Free Pearson Edexcel International A Level Physics lesson
Resonance, damping and measurement
Follow energy transfer into and out of an oscillator.
What you’ll learn
- Track mechanical stores and distinguish their sum from the energy of the wider system.
- Separate free motion, forced frequency and the amplitude response of a damped system.
- Calibrate the complete oscillator, resolve the resonance peak and interpolate with uncertainty.
Energy moves between kinetic and potential forms
For an undamped linear spring oscillator, elastic potential energy about equilibrium is ½kx² and kinetic energy is ½mv². The total mechanical energy is ½kA². At the extremes it is all potential energy in this model; at equilibrium it is all kinetic. For a vertical spring, the combined spring and gravitational energy about the loaded equilibrium has the same quadratic change.
Conservation gives ½mv² = ½k(A² − x²), so v² = ω²(A² − x²). This determines speed magnitude, while the direction must be inferred from the phase or stated motion. A pendulum exchanges kinetic and gravitational potential energy; at small angles the quadratic approximation explains its similar harmonic behaviour.
Damping transfers mechanical energy into thermal energy, sound or other modes. Total energy of a sufficiently broad system remains conserved even as the measured oscillator’s mechanical energy falls. At unchanged k, halving amplitude leaves one quarter of the mechanical energy. A force–deformation hysteresis loop represents work transferred out of the recoverable mechanical store during a cycle.
E = ½kA² Ep = ½kx² Ek = ½k(A² − x²) v² = ω²(A² − x²)
Symbols and units
- E is total mechanical energy, Ep is potential energy measured relative to equilibrium and Ek is kinetic energy, all in J.
- k is spring stiffness in N m⁻¹; A is nonnegative amplitude and x is signed equilibrium displacement, both in m.
- m is oscillating mass in kg, v is velocity in m s⁻¹ and ω = √(k/m) is angular frequency in rad s⁻¹.
- For a vertical spring, Ep = ½kx² represents the change in combined spring and gravitational potential energy about loaded equilibrium, not spring energy alone.
How the equation explains the idea
- At an undamped turning point, v = 0 and x = A, so E = ½kA² = ½ × 40 × 0.050² = 0.050 J.
- At x = 0.030 m, Ep = ½kx² = ½ × 40 × 0.030² = 0.018 J. Conservation then gives Ek = E − Ep = 0.032 J.
- Equating ½mv² to ½k(A² − x²) and using k/m = ω² gives v² = ω²(A² − x²). A square root yields speed magnitude; phase is still needed to select the sign of velocity.
- If damping later halves A at the same k, the turning-point energy ratio is (1/2)² = 1/4. The missing mechanical energy has transferred to other stores, not disappeared.
Conditions. Ideal undamped linear oscillator: k = 40 N m⁻¹, A = 0.050 m, and E = ½kA² = 0.050 J. The potential-energy zero is the equilibrium state; only accessible displacements −A ≤ x ≤ A are shown.
Displacement from equilibrium / m; Mechanical energy / J.
At x = 0, all 0.050 J is kinetic. At either extreme all is potential. At half-amplitude, potential energy is one quarter of the total, not one half. For a vertical spring, this potential curve includes the combined spring and gravitational change about equilibrium.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
Energy away from equilibrium
- k = 40 N m⁻¹ and A = 0.050 m give total E = 0.050 J.
- At x = 0.030 m, Ep = 0.018 J.
- Ek = 0.050 − 0.018 = 0.032 J.
Subtract energy stores using the same equilibrium reference.
Keep this idea. Track mechanical stores and distinguish their sum from the energy of the wider system.
Driving and damping shape the response
A free oscillator moves after an initial disturbance without a sustained periodic driving force. A forced oscillator receives a periodic input and, after transients decay, responds at the driving frequency. Its amplitude depends on driving frequency, drive strength and damping, so observing repeated motion alone does not distinguish free from forced motion.
Resonance is a large response when the drive is near the natural frequency and energy transfer from the driver is effective. On an amplitude-versus-driving-frequency graph, low damping produces a high narrow peak; greater damping produces a lower broader peak. For displacement response the maximum can shift slightly below the undamped natural frequency, so exact equality should not be imposed at all damping strengths.
A damper removes mechanical energy each cycle, reducing free amplitude and limiting a forced response. Ductile materials can also dissipate energy through plastic deformation: work goes into irreversible internal changes and heating rather than being fully returned on unloading. Earthquake-resistant structures can use these principles, but a simple oscillator does not establish that a particular building is safe.
Steady forced motion: response frequency = drive frequency input per cycle balances dissipation
Symbols and units
- Drive frequency and steady response frequency are measured in Hz. The natural frequency describes free motion of the corresponding oscillator, not a separate steady response frequency.
