Simple Harmonic Motion

Many systems oscillate — a mass on a spring, a swinging pendulum, a vibrating guitar string, atoms in a solid. Simple harmonic motion (SHM) is the special, mathematically clean kind of oscillation that all of these approximate, and it is defined by a single condition on the acceleration:

$a = -\omega^2 x$

In words, an object moves with SHM if its acceleration is:

1. directly proportional to its displacement $x$ from a fixed equilibrium position ($a\propto x$), and
2. always directed towards that equilibrium position — opposite in direction to the displacement.

The minus sign carries condition 2. When the object is displaced to the right ($x>0$), the acceleration is to the left (negative); when displaced left, the acceleration is to the right. The acceleration always pulls the object back towards the centre, which is why the motion oscillates. The constant of proportionality is written $\omega^2$ (always positive), where $\omega$ is the angular frequency:

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

with $f$ the frequency (in Hz) and $T$ the period (in s). $\omega$ is measured in $\text{rad s}^{-1}$, so every SHM calculation is done in radians — keep your calculator in radian mode.

The physical cause is a restoring force that is proportional to displacement. For a spring, Hooke's law $F=-kx$ gives exactly this, and combining it with $F=ma$ shows $a=-\dfrac{k}{m}x$, so $\omega^2=\dfrac{k}{m}$. Any system with a linear restoring force oscillates with SHM.

Solving $a=-\omega^2 x$ gives a sinusoidal displacement. Which trig function you use depends on where the object starts the clock:

• Timing from maximum displacement (the amplitude): $x=A\cos(\omega t)$.
• Timing from the equilibrium position: $x=A\sin(\omega t)$.

Here $A$ is the amplitude — the maximum displacement. Differentiating $x=A\cos(\omega t)$ once gives the velocity, and again gives the acceleration:

$x = A\cos(\omega t)$ $v = -A\omega\sin(\omega t)$ $a = -A\omega^2\cos(\omega t) = -\omega^2 x$

The last line recovers the defining equation — a useful check. From these, the maximum values are read off directly:

• Maximum speed $v_{max}=\omega A$ (when $\sin=\pm 1$), occurring at the equilibrium position.
• Maximum acceleration $a_{max}=\omega^2 A$ (when $\cos=\pm 1$), occurring at the amplitude.

Phase relationships. All three quantities oscillate at the same angular frequency $\omega$, but they are shifted:

• The velocity leads the displacement by $\tfrac{\pi}{2}$ (a quarter cycle, $90°$): when $x=0$ (at equilibrium) the speed is greatest; when $x=\pm A$ the speed is zero.
• The acceleration is in antiphase with the displacement (a phase difference of $\pi$, $180°$): the minus sign means whenever $x$ is positive, $a$ is negative.

You often need the speed at a particular displacement without knowing the time. The relation is:

$v = \pm\omega\sqrt{A^2 - x^2}$

This follows from energy conservation (kinetic + potential = constant) and captures the whole shape of the motion:

• At equilibrium ($x=0$): $v=\pm\omega\sqrt{A^2}=\pm\omega A$ — the maximum speed.
• At the amplitude ($x=\pm A$): $v=\pm\omega\sqrt{A^2-A^2}=0$ — momentarily at rest before turning back.
• In between, the speed falls smoothly as $x$ increases.

The $\pm$ reminds you that the object passes each point once moving each way; usually you quote the magnitude (the speed).

The single biggest arithmetic trap is treating $\sqrt{A^2-x^2}$ as $A-x$. The square root does not distribute over subtraction. Always square $A$ and $x$ first, subtract, then take the root. For $A=0.12\ \text{m}$ and $x=0.060\ \text{m}$: $\sqrt{A^2-x^2}=\sqrt{0.0144-0.0036}=\sqrt{0.0108}=0.104\ \text{m}\quad(\text{not }0.060\ \text{m}).$

Notice that at $x=\tfrac{A}{2}$ the speed is $\omega\sqrt{A^2-A^2/4}=\omega A\sqrt{3}/2\approx 0.87\,v_{max}$ — halfway out in displacement the object is still moving at 87% of its top speed, because speed falls off slowly near the centre and fast near the ends.

