Oscillations · Lesson 2 of 8
Simple Harmonic Motion, Phase and Reference Circles
“Describe SHM with amplitude and phase, determine initial conditions, and construct circular projections with the correct rotational sense.”
• Interpret amplitude, angular frequency, phase and the phase constant in an SHM equation. • Use both initial position and initial velocity to remove phase ambiguity. • Show why a uniform circular projection is SHM, including clockwise motion. • Classify combinations and shifted-equilibrium sinusoidal functions. • Construct reference circles and distinguish a circular path from its linear projection.
One sinusoid describes the entire motion
Imagine a particle moving along a straight line between −A and +A, with equilibrium at x = 0. It executes simple harmonic motion when its displacement is x(t) = A cos(ωt + φ), with A and ω positive constants. A is the amplitude: the magnitude of the greatest displacement from equilibrium. The full extreme-to-extreme separation is 2A. For an ideal nonzero SHM, the particle cannot move outside these limits. The motion is defined relative to equilibrium; a coordinate measured from some other origin may contain a constant offset.
| Symbol | Meaning | What it changes |
|---|---|---|
| A | Positive amplitude | Vertical size of the displacement graph |
| ω | Angular frequency | How quickly the cycle is completed |
| ωt + φ | Instantaneous phase | Where the oscillator is within its cycle |
| φ | Phase constant at the chosen t = 0 | Starting state at that time origin |
| T = 2π/ω | Period | Full repeat interval, independent of φ |
Phase is an angle that labels progress through the cycle, rather than a physical angle necessarily swept by the linearly moving body. Increasing the phase by 2π gives the same displacement and velocity. The phase constant φ is its value at the chosen time zero. Moving the stopwatch’s zero changes φ without changing the physical oscillator, its amplitude or its frequency. A graph shifted horizontally may therefore represent the same oscillation with a different time origin.
During one full cycle the phase advances by 2π. Requiring A cos[ω(t + T) + φ] = A cos(ωt + φ) for all t gives ωT = 2π and therefore ω = 2π/T = 2πf. Angular frequency is not frequency in hertz: one cycle corresponds to 2π radians of phase. A larger ω compresses the graph horizontally. A larger A stretches it vertically. Neither change should be confused with a phase shift.
| Time for x = A cos(ωt) | Position | Direction immediately after that instant |
|---|---|---|
| 0 | +A | Toward equilibrium (left) |
| T/4 | 0 | Left |
| T/2 | −A | Toward equilibrium (right) |
| 3T/4 | 0 | Right |
| T | +A | Left again |
The snapshots sample a continuous cosine graph; the particle does not jump between them. A quarter-period takes it between an extreme and the centre. The centre is passed twice per cycle with opposite velocities. In one full cycle it travels a distance 4A while its net displacement is zero. Speed is greatest at the centre and zero at each extreme; the following lesson derives these facts mathematically.
Equivalent forms and shifted equilibrium
Problem
For ω > 0 examine q₁ = sin(ωt) − cos(ωt) and q₂ = sin²(ωt). Decide whether each is harmonic and identify its period and equilibrium.
- 1.Use sin(θ − π/4) = [sin θ − cos θ]/√2. Thus q₁ = √2 sin(ωt − π/4). It is one sinusoid, with amplitude √2 and period 2π/ω.
- 2.This −π/4 is the phase constant in the sine convention. In the positive-amplitude cosine convention, the same q₁ is √2 cos(ωt − 3π/4). The numerical phase constant depends on which convention is written.
- 3.Use sin²θ = (1 − cos 2θ)/2. Therefore q₂ = 1/2 − (1/2) cos(2ωt). The coordinate oscillates about 1/2, with amplitude 1/2 and period π/ω.
- 4.Its equilibrium displacement is y = q₂ − 1/2 = −(1/2) cos(2ωt), which is SHM. The coordinate q₂ measured from zero does not obey q₂″ = −4ω²q₂; it obeys q₂″ = −4ω²(q₂ − 1/2). Always state the centre.
A sinusoidal combination at one frequency is harmonic, but the word “periodic” includes more. For instance, sin³(ωt) = [3 sin(ωt) − sin(3ωt)]/4 has a fundamental period 2π/ω but contains two frequencies, so it is not one SHM. The sum cos(ωt) + cos(3ωt) + cos(5ωt) also has period 2π/ω and is not a single sinusoid. In contrast, 3 cos(π/4 − 2ωt) is SHM because cosine is even: it is 3 cos(2ωt − π/4), with period π/ω. A Gaussian exp(−ω²t²) and the growing polynomial 1 + ωt + ω²t² are nonperiodic, rather than bounded oscillations.
