5.3.1 - Simple harmonic motion

Defining SHM

An oscillation is repeated motion about an equilibrium position. For SHM, the word "simple" has a strict meaning: the acceleration is always towards equilibrium and its magnitude is directly proportional to the displacement from equilibrium.

Take displacement $x$ as positive to the right of equilibrium. If the oscillator is to the right, $x$ is positive, so the acceleration must be to the left and therefore negative. If the oscillator is to the left, $x$ is negative, so the acceleration must be to the right and therefore positive. That sign reversal is the heart of the model.

Here $a$ is acceleration in $\mathrm{m}\,\mathrm{s}^{-2}$, $x$ is displacement from equilibrium in $\mathrm{m}$, and $\omega$ is angular frequency in $\mathrm{rad}\,\mathrm{s}^{-1}$ or $\mathrm{s}^{-1}$. The negative sign does not mean "slowing down"; it means the acceleration is in the opposite direction to the displacement.

A useful OCR test is: can the acceleration be written as "a constant multiplied by $-x$"? If yes, the motion satisfies the SHM defining equation, and that constant is $\omega^2$.

Worked example: a glider has acceleration

where $a$ is in $\mathrm{m}\,\mathrm{s}^{-2}$ and $x$ is in $\mathrm{m}$.

Compare this with $a=-\omega^2x$, so

and

The equation also tells you the acceleration direction. At $x=+0.050\,\mathrm{m}$, the acceleration is

The glider is to the positive side of equilibrium, but its acceleration is negative, back towards equilibrium.

[DIAGRAM: asset_slug: 5_03_1_simple_harmonic_motion_kinematics__diagram_01; description: In simple harmonic motion, acceleration is proportional to displacement and directed towards equilibrium.; caption: In simple harmonic motion, acceleration is proportional to displacement and directed towards equilibrium.]

Timing and phase

SHM repeats, so OCR expects you to handle the quantities that describe repeated motion.

Angular frequency is the rate of phase change. One full cycle is $2\pi$ radians, so one period corresponds to $2\pi$ radians of phase.

Phase difference tells you how far apart two oscillations, or two points in the same oscillation, are within the cycle. A quarter-cycle phase difference is $\pi/2$ rad; a half-cycle phase difference is $\pi$ rad. If two identical oscillators have the same $T$ but one reaches a maximum $0.20T$ later than the other, their phase difference is

For PAG10-style period measurement, the practical aim is usually to measure $T$ or $f$ accurately for a mass-spring oscillator or a pendulum. The common good practice is to time many complete oscillations and divide by the number timed:

For a mass on a spring, a student might use a fiducial marker at the equilibrium position and start/stop the timer each time the mass passes the marker in the same direction. For a pendulum, the amplitude should be small, and the bob should pass a fixed marker cleanly. A ruler, set square or plumb line helps measure displacement or amplitude without parallax. Repeating the timing, using a digital timer, light gate, motion sensor or data logger where available, and timing 10 or 20 oscillations reduces the percentage uncertainty in $T$.

Worked example: a student times 20 oscillations of a mass-spring system. The time is $16.8\,\mathrm{s}$.

The final answers are sensible to 3 significant figures because the timing is given to 3 significant figures.

[DIAGRAM: asset_slug: 5_03_1_simple_harmonic_motion_kinematics__diagram_03; description: Time many spring or pendulum oscillations past a fixed reference and divide by the number of complete cycles.; caption: Time several spring or pendulum oscillations past a fixed reference to determine the period.]

Sinusoidal displacement

The defining equation $a=-\omega^2x$ has sinusoidal solutions. OCR gives the common forms:

The two forms describe the same kind of motion. The difference is the choice of when $t=0$ is started.

If the oscillator starts at maximum positive displacement and is momentarily at rest, then $x=A\cos(\omega t)$ is convenient because $\cos(0)=1$. If the oscillator starts at equilibrium and moves in the positive direction, then $x=A\sin(\omega t)$ is convenient because $\sin(0)=0$ and then increases.

The argument $\omega t$ is a phase angle, so it is measured in radians. That is why $\omega$ must have units of per second: multiplying by time gives a dimensionless angle in radians.

Worked example: an oscillator has amplitude $0.060\,\mathrm{m}$ and period $2.0\,\mathrm{s}$. It starts at maximum positive displacement, so

First find angular frequency:

At $t=0.50\,\mathrm{s}$,

The phase is exactly $\pi/2$ because this is a quarter of the period.

After a quarter of a period, the oscillator has moved from maximum displacement to equilibrium.