IB Physics HLTopic 3 — Oscillations & WavesPaper 1 & 2x, v, a equations~13 min read
Equations for Simple Harmonic Motion
So far SHM has been about ideas and graphs. Now we put real formulas to work — equations that tell you the exact displacement, velocity and acceleration of an oscillator at any moment. There are two sets, and the only thing that decides which to use is a simple question: where did the object start? Get that straight and the rest is just careful substitution.
📘 What you need to know
Which equations you use depends on the starting position at t = 0
Starting at equilibrium (x = 0): use sine for displacement — x = x0 sin ωt
Starting at the amplitude (x = x0): use cosine — x = x0 cos ωt
Velocity is largest at the centre; maximum speed = ωx0
Acceleration is largest at the extremes; maximum acceleration = ω²x0
Velocity at any displacement: v = ±ω√(x0² − x²) — no time needed
Two starting points, two sets of equations
An oscillator’s equations depend on where it is when the clock starts (t = 0). There are two standard choices, and they differ only by whether displacement follows a sine or a cosine curve.
Blue starts at equilibrium (sine); red starts at the amplitude (cosine). Same amplitude, same period — just shifted a quarter-cycle. The start point decides which formula to reach for.
Starting from equilibrium (sine equations)
If the object begins at the centre (x = 0 when t = 0), the displacement grows as a sine curve from zero. The three equations are:
Object starts at equilibriumx = x0 sin ωtv = ωx0 cos ωta = −ω²x0 sin ωt
Each one follows from the one above it — velocity is the rate of change of displacement, and acceleration is the rate of change of velocity. Notice the acceleration equation is just −ω² times the displacement equation, which is exactly the SHM rule a = −ω²x in disguise.
WE 1
An object starts oscillating from equilibrium with amplitude 0.08 m and frequency 1.5 Hz. Find its displacement 0.10 s after it is released.
Step 1 — starts at equilibrium, so x = x₀ sin(ωt)Step 2 — find ωω = 2πf = 2π × 1.5 = 9.42 rad s⁻¹Step 3 — substitute (calculator in radians!)x = 0.08 × sin(9.42 × 0.10) = 0.08 × sin(0.942)x = 0.08 × 0.809x = 0.065 m (6.5 cm)The angle ωt is in radians — sin(0.942 rad), not sin(0.942°). This is the classic trap.
Starting from the amplitude (cosine equations)
If instead the object is released from a maximum displacement (x = x0 when t = 0) — like pulling a mass down and letting go — the displacement starts at its peak and follows a cosine curve:
Object starts at the amplitudex = x0 cos ωtv = −ωx0 sin ωta = −ω²x0 cos ωt
Here’s the memory hook: sine starts at zero, cosine starts at the top. If the object is let go from rest at the edge of its swing (the most common lab setup — pull and release), it starts at maximum displacement, so use cosine. If it’s pushed off from the middle, it starts at zero, so use sine. Match the graph’s starting height to sin(0)=0 or cos(0)=1.
Maximum speed and maximum acceleration
The equations reveal two very useful “peak” values. Since sine and cosine never exceed 1, the biggest that velocity and acceleration can get are:
Maximum valuesmaximum speed: vmax = ωx0maximum acceleration: amax = ω²x0
The maximum speed happens at the centre (where the object rushes through equilibrium), and the maximum acceleration happens at the extremes (where it’s flung back hardest). These pop up constantly in exam questions.
WE 2
A mass oscillates in SHM with amplitude 0.05 m and time period 0.40 s. Calculate its maximum speed and maximum acceleration.
Step 1 — find ω from the periodω = 2π/T = 2π/0.40 = 15.71 rad s⁻¹Step 2 — maximum speed = ωx₀v_max = 15.71 × 0.05 = 0.785 m s⁻¹Step 3 — maximum acceleration = ω²x₀a_max = (15.71)² × 0.05 = 12.3 m s⁻²v_max = 0.79 m s⁻¹, a_max = 12.3 m s⁻²Both use the same ω — find it once from the period, then reuse it.
Velocity without time: the v–x relation
Sometimes you know where the object is but not when. There’s a neat equation that gives velocity directly from displacement, with no time involved:
Velocity–displacement relationv = ±ω√(x0² − x²)
The ± sign is there because the object passes each point twice — once going each way — so the velocity can be positive or negative. Reading the equation: at the centre (x = 0) the speed is maximum (ωx0); at the extremes (x = x0) the square root is zero, so the object is momentarily still.
Plotting v against x gives an ellipse: fastest at the centre (x = 0), stationary at the extremes (x = ±x0). Two branches because the object passes each point both ways.
WE 3
An oscillator has amplitude 0.12 m and angular frequency 8.0 rad s⁻¹. Find its speed when it is 0.06 m from equilibrium.
Step 1 — use v = ω√(x₀² − x²)Step 2 — substitute x₀ = 0.12 m, x = 0.06 mv = 8.0 × √(0.12² − 0.06²) = 8.0√(0.0144 − 0.0036)v = 8.0 × √0.0108 = 8.0 × 0.104v = 0.83 m s⁻¹At half the amplitude the speed is still 87% of the maximum (0.96 m s⁻¹) — the object is fast for most of its swing.
The equations summarised
Quantity
Starts at equilibrium
Starts at amplitude
Displacement
x = x0 sin ωt
x = x0 cos ωt
Velocity
v = ωx0 cos ωt
v = −ωx0 sin ωt
Acceleration
a = −ω²x0 sin ωt
a = −ω²x0 cos ωt
Velocity at any x
v = ±ω√(x0² − x²)
Displacement x
rate of change (differentiate)
Velocity v
rate of change (differentiate)
Acceleration a
💡 Top tips
Sine starts at 0, cosine starts at the top. Match the equation to where the object begins.
Calculator in radians whenever you use sin ωt or cos ωt — ωt is in radians.
vmax = ωx0 at the centre; amax = ω²x0 at the extremes.
Use v = ±ω√(x0² − x²) when a question gives you displacement but not time.
Find ω first (from f or T) — almost every equation needs it.
⚠ Common mistakes
Using sine equations when the object starts at the amplitude (or vice versa) — check the start point
Leaving the calculator in degrees — ωt must be evaluated in radians
Forgetting the ± in the velocity–displacement relation
Squaring only part of the bracket — it’s x0² − x², both squared
Mixing up where max speed and max acceleration occur — speed peaks at the centre, acceleration at the extremes
Quick recap: Which SHM equations you use depends on the start. From equilibrium: x = x0 sin ωt (and its v, a); from the amplitude: x = x0 cos ωt. Maximum speed ωx0 is at the centre, maximum acceleration ω²x0 is at the extremes, and v = ±ω√(x0² − x²) gives velocity from displacement without needing time.
These equations all assumed the object starts neatly at x = 0 or x = x0. But what if it starts somewhere in between? Then we need a phase angle to shift the curve along. That’s the final piece of the SHM toolkit, and it’s the next page: phase angles in SHM.
Want the SHM equations to feel automatic?
Book a free meeting and we’ll drill the sine/cosine choice, the v–x relation and past-paper questions together.