A circular motion casts a simple harmonic projection
Imagine a small ball moving at constant angular speed in a horizontal circle. Viewed exactly edge-on, or projected by parallel light onto a line, its sideways coordinate moves between the two extreme edges. The ball’s actual speed around the circle stays constant, yet the projected point moves slowly near the edges and quickly near the centre. This provides a geometrical model, not an assertion that the linearly oscillating object secretly moves in a circle. A controlled rotating demonstration or a diagram is enough; swinging a loose object near people is unnecessary.
Let the reference circle have radius A, centre O and initial angle φ. For anticlockwise angular speed ω, its angle at time t is θ = ωt + φ. A right triangle gives the x-coordinate A cos θ, exactly the displacement law for SHM. Projection onto the y-axis gives A sin θ = A cos(θ − π/2). Both projections have amplitude A and period 2π/ω, while their cosine phases differ by π/2. This result holds for projection onto any fixed diameter, with an appropriate initial phase.
Problem
A radius A initially lies at 45° and rotates anticlockwise with period 4 s. A second radius B initially points along +y and rotates clockwise with period 30 s. Find both x-projections.
- 1.For the first motion θ₀ = π/4 and ω = 2π/4 = π/2 rad s⁻¹. Anticlockwise rotation increases the angle: x₁ = A cos[(π/2)t + π/4].
- 2.Initially x₁ = A/√2; a little later the first-quadrant cosine decreases. This checks the rotation sign.
- 3.For the second motion θ₀ = π/2 and ω = π/15 rad s⁻¹. Clockwise rotation gives x₂ = B cos[π/2 − (π/15)t] = B sin[(π/15)t].
- 4.The second motion can also be written B cos[(π/15)t − π/2]. It starts at x = 0 and moves toward positive x, as the clockwise reference point leaves the top of the circle toward the right.
Initial position and velocity determine the state
A given amplitude and one initial position usually allow two phases: cos φ = x₀/A has two solutions in a cycle. At that same position the particle may be moving left or right. The initial velocity removes the ambiguity. The velocity law, derived next, is v = −Aω sin(ωt + φ), so v₀ = −Aω sin φ. Squaring x₀ = A cos φ and v₀/ω = −A sin φ and adding determines A. The two signed ratios then select the quadrant of φ. Knowing energy alone determines A when the stiffness is known, but still does not select a unique phase.
Problem
At t = 0, x₀ = 1 cm and v₀ = ω cm s⁻¹, with ω = π rad s⁻¹. Find amplitude and initial phase in x = A cos(ωt + φ) and x = B sin(ωt + α).
- 1.Here v₀/ω = 1 cm, so A = √(1² + 1²) = √2 cm. The initial speed is π cm s⁻¹, not 1 cm s⁻¹.
- 2.cos φ = 1/√2 and sin φ = −1/√2. Thus φ = −π/4, equivalently 7π/4. The fourth quadrant gives positive initial velocity through the minus sign in v₀.
- 3.Since cos z = sin(z + π/2), B = √2 cm and α = φ + π/2 = π/4.
- 4.Check directly: x₀ = √2 sin(π/4) = 1 cm and v₀ = √2 ω cos(π/4) = ω cm s⁻¹. Both conventions represent the same motion, not different phases of the actual particle.
Problem
A spring oscillator has amplitude 2 cm and angular frequency 20 rad s⁻¹. Write x(t) if time zero is chosen at a centre crossing, at the right extreme, or at the left extreme.
- 1.At a centre crossing toward the right, x = 2 sin(20t) cm. A crossing toward the left would instead give −2 sin(20t) cm; the direction must be specified.
- 2.At the right extreme, x = 2 cos(20t) cm. It has zero initial velocity and then moves left.
- 3.At the left extreme, x = −2 cos(20t) cm = 2 cos(20t + π) cm. It has zero initial velocity and then moves right.
- 4.All three have amplitude 2 cm and period 2π/20 = π/10 s. Only the phase associated with the chosen starting state differs.
Problem
One circle has R = 3 cm, T = 2 s, starts at the bottom and rotates clockwise. Another has R = 2 m, T = 4 s, starts at the left edge and rotates anticlockwise. Write each x-projection.
- 1.For the first use θ₀ = −π/2 and θ = −π/2 − πt. Then x = 3 cos(−π/2 − πt) cm = −3 sin(πt) cm. Its initially negative velocity matches clockwise motion at the bottom.
- 2.For the second use θ₀ = π and ω = π/2. Then x = 2 cos(π + πt/2) m = −2 cos(πt/2) m.