- Displacement amplitude is the maximum displacement magnitude from equilibrium in m; the example uses mm, with 1 mm = 10⁻³ m.
- Energy supplied per cycle and energy dissipated per cycle are both in J. Their equality describes a periodic steady state after transients have decayed.
How the equation explains the idea
- Choose one complete cycle of the steady response. Its stored mechanical energy returns to the same value, so energy conservation gives 0 = energy supplied − energy dissipated per cycle.
- For a linear oscillator driven sinusoidally, the sustained response repeats at the drive frequency. Varying that frequency changes amplitude; it does not create a second freely chosen steady frequency.
- Compare the same drive and response measure: the peak changes from 12 mm to 4 mm, an amplitude ratio 4/12 = 1/3, and becomes broader. Together these observations support increased damping.
- The energy balance explains why the driver must replace losses. It does not by itself calculate the peak height or make the displacement-resonance frequency exactly equal to the undamped natural frequency.
Conditions. Graph-generation model (extension): the steady solution of mx″ + bx′ + kx = F₀ cos(ωdt), with the same m, k and force amplitude F₀. Here ω₀ = √(k/m), ζ = b/(2√(km)), and A/(F₀/k) = [(1 − r²)² + (2ζr)²]⁻¹ᐟ². Only damping changes; transients have decayed. This formula explains the computed curves and is not an additional Unit 5 memorization requirement.
Driving frequency ratio r = ωd/ω₀; Amplitude ratio A/(F₀/k).
Greater damping reduces and broadens the peak. The displacement maximum occurs at r = √(1 − 2ζ²), about 0.994 for ζ = 0.08 and 0.935 for ζ = 0.25, slightly below the undamped natural frequency. At r = 0 the value is the static-displacement limit.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
Interpret two response curves
- The same drive gives a narrow 12 mm peak with little damping and a broad 4 mm peak with more damping.
- The lower broader peak is evidence of greater mechanical energy dissipation.
- Both steady responses follow the selected drive frequency, not a separate freely chosen frequency.
Compare curves with drive strength and measured response quantity held fixed.
Keep this idea. Separate free motion, forced frequency and the amplitude response of a damped system.
Core practical 16: calibrate an unknown mass
Mount a spring oscillator securely and drive it with a small periodic motion. For several known attached masses, vary the driving frequency slowly and identify the steady maximum amplitude, allowing transients to settle. Keep the same spring, attachments, drive amplitude and damping conditions. Use modest amplitudes within the linear range and secure masses against falling.
For light damping, resonant frequency approximates f = (1/2π)√(k/meffective), so 1/f² is proportional to effective mass. Plot 1/f² vertically against known added mass. An intercept can account for the moving holder and effective spring mass. Determine an unknown added mass by interpolation on that same calibration rather than assuming the intercept is zero.
Take closely spaced frequencies near a broad peak and quote an uncertainty in the peak location rather than claiming the exact frequency from one reading. Repeat the sweep in both directions and stay inside the calibrated mass range. Because 1/f² magnifies fractional frequency uncertainty by approximately a factor of two, poor peak identification can dominate the inferred mass uncertainty.
1/f² = (4π²/k)(madded + meffective offset) infer unknown mass from the calibration
Symbols and units
- f is the measured resonant frequency in Hz; 1/f² therefore has units s².
- madded is the known or unknown added mass in kg; meffective offset is the additional effective moving mass of holder and spring in kg.
- k is spring stiffness in N m⁻¹. In 1/f² = slope × madded + intercept, the slope has units s² kg⁻¹ and the intercept has units s².
- In the worked calibration, m is added mass in kg, the slope is 0.80 s² kg⁻¹ and the intercept is 0.040 s².
How the equation explains the idea
- For light damping, f ≈ (1/2π)√(k/(madded + meffective offset)). Squaring and inverting gives 1/f² ≈ (4π²/k)(madded + meffective offset), a straight calibration with a possible intercept.
- For the unknown at f = 2.00 Hz, calculate the plotted quantity first: 1/f² = 1/(2.00)² = 0.250 s².
- Rearrange the fitted line before substitution: m = ((1/f²) − intercept)/slope = (0.250 − 0.040)/0.80 = 0.2625 kg.
- Subtracting the intercept removes the offset inertia from the unknown mass. Ignoring it would give 0.3125 kg, a 0.0500 kg overestimate; use the unchanged setup and a validated interpolation range.
Worked example
Use an intercept
- A fitted calibration is 1/f² = 0.80m + 0.040, with m in kg and time in s.
- For unknown mass f = 2.00 Hz, 1/f² = 0.250 s².
- m = (0.250 − 0.040)/0.80 = 0.2625 kg.
The intercept prevents the holder from being counted as unknown sample mass.
Keep this idea. Calibrate the complete oscillator, resolve the resonance peak and interpolate with uncertainty.
Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.
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