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GRADE 9 · MECHANICAL OSCILLATIONS

Harmonic, damped and forced oscillations

Harmonic law, energy loss and resonance.

26 minutes6 theory cards
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Read the full lesson theory here without an account. In the interactive lesson, the same theory is presented as cards, followed by questions with feedback and practice problems. Your progress is saved.

Understanding step by step

Small pushes can build a large swing, while a badly timed push can slow it down. How does the rhythm of the drive change energy transfer, and why do real oscillations weaken without further pushes?

Push along the motionFThe seat moves rightThe push does positive workThe external force supplies energy
The swing seat passes through its lowest point moving right. The push acts along the motion and does positive work. Repeated pushes at suitable moments supply energy and increase the amplitude; resistance limits this growth.

Ideal repetition

An ideal mass on a spring slows smoothly near the extremes and crosses the middle quickly. A sine or cosine describes this motion. Oscillations with constant amplitude and frequency whose displacement follows this law are called harmonic. Ideal springs and small-angle pendulums behave approximately this way when losses are negligible. Periodic motion does not always have this shape.

Constant amplitude and periodx, смt, с-40401234T = 2 сA = 4 см; neglect losses
The graph x = 4 cos(πt) uses centimetres and seconds. Its amplitude remains 4 cm and its period is 2 s. Two complete cycles occur in four seconds. This describes free harmonic oscillations when energy losses can be neglected.

Position within a cycle

Two identical pendulums may share a period while one is on the right and the other on the left. Predicting position also requires phase φ, the stage within the cycle. Phase is measured in radians; one full cycle changes it by 2π. It describes both displacement and motion direction within the cycle. Adding 2π repeats the state, like a clock hand after a full turn.

Identical pendulums, same instantxx−A+A0v = 00v = 0Δφ = π — half a cycleFull cycle: Δφ = 2π
Two identical pendulums are displaced through equal small angles in opposite directions. At this instant, one is at −A and the other at +A, both with zero velocity. Their periods are equal and their phase difference is π: half a full cycle.

How fast does phase change?

Phase increases by 2π during period T. Its rate of change is therefore ω = 2π/T = 2πν, the angular frequency in rad/s. Frequency ν counts cycles per second; ω counts phase radians per second. Doubling ν doubles ω and halves T. Their numerical values differ by 2π, so never substitute hertz for radians per second without checking which symbol a formula uses.

ω = 2π/T = 2πν2π рад in Tν — cycles per secondГц = 1/сω — radians per secondрад/сThe circle models phase change
One full turn of the phase radius represents a phase increase of 2π radians in period T. Frequency ν counts cycles per second, while angular frequency ω counts radians per second: ω = 2πν = 2π/T. The circle represents phase, not the path of the mass.

Write the motion law

Combine the description: x = A cos(ωt + φ₀). Cosine repeats between −1 and +1, and A scales this to displacements from −A to +A in metres. ω sets the repetition rate, t is in seconds and φ₀ sets the initial state. The current phase is ωt + φ₀, not φ₀ alone. Increasing A stretches the graph vertically; increasing ω compresses it in time.

Same A and T, different φ₀x, смt, с-40401234T = 2 сφ₀ = 0φ₀ = π/2x = A cos(ωt + φ₀)
Both exact cosine curves have amplitude 4 cm and period 2 s. The blue curve starts at +4 cm with initial phase 0; the red curve starts at 0, moving in the negative direction, with initial phase π/2. The current phase of each oscillation is ωt + φ₀; it increases by 2π in one period.

Cosine describes the state

Check a familiar release: pull the mass to +A and let go without a push. Then φ₀ = 0 because cos 0 = 1. After T/4, phase is π/2 and x = 0; after T/2, phase is π and x = −A. The same x = 0 occurs twice per cycle with opposite motion directions. Choose initial phase from the full starting state, not just a coordinate value.

x = A cos(ωt), φ₀ = 00T/4T/2T+A0−Ax, мt, сAt T/4, the mass moves towards −A
One full cosine period starts at x = +A. After T/4, the coordinate is zero and decreasing; after T/2, the mass reaches −A; after T, it returns to +A. The marked points lie on the exact cosine graph.