- 3.The second amplitude is 2 metres, not 2 centimetres. Its initial x = −2 m is an extreme; its initial x-velocity is zero. Both unit and sign must be carried from the diagram.
| Given x (cm), t (s) | Anticlockwise cosine form | Radius; angular speed; initial angle |
|---|---|---|
| −2 sin(3t + π/3) | 2 cos(3t + 5π/6) | 2 cm; 3 rad s⁻¹; 5π/6 |
| cos(π/6 − t) | cos(t − π/6) | 1 cm; 1 rad s⁻¹; −π/6 |
| 3 sin(2πt + π/4) | 3 cos(2πt − π/4) | 3 cm; 2π rad s⁻¹; −π/4 |
| 2 cos(πt) | Already a positive-amplitude cosine | 2 cm; π rad s⁻¹; 0 |
To draw any row’s reference circle, draw x and y axes, mark a circle of the listed radius, measure the initial angle anticlockwise from +x, and mark P₀ there. Add an anticlockwise arrow labelled with the listed angular speed, then drop a perpendicular from P₀ to x. Negative angles can be replaced by angles plus 2π. The first row uses −sin z = cos(z + π/2); the second uses cos(−z) = cos z. These identities make the speed positive while preserving the physical x(t).
There is also a physical two-dimensional oscillator with force vector F = −mω²r. If its coordinates are x = A sin(ωt) and y = A cos(ωt), then x² + y² = A²: the path is a circle. At t = 0 it has position (0, A) and velocity (ωA, 0), so this example is clockwise. Equal amplitudes and a phase difference π/2 produce circular motion; other initial conditions can produce an ellipse. Circular motion can also be maintained by an inverse-square force at a suitable orbit. A circular path alone therefore does not identify the underlying force law, and a reference-circle construction does not give the real linear oscillator a rotating force.
Quick Quiz
Quiz
An oscillator moves between x = −4 cm and +4 cm. What is its amplitude?
Amplitude is maximum displacement from equilibrium, half the extreme-to-extreme separation.
A positive-amplitude cosine oscillator starts at x = 0 with positive velocity. Which phase works?
x₀ = A cos φ = 0 and v₀ = −Aω sin φ > 0 require sin φ < 0.
A reference point starts at +y and rotates clockwise. Its x-projection initially does what?
Clockwise rotation from the top moves the reference point to the right.
What is the centre and period of q = sin²(ωt)?
q = 1/2 − (1/2) cos(2ωt), so the oscillating displacement must be measured about 1/2.
Why does x₀ alone usually not determine φ when A is known?
Both the sine and cosine of phase, hence initial direction as well as position, fix the phase quadrant.
Practice Problems
- For x = 0.06 cos(5t + π/3) m, find amplitude, period, frequency, initial position and initial direction of motion. Use the signed velocity law to choose the direction.
- An oscillator has ω = 4 rad s⁻¹, x₀ = 3 cm and v₀ = −16 cm s⁻¹. Determine A and a quadrant-correct cosine phase. Express the same motion as a sine.
- Classify sin(ωt) − cos(ωt), sin³(ωt), 3 cos(π/4 − 2ωt), cos(ωt) + cos(3ωt) + cos(5ωt), exp(−ω²t²), and 1 + ωt + ω²t². State periods where they exist.
- For q = 2 + 3 sin²(ωt), identify the equilibrium coordinate, amplitude and period. Write the displacement from equilibrium as one sinusoid.
- Draw a reference circle for each: x = −2 sin(3t + π/3), x = cos(π/6 − t), x = 3 sin(2πt + π/4), and x = 2 cos(πt), with x in cm and t in s. Label radius, P₀, initial angle and positive anticlockwise speed.
- A 3 cm reference radius starts at the bottom and turns clockwise in 2 s; a 2 m radius starts at the left and turns anticlockwise in 4 s. Draw both circles and derive their x-projection equations, checking units and initial directions.
- For amplitude 2 cm and ω = 20 rad s⁻¹, give equations starting at the right extreme, left extreme and a centre crossing in each direction. Explain which parameters change.
- Show that x = A sin(ωt), y = A cos(ωt) describes a circle and satisfies F = −mω²r. Explain why this does not make the actual one-dimensional projection a circular motion.
Key Takeaways
• A is the maximum equilibrium displacement, while ω sets period and φ sets the starting state. • Always state sine or cosine convention when quoting a phase constant. • Position and velocity together select phase; a position alone usually leaves two possibilities. • A uniform circular projection is SHM even though the actual circular motion is not linear oscillation. • Constant coordinate offsets shift equilibrium, and mixed harmonics need not form a single SHM.