Two standard SHM systems appear again and again, each with its own period formula.

Mass–spring system. A mass $m$ on a spring of spring constant $k$ oscillates with period: $T = 2\pi\sqrt{\frac{m}{k}}$ A heavier mass oscillates more slowly (larger $T$); a stiffer spring (larger $k$) oscillates faster. This works for a horizontal spring or a vertical one — for a vertical spring gravity just shifts the equilibrium position, it does not change the period. Rearranged, this formula lets you find an unknown mass ($m=\dfrac{kT^2}{4\pi^2}$) or an unknown spring constant ($k=\dfrac{4\pi^2 m}{T^2}$).

Simple pendulum. A small bob of mass on a light string of length $l$ swings with period: $T = 2\pi\sqrt{\frac{l}{g}}$ Two striking features: the period depends on the length and $g$ only — not on the mass of the bob — and the formula is valid only for small angular amplitudes (below about $10°$), where the restoring force is approximately proportional to displacement. Beyond that the motion is still oscillatory but no longer simple harmonic. Because $T$ depends on $g$, a pendulum is a neat way to measure $g$: rearranging gives $g=\dfrac{4\pi^2 l}{T^2}$.

Both periods are independent of amplitude — this is the isochronous property that made pendulums the basis of accurate clocks. Increasing the amplitude makes the object travel further each cycle, but $v_{max}=\omega A$ shows it also moves proportionally faster, so the time per cycle is unchanged.

As an object oscillates it continuously interchanges kinetic energy (KE) and potential energy (PE), but (with no damping) their sum — the total energy — is constant.

• At the equilibrium position the object moves fastest, so KE is a maximum and PE is zero. This maximum kinetic energy equals the total energy of the oscillation: $E_{k,max} = \tfrac{1}{2}m v_{max}^2 = \tfrac{1}{2}m\omega^2 A^2$
• At the amplitude the object is momentarily at rest, so KE is zero and PE is a maximum (equal to the total energy).
• In between, KE and PE share the total. At displacement $x$: $E_k = \tfrac{1}{2}m\omega^2(A^2 - x^2), \qquad E_p = \tfrac{1}{2}m\omega^2 x^2.$

Because the total energy is $E=\tfrac{1}{2}m\omega^2 A^2$, it is proportional to the square of the amplitude, $E\propto A^2$. Doubling the amplitude quadruples the energy.

If the oscillation is damped, energy is gradually removed (usually as heat), so the amplitude decays and, because $E\propto A^2$, the energy falls with the square of the shrinking amplitude.

Core Practical 16 uses SHM to determine an unknown mass or a spring constant from the period of a mass–spring oscillator.

Method. Hang a known mass $m$ from a vertical spring, pull it down a small distance and release it so it oscillates vertically in SHM. Time a large number of oscillations (say 20) with a stopwatch and divide to get the period $T$. Repeat for a range of masses.

Linearising the equation. Starting from $T=2\pi\sqrt{\dfrac{m}{k}}$ and squaring both sides removes the awkward square root: $T^2 = \frac{4\pi^2}{k}\,m$ Comparing with $y=(\text{gradient})x$ (here $y=T^2$, $x=m$), a graph of $T^2$ (y-axis) against $m$ (x-axis) is a straight line through the origin whose gradient is $\dfrac{4\pi^2}{k}$. Therefore: $\boxed{k = \frac{4\pi^2}{\text{gradient}}}$ Reading the gradient from a large triangle and substituting gives the spring constant. (To find an unknown mass instead, read its $T^2$ value off the same line, or use $m=\dfrac{kT^2}{4\pi^2}$ once $k$ is known.)

Reducing and judging uncertainty.

• The dominant random error is the timing (reaction time). Time many oscillations (20) and divide, and repeat each reading and take a mean.
• Use a fiducial marker at the equilibrium position (where the mass moves fastest) to start and stop timing consistently.
• Keep the amplitude small so the motion stays simple harmonic, and avoid overloading the spring beyond its limit of proportionality.
• A line through the origin confirms $T^2\propto m$; a small positive intercept usually reveals the effective mass of the spring itself contributing to the oscillation.