Why does amplitude decrease?

Real swings eventually stop without deliberate braking. Resistance does negative work, turning some mechanical energy into internal energy of objects and air. Free oscillations started by one displacement or push become damped: amplitude decreases. Removing losses in thought would stop this decrease. A formula with one constant A cannot exactly describe the whole damped motion.

Amplitude decreases on both sidesx, смt, с-4040246Free oscillations: Eмех → U
In this example, x = 4e^(−t/3)cos(πt), with x in centimetres and t in seconds. The two grey envelopes show the amplitude decreasing on both sides of equilibrium. Resistance converts mechanical energy into internal energy; without external energy input, the oscillations decay.

An external rhythm

To keep a swing moving, supply energy periodically. This produces forced oscillations. Natural frequency characterises the system's free motion, while the source chooses the driving frequency. For a linear system driven sinusoidally, once the initial transient oscillations decay, steady motion has the force's frequency. The two frequencies can differ, and the response depends on how close they are.

After the transient has died awayF, Нx, см012t, с1−11−1Force and response: same period, 1 с
After the transient motion dies away, the force and displacement have the same period: 1 s. In this example, the driving frequency equals the natural frequency, and displacement lags behind force by a quarter-cycle. In general, the driving frequency may differ from the natural frequency; the steady response still follows the force's frequency.

Resonant response

Try pushing a swing at different rhythms. With weak losses, amplitude becomes especially large when the driving frequency is close to the natural frequency: energy transfer is efficient from cycle to cycle. This enhanced response is resonance. Frequencies do not create energy; the person pushing does work. An ideal lossless model permits growth, but in a real system growth changes the balance of energy supplied and lost.

Amplitude versus driving frequencyA, смν, Гц0246012Small lossesν₀ = 1 ГцThe external force supplies energy
This calculated frequency response is for a linear system with natural frequency 1 Hz and weak resistance. For a fixed driving-force amplitude, the response has a smooth maximum near this frequency. The external force supplies energy, while losses limit the amplitude.

Losses limit growth

Why can a real swing not reach infinite amplitude? As motion grows, friction and resistance limit the response, and the linear model may cease to apply. A damper dissipates mechanical energy and reduces resonant vibration in machines. In a musical instrument, enhanced response in a useful frequency range is desirable. The same mechanism has different consequences depending on system parameters, driving frequency and losses.

Same m, k and driving forceA, смν, Гц0246012smaller losseslarger lossesν₀ = 1 Гц — natural frequency
Both frequency responses are calculated for the same linear system, with natural frequency 1 Hz and the same driving-force amplitude. Smaller losses produce a taller, sharper peak; larger losses produce a lower, broader peak. With damping, the frequency of the maximum can differ slightly from the natural frequency.

Coordinate after half a period

A mass oscillates according to x = 0,04 cos(πt), with x in metres and t in seconds. Find amplitude, period and x after 1 s.

The motion law is givenx = 0,04 cos(πt)−0,040+0,04x, мt = 0One mass on a smooth horizontal surfaceA, T and x(1 с) — ?t is measured in seconds
A single mass moves along a smooth horizontal surface. The coordinate origin is at equilibrium; initially, the mass centre is at +0.04 m. The motion law uses metres and seconds. Find the amplitude, period and coordinate after one second.

Let's look at an example · 1

Compare x = 0,04 cos(πt) with x = A cos(ωt + φ₀). The cosine coefficient gives A = 0,04 м = 4 см; ω = π рад/с, φ₀ = 0.

Compare with x = A cos(ωt + φ₀)x = 0,04 cos(πt)−0,040+0,04x, мt = 0One mass on a smooth horizontal surfaceA = 0,04 м; ω = π рад/сφ₀ = 0 — starts at +A
Keep the same axis and initial mass position. Comparing with x = A cos(ωt + φ₀), the prefactor gives A = 0,04 м, the coefficient of time gives ω = π рад/с, and the initial phase is zero.

Let's look at an example · 2

A full-cycle phase increase of 2π takes T = 2π/ω = 2 с. Thus 1 s is half a period.

