C.1 Simple harmonic motion

UNIT SYLLABUS

SL/HL 3 hours SL + 4 hours HL

Simple harmonic motion is nature's default oscillation: whenever a system is disturbed and pulled back by a force proportional to displacement, you get SHM. The defining equation $a = -\omega^2 x$ says it all — acceleration always points home and grows with distance. Two model systems dominate: the mass–spring ($T = 2\pi\sqrt{m/k}$, gravity-independent) and the pendulum ($T = 2\pi\sqrt{l/g}$, mass-independent). At HL you add the full time-dependent machinery — sine solutions, phase angle, and energy sloshing between kinetic and potential twice per cycle. SHM is also the gateway to waves: every wave is SHM passed from particle to particle.

Guiding Questions

? What distinguishes simple harmonic motion from other kinds of oscillation?
? How do the energy transformations within an oscillating system evolve over one cycle?

Displacement, velocity and acceleration in SHM: v leads x by a quarter cycle, and a is exactly out of phase with x (a = −ω²x).

✓ What the IB expects you to master

State the conditions for SHM: restoring force (and hence acceleration) proportional to displacement and directed towards equilibrium.
Apply the defining equation $a = -\omega^2 x$ .
Describe oscillations with period $T$ , frequency $f$ , angular frequency $\omega = 2\pi/T$ , amplitude and displacement.
Use $T = 2\pi\sqrt{m/k}$ for a mass–spring system and $T = 2\pi\sqrt{l/g}$ for a simple pendulum.
Describe the kinetic–potential energy interchange over one cycle qualitatively (SL).
HL: describe motion with phase angle using $x = x_0\sin(\omega t + \phi)$ and $v = \omega x_0\cos(\omega t + \phi)$ .
HL: use $v = \pm\omega\sqrt{x_0^2 - x^2}$ , total energy $E_T = \frac{1}{2}m\omega^2 x_0^2$ and potential energy $E_p = \frac{1}{2}m\omega^2 x^2$ .

1 Key Formulas

Defining equation of SHM

a = -\omega^{2}x

Period and frequency

T = \frac{1}{f} = \frac{2\pi}{\omega}

Mass–spring period

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

Simple pendulum period

T = 2\pi\sqrt{\frac{l}{g}}

Displacement (HL)

x = x_{0}\sin(\omega t + \phi)

Velocity (HL)

v = \omega x_{0}\cos(\omega t + \phi)

Velocity–displacement (HL)

v = \pm\,\omega\sqrt{x_{0}^{2} - x^{2}}

Total energy (HL)

E_{T} = \tfrac{1}{2}m\omega^{2}x_{0}^{2}

Potential energy (HL)

E_{p} = \tfrac{1}{2}m\omega^{2}x^{2}

2 Exam Preparation & Topic Explanations

The SHM graph triangle

Displacement, velocity and acceleration in SHM are three sinusoids locked in phase relationships: $v$ leads $x$ by 90°, and $a$ is inverted relative to $x$ . Given any one graph you can construct the other two — and exams test exactly that.

Energy graphs are parabolas against displacement: $E_p$ opens upward from zero at equilibrium, $E_k$ is its mirror, and their sum is the flat line $E_T$ .

Pro Exam Strategy

Check the phase: if $x = x_0\sin\omega t$ , then $v = \omega x_0\cos\omega t$ — differentiate, don't memorise blindly.
The pendulum formula fails for large amplitudes (>~10°) — the restoring force is only approximately proportional to displacement.
Frequency of the ENERGY oscillation is twice the motion frequency — kinetic energy peaks twice per cycle.
Angular frequency $\omega$ is in rad s $^{-1}$ ; forgetting to convert from Hz costs a factor of $2\pi$ .

3 MCQ Practice

Q1. A particle performs SHM. Its acceleration is greatest when:

Its speed is greatest
It passes through the equilibrium position
Its displacement is greatest
Its kinetic energy is greatest

Q2. A pendulum clock is taken from Earth to the Moon ( $g_{Moon} \approx g/6$ ). Its period will:

Decrease by a factor of 6
Increase by a factor of 6
Increase by a factor of $\sqrt{6}$
Stay the same

Q3. In SHM, at what displacement is the kinetic energy equal to the potential energy? ( $x_0$ = amplitude)

$x_0/4$
$x_0/2$
$x_0/\sqrt{2}$
$x_0$

4 Short Answer Questions

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