Oscillations: SHM, damping and resonance
Key idea: Connect measured periodic motion to the defining SHM model, its phase and energy relationships, then distinguish damping from driven resonance.
Before you start: Circular Motion objective chainEnergy & Fields objective chain
By the end, you can
- Describe and investigate free oscillations, and use amplitude, period, frequency, angular frequency, phase and phase difference.
- Recognise SHM from a = −ω²x and use its displacement, velocity, acceleration and graphical relationships.
- Describe kinetic–potential energy interchange and compare light, critical and heavy damping in practical systems.
- Interpret forced-response curves, damping effects and useful or harmful resonance.
Starting-point self-check
1. Check your starting point
Attempt all four groups without notes and mark the first definition, phase sign, energy location or response-curve feature you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
SHM definition, equations, graphs and phase 9(d)–(g)
Question 1
An oscillator follows x = 0.080 sin(5.0t) in SI units. At t = 0.20 s, calculate x, v and a and state how the acceleration demonstrates SHM.
Check the model response
Here ωt = 1.00 rad. x = 0.0673 m, v = 0.080(5.0)cos1.00 = 0.216 m s⁻¹ and a = −ω²x = −1.68 m s⁻². Acceleration is proportional and opposite to displacement, so it points towards equilibrium.
repair
2. Repair the eight common breaks
Use only the repair matching an error, then redraw the equilibrium line, phase cycle, energy bars or response axes before retrying.
SHM definition, equations, graphs and phase 9(d)–(g)
Check this idea
Misconception: Every periodic motion is simple harmonic motion.
Repair: SHM specifically requires a = −ω²x: acceleration must be proportional to displacement and directed towards equilibrium.
Check this idea
Misconception: Velocity and acceleration are both greatest at equilibrium.
Repair: Speed is greatest at equilibrium where acceleration is zero. Acceleration magnitude is greatest at the extreme displacements where velocity is zero.
worked example
3. Follow four worked models
Follow how each solution fixes the timing reference, equilibrium direction, energy boundary and driving frequency before calculating.
SHM definition, equations, graphs and phase 9(d)–(g)
Model 1
A particle performs SHM with amplitude 0.050 m and ω = 6.0 rad s⁻¹. At x = +0.030 m while moving towards equilibrium, find acceleration and velocity.
Check the model response
a = −ω²x = −36(0.030) = −1.08 m s⁻². The speed is ω√(A² − x²) = 6.0√(0.050² − 0.030²) = 0.240 m s⁻¹. Motion towards equilibrium from positive x makes v = −0.240 m s⁻¹.
guided practice
4. Guided practice
Use each hint only to select a time-to-phase conversion, SHM equation, energy location or response-curve comparison.
SHM definition, equations, graphs and phase 9(d)–(g)
Question 1
For x = 0.040 sin(10t), find maximum speed, acceleration at x = −0.020 m and the first time after t = 0 at which x is maximum.
Hint: Use vmax = ωA, a = −ω²x and set sinωt = 1.
Check the model response
vmax = 10(0.040) = 0.400 m s⁻¹. At x = −0.020 m, a = −100(−0.020) = +2.00 m s⁻². The first maximum occurs when 10t = π/2, so t = 0.157 s.
independent practice
5. Independent practice
Solve without repair notes and state the equilibrium reference, phase convention and energy or driving assumptions used.
SHM definition, equations, graphs and phase 9(d)–(g)
Question 1
At x = 0.025 m, an oscillator has a = −0.900 m s⁻². If its amplitude is 0.060 m, find ω, maximum speed and speed at that position.
Check the model response
ω = √(|a|/|x|) = √(0.900/0.025) = 6.00 rad s⁻¹. vmax = ωA = 0.360 m s⁻¹ and |v| = 6.00√(0.060² − 0.025²) = 0.327 m s⁻¹.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
SHM definition, equations, graphs and phase 9(d)–(g)
Question 1
Show that motion with x = 0.030 sin(8.0t) is SHM. Find its maximum speed and the speed and acceleration at x = +0.018 m while moving away from equilibrium.
Check the model response
Differentiating twice gives a = −8.0²x = −64x, the SHM definition. vmax = ωA = 0.240 m s⁻¹. |v| = 8.0√(0.030² − 0.018²) = 0.192 m s⁻¹; moving away from positive equilibrium makes v positive. a = −64(0.018) = −1.15 m s⁻².
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
SHM definition, equations, graphs and phase 9(d)–(g)
Question 1
An SHM particle has amplitude 0.070 m and maximum speed 0.56 m s⁻¹. Find ω, acceleration at x = −0.035 m and both possible velocities there.
Check the model response
ω = vmax/A = 0.56/0.070 = 8.0 rad s⁻¹. a = −ω²x = +2.24 m s⁻². v = ±8.0√(0.070² − 0.035²) = ±0.485 m s⁻¹, with sign set by travel direction.