every SHM period is set by the same competition between an "inertia" quantity and a "stiffness" quantity.

Mass–spring: reasoning from Newton's second law. For a mass $m$ on a spring of spring constant $k$, Hooke's law gives a restoring force $F = -kx$ (the minus sign again: force opposes displacement). Newton's second law says $F = ma$, so:

$ma = -kx \quad\Rightarrow\quad a = -\frac{k}{m}\,x$

Compare this directly with the defining equation $a = -\omega^2 x$. The two are identical if

$\omega^2 = \frac{k}{m} \quad\Rightarrow\quad \omega = \sqrt{\frac{k}{m}}$

Now use $\omega = \dfrac{2\pi}{T}$ and rearrange for the period:

$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}$

That is the mass–spring period equation from the data booklet — and you have just seen why a bigger mass (more inertia) lengthens the period while a stiffer spring (bigger $k$) shortens it.

Simple pendulum: the same logic. For a bob of mass $m$ on a string of length $L$, the sideways restoring force for a small angle works out (using the geometry of the swing) to be proportional to the displacement, giving

$a = -\frac{g}{L}\,x$

Matching to $a = -\omega^2 x$ gives $\omega^2 = \dfrac{g}{L}$, so $\omega = \sqrt{\dfrac{g}{L}}$ and therefore

$T = 2\pi\sqrt{\frac{L}{g}}$

The punchline for the pendulum: the mass $m$ cancels out — it never appears in the period. That is why the period of a simple pendulum depends only on its length $L$ and the local gravitational field strength $g$, and not on the mass of the bob or (for small swings) the amplitude. This is one of the most tested facts in the whole topic.

The universal pattern. Every SHM period has the form $T = 2\pi\sqrt{\dfrac{\text{inertia}}{\text{stiffness}}}$. For the spring it is mass over spring constant; for the pendulum it is (effectively) length over $g$. Recognising this pattern means you can never mix up which quantity goes on top.

Both period equations are in the data booklet, so the marks come from choosing the right one, substituting correctly (watch the square root) and interpreting the result.

Mass–spring system $T = 2\pi\sqrt{\frac{m}{k}}$

• $m$ = oscillating mass (kg), $k$ = spring constant / force constant (N m⁻¹).
• A larger mass → longer period (more inertia, slower to turn around).
• A stiffer spring (larger $k$) → shorter period (stronger restoring force).
• Notice the square root: to double the period you must quadruple the mass, not just double it.

Simple pendulum (small swings) $T = 2\pi\sqrt{\frac{L}{g}}$

• $L$ = length from pivot to the centre of the bob (m), $g$ = gravitational field strength (m s⁻² or N kg⁻¹).
• Longer string → longer period. To double the period, make the pendulum four times as long.
• Higher $g$ → shorter period (a pendulum runs faster where gravity is stronger, e.g. it swings more slowly on the Moon where $g \approx 1.6\,\text{m s}^{-2}$).
• The period does NOT depend on the mass of the bob or the amplitude (for small angles). Two pendulums of the same length keep the same time whether their bobs are light or heavy.

Rearranging is a common exam move. Because these equations often appear as "determine $k$" or "determine $g$", get comfortable squaring both sides to remove the root. For the pendulum, squaring gives $T^2 = 4\pi^2\dfrac{L}{g}$, so

$g = \frac{4\pi^2 L}{T^2}$

This is the basis of the classic "measure $g$ with a pendulum" experiment: time many oscillations to get an accurate $T$, then plot $T^2$ against $L$ and use the gradient $\left(=\dfrac{4\pi^2}{g}\right)$.

An oscillator continually converts energy between two forms without (in the ideal case) losing any:

• Kinetic energy (KE) — energy of the movement of the mass.
• Potential energy (PE) — energy stored because of the displacement (elastic PE in a spring, gravitational PE for a pendulum).

The key picture:

• At the amplitude ($x = \pm A$) the object is momentarily at rest ($v = 0$), so KE = 0 and PE is at its maximum. All the energy is stored.
• At the equilibrium position ($x = 0$) the object is moving at its maximum speed, so KE is at its maximum and PE = 0 (taking equilibrium as the zero of PE). All the energy is kinetic.
• Everywhere in between, the energy is part kinetic and part potential.

The total energy stays constant (energy is conserved when there is no friction or drag):

$E_{\text{total}} = \text{KE} + \text{PE} = \text{constant}$

So as the object moves from the amplitude towards the middle, PE falls and KE rises by exactly the same amount; moving from the middle back out to the amplitude, KE falls and PE rises. Energy sloshes back and forth twice per oscillation (KE peaks each time the object passes through the middle, going either way).

At SL you describe this interchange qualitatively and graphically — you are not expected to use algebraic energy formulae. If you ever need the maximum kinetic energy numerically, use the ordinary $\text{KE} = \tfrac{1}{2}mv^2$ with the maximum speed (the speed at the equilibrium position); that maximum KE equals the total energy of the oscillation.

Sketching or reading the three time-graphs of an oscillator is examined in every session. Take an object released from its maximum displacement at $t = 0$, so the displacement–time graph is a cosine-shaped curve starting at $+A$.

Displacement (x–t): a smooth wave starting at the amplitude $+A$, falling through zero at equilibrium, down to $-A$, and back — one full wave per period $T$.

Velocity (v–t): the velocity is the gradient of the displacement graph.

• At the amplitude ($x = \pm A$) the displacement graph is momentarily flat (turning point), so the velocity is zero.
• At equilibrium ($x = 0$) the displacement graph is steepest, so the velocity has its maximum magnitude.
• The velocity curve therefore leads the displacement curve by a quarter of a period (a phase difference of 90°). When $x$ is at a maximum, $v$ is zero; a quarter-period earlier, $v$ was at its peak.

Acceleration (a–t): from the defining equation $a = -\omega^2 x$, the acceleration is just the displacement "flipped and scaled".

• Wherever the displacement is at $+A$, the acceleration is at its most negative ($-\omega^2 A$), and vice versa.
• So the acceleration is exactly antiphase with the displacement — a phase difference of 180° (half a period). The two curves are perfect mirror images through the time axis.

A quick way to check any point: pick a moment and reason physically. At the extreme of the swing the object is stationary (v = 0) but being pushed back hardest (a maximum, opposite to x). As it races through the middle it is fastest (v maximum) but weightless of net force (a = 0). Those two checkpoints alone let you place all three curves correctly.