Find the time for a full cyclex = 0,04 cos(πt)−0,040+0,04x, мt = 0One mass on a smooth horizontal surfaceT = 2π/ω = 2π/π = 2 с1 с = T/2
The same mass starts at +0.04 m. Over a full cycle, the phase increases by 2π radians; at ω = π рад/с, this takes two seconds. One second is therefore half a period.

Let's look at an example · 3

x(1) = 0,04 cos π = −0,04 м = −4 см.

Substitute t = 1 сx = 0,04 cos(πt)−0,040+0,04x, мt = 1 сt = 0One mass on a smooth horizontal surfacex(1) = 0,04 cos π = −0,04 мcos π = −1
After one second, the centre of the same mass is shown at −0.04 m on the original axis; its initial position, +0.04 m, remains as a faint outline. Substitution gives cos π = −1, so the coordinate is negative.

Let's look at an example · 4

The mass starts at the right extreme +A and reaches the left extreme −A after half a cycle. This agrees with the calculated coordinate, whose magnitude does not exceed amplitude.

Half a cycle — opposite extremex = 0,04 cos(πt)−0,040+0,04x, мt = 1 сt = 0One mass on a smooth horizontal surface+4 см → 0 → −4 см in T/2|x| ≤ A; at the extremes v = 0
The faint outline shows the mass initially at +4 cm; the coloured mass is the same object one second later at −4 cm. In half a period, it passes through equilibrium to the opposite extreme. At both extremes, velocity is zero and the magnitude of the coordinate equals the amplitude.

Check the result

A = 4 см, T = 2 с, x(1 с) = −4 см. In one second the mass travels from the positive extreme to the opposite extreme.

Key point · 1

The harmonic law separates amplitude A, rate ω and initial state φ₀; current phase changes with time.

Key point · 2

Damping transfers mechanical energy to other forms; maintaining motion requires an energy supply.

Key point · 3

Resonance enhances the response to an external force at a suitable frequency; the energy source remains external.

What comes next

When oscillations are passed to neighbours, a wave forms.

Basic Concepts

3 concepts

Damped and forced oscillations

Damped oscillations decrease in amplitude because of resistance; forced oscillations are maintained by an external periodic force.

After the transient dies away, the driven system oscillates at the driving frequency.

Harmonic oscillations

Oscillations whose coordinate follows a sine or cosine law at constant frequency.

Phase specifies the oscillator's state at a chosen instant and allows different oscillations to be compared.

Resonance

A strong increase in forced-oscillation amplitude when driving and natural frequencies are close.

Resonance is useful in musical instruments and radio technology but can endanger bridges and machines without damping.

Physical quantities

2 quantities

Phase

\varphi

φ is the current phase; φ₀ is initial phase at t = 0. In harmonic motion, φ = ωt + φ₀; phase gives the stage of oscillation within a period.

Unit: radian · \mathrm{рад}

Angular frequency

\omega

The rate at which the phase of harmonic oscillation changes.

Unit: radians per second · \frac{\mathrm{рад}}{\mathrm{с}}

Lesson formulas

1 formula

Harmonic law

x=A\cos(\omega t+\varphi_0)

Displacement from equilibrium equals amplitude times the cosine of current phase ωt + φ₀; φ₀ is initial phase.

Conditions: Harmonic oscillations with constant amplitude and frequency, with coordinate zero at equilibrium.

Test the relationship

Make a prediction first. The button changes the stated quantity; then compare the result and explore other values.

Resonance and damping

A = F₀/√[(k − mω²)² + b²ω²]; ω = 2πν

Linear oscillator: m = 1 kg, k = 4π² N/m, F₀ = 1 N, resistive force −bv. Only steady-state response is shown, without the initial transient.

At ν = 1 Hz, damping increases from 0.5 to 1 kg/s. How will the steady-state amplitude change?

Your prediction

Initial values

0-0.03820.50.0604810.15921.50.257820.3565A, мν, Гц
A

Calculated result

A
0.3183 м
Explaining the result

Damping limits amplitude and broadens the peak. With damping, the peak lies slightly below the natural frequency of 1 Hz; resonance does not mean infinite amplitude.

Related formulas

The same formulas in the reference, with symbols and units.

Problems on this topic

Statements, answer options and the walkthrough.