Phase Relationships & the Three SHM Graphs
- Relate the displacement, velocity and acceleration graphs of an oscillator by their quarter-cycle phase shifts
- Locate where speed and acceleration are maximum and where each is zero
- Connect the restoring-force condition F = −kx to the sinusoidal shape of the motion
What makes motion simple harmonic
An oscillation is simple harmonic when the restoring force is proportional to the displacement and directed back toward equilibrium: F = −kx. That single condition is what produces a sinusoidal graph, a period independent of amplitude, and every other property of the unit. If the restoring force grows faster or slower than linearly with displacement — a large-angle pendulum, a bouncing ball — the motion is periodic but not simple harmonic, and none of the SHM formulas apply.
Acceleration is the displacement graph flipped
Because a = −(k/m)x, the acceleration graph is the displacement graph inverted and scaled — they are 180° out of phase. Where displacement is at its positive maximum, acceleration is at its negative maximum. Where displacement is zero, acceleration is zero. This is the easiest of the three relationships to see and the one most often stated backward.
Velocity leads displacement by a quarter cycle
Velocity is the slope of the displacement graph, and the slope of a sine is a cosine — so the velocity graph is a quarter period (90°) ahead of the displacement. Concretely: at the extremes, x is maximum, v is zero, and a is maximum; at equilibrium, x is zero, v is maximum, and a is zero. Speed and acceleration never peak at the same moment. That alternation is the whole story of the three graphs, and it is worth being able to say without drawing them.
Energy trades at twice the frequency
Total mechanical energy is constant at E = ½kA², set entirely by the amplitude. It sloshes between elastic potential (maximum at the extremes) and kinetic (maximum at equilibrium). A detail the exam likes: the energy exchange happens twice per cycle, since the oscillator passes through equilibrium twice and reaches an extreme twice. So an energy-versus-time graph oscillates at double the frequency of the displacement graph.
A 0.50 kg mass on a spring with k = 200 N/m oscillates with amplitude 0.10 m. Find the maximum speed and the maximum acceleration.
- 1.Angular frequency: ω = √(k/m) = √(200 ÷ 0.50) = √400 = 20 rad/s.
- 2.Maximum speed at equilibrium: v_max = ωA = (20)(0.10) = 2.0 m/s.
- 3.Maximum acceleration at the extremes: a_max = ω²A = (400)(0.10) = 40 m/s².
- 4.Cross-check with energy: ½kA² = ½(200)(0.10)² = 1.0 J = ½mv² gives v = √(2 × 1.0 ÷ 0.50) = 2.0 m/s. Consistent.
At the extreme of an oscillation the mass is momentarily at rest but its acceleration is at its largest. "Momentarily at rest, therefore zero net force" is wrong — it is precisely the large restoring force there that turns the motion around.
At the equilibrium position of a mass on a spring, which is true?
If the amplitude of an oscillating spring–mass system is doubled, the maximum speed:
The displacement of an oscillator is graphed as a sine curve. The acceleration graph is:
Graph questions often supply one curve and ask you to sketch another. Mark the moments where the given curve is zero and where it peaks, then use the rule that peaks in one graph line up with zeros in the next. Sketching from those anchor points is far more reliable than trying to draw a sinusoid freehand.
Answer the 3 checkpoints as you read.
Sign in to save your progress