Oscillations and waves

6 visual steps · 3 worked examples · 6 exercises with solutions

First, let’s see the idea.

How far does a wave's shape travel in one period?

Move forward one step at a time. Change the values and observe what happens.

The wave travels; the medium oscillatesx (m)A = 0,3 mSpatial profile; the amber point has fixed x; dashed = initial instant.
Drawing description

Mark a point on the string. We imagine a regular wave on an ideal string: the disturbance propagates horizontally and the point oscillates vertically. Amplitude A is measured from equilibrium to one extreme, not from bottom to top. The lilac arrow runs from the equilibrium line to the amber point at the upper extreme. Source frequency: 1 Hz. Speed in the medium: 3 m/s. Amplitude: 0,3 m.

Step 1 of 6

Mark a point on the string

The lilac arrow runs from the equilibrium line to the amber point at the upper extreme.

We imagine a regular wave on an ideal string: the disturbance propagates horizontally and the point oscillates vertically. Amplitude A is measured from equilibrium to an extreme, rather than from bottom to top.

Source frequency
1 Hz
Speed in the medium
3 m/s
Amplitude
0,3 m

What if you change this?

0,5 Hz3 Hz
1 m/s6 m/s
0,1 m0.5 m

Before moving on

A wave has f = 2 Hz and v = 6 m/s. What is its period and how far does the shape travel in one cycle?

Compare with the explanation

T = 0.5 s and λ = 3 m. In one period, the shape travels 3 m; a point on the string oscillates locally and does not travel those 3 m.

You decide when to mark this lesson as studied.

The wave travels; the medium oscillates

Slide the drawing to see the details.

Explore the full explanationThe foundations, principles, worked examples and all the exercises, at your own pace.

What you will understand

  • Characterise oscillations by amplitude, period, frequency and phase.
  • Relate restoring force to simple harmonic motion.
  • Distinguish motion of the medium from wave propagation.
  • Apply v = λf and interpret superposition and standing waves.
  • Explain damping and resonance without predicting unlimited amplitudes.

Before you start

You do not need to know everything by heart. The explanation revisits the necessary ideas and shows how they connect. If a foundation is still unclear, you can return to these lessons:

Equilibrium and repetition

A system is in equilibrium when the relevant resultant is zero. In stable equilibrium, a small displacement produces a tendency to return. If the system repeatedly passes from one side of this position to the other, it describes an oscillation. Not every periodic motion is a straight-line oscillation: uniform rotation also repeats, but does not cross an equilibrium position in the same way.

Amplitude A is the maximum displacement from equilibrium, in m when describing position. Period T is the duration of one complete cycle, in s. Frequency f = 1/T measures cycles per second, in Hz. Going from one extreme to the other is half a cycle in a symmetric oscillator, not an entire period.

Phase indicates the stage of the cycle: two oscillators can have the same frequency and amplitude but reach maxima at different instants. Knowing the period alone is not enough to know position at a given instant. The initial condition, such as position and direction of motion, completes the description.

Force proportional to displacement

For an ideal spring, F = −kx. x is deformation relative to equilibrium and k measures stiffness in N/m. The negative sign shows that force points opposite to displacement. With mass m and no other net forces along that axis, the second law gives a = −(k/m)x.

This proportionality between acceleration and position, with opposite directions, characterises simple harmonic motion. One representation is x(t) = A cos(ωt + φ), with ω = √(k/m) and φ the initial phase. T = 2π√(m/k) and f = 1/T. The trigonometric function’s argument must be an angle in radians.

At equilibrium, x = 0 and a = 0, but speed is greatest. At the extremes, speed is zero and acceleration magnitude is greatest. Do not confuse “equilibrium position” with “being at rest”. The spring relationship requires the linear elastic regime; large deformations may invalidate the proportionality and predicted period.

Energy and another model: the pendulum

For an ideal mass–spring oscillator, E = mv²/2 + kx²/2 is constant. At the extremes, all mechanical energy is elastic potential energy: E = kA²/2. At equilibrium it is kinetic: E = m maximum v²/2. Thus maximum v = A√(k/m) = Aω. Doubling A quadruples energy while retaining the same period in the linear model.

A simple pendulum is a point mass attached to a light, inextensible string of length L. For small angles and neglecting resistance, its motion approximates simple harmonic motion with T ≈ 2π√(L/g). We use g = 9,8 m/s² in this lesson. Length runs from the suspension point to the centre of the mass.

The pendulum’s approximate period does not depend on mass, but does depend on L and g. The small-angle condition is essential: the approximation uses sin θ ≈ θ when θ is in radians. At large angular amplitudes, the actual period depends more on amplitude and the simple formula loses accuracy.

A disturbance propagates

A wave carries a disturbance and energy between regions. On a string, each small section can move up and down while the pulse advances horizontally. Local matter motion is not propagation speed. In the small-wave model, the entire string does not travel along with the pulse.

Mechanical waves need a material medium. Electromagnetic waves, such as light, can propagate through a vacuum. In a transverse wave, the disturbance is perpendicular to propagation direction; in a longitudinal wave it is parallel. Sound in air is longitudinal, with pressure compressions and rarefactions.

Wavelength λ is the spatial separation between consecutive points in phase, such as two crests. Period T is the time between two crests passing the same point. During one period, the shape advances λ, so v = λ/T = λf. Amplitude must not be confused with λ: one measures the strength of disturbance, the other repetition in space.

λ: distance between crestsApropagation →

Spatial diagram at an instant. A graph of oscillation at one point would use time on the horizontal axis and show T instead of λ.

The medium governs propagation

In many linear models, propagation speed depends on the medium’s properties rather than the frequency chosen by the source. For an ideal stretched string, v = √(tension F/μ), where μ is mass per length in kg/m. Greater tension increases speed; greater mass per length decreases it.

If a wave crosses into another medium, the source-imposed frequency stays the same at a stationary interface. Since v changes, λ changes according to v = λf. This distinction prevents the idea that a wave “always carries the same length” into any material.

Energy carried depends on amplitude and system properties. For linear harmonic waves, holding other conditions fixed, energy or intensity is proportional to A². Doubling amplitude can quadruple intensity. This does not mean propagation speed should double: speed and energy per wave are different properties.

Adding disturbances and forming patterns

In the linear regime, when two waves meet, the resultant displacement at a point is the sum of the displacements each would produce separately. This is the superposition principle. Two in-phase crests can reinforce each other; a crest and a trough of equal magnitudes can cancel displacement at that point and instant.

Local cancellation does not destroy total energy. After crossing, ideal pulses continue propagating. Energies and fluxes must be analysed together; total energy is not obtained simply by squaring the sum of amplitudes without considering interference and the region.

Two identical waves travelling in opposite directions can form a standing wave. Nodes remain at zero displacement and antinodes have maximum amplitude. For a string fixed at both ends, L = nλ/2, with positive integer n. Thus the allowed frequencies are fₙ = nv/(2L). The distance between consecutive nodes is λ/2, not λ.

Real oscillations and resonance

Friction and resistance remove mechanical energy from the oscillator, transforming it into internal energy: damping occurs. With weak damping, amplitude decreases over successive cycles. With very strong damping, the system may return to equilibrium without oscillating. The critical condition separates regimes and corresponds to a rapid return without oscillations in this model.

An external periodic force can sustain oscillations by supplying energy. Response amplitude depends on driving frequency, natural frequency and damping. Near a resonant frequency, energy transfer may be more effective and amplitude may increase.

Resonance does not mean infinite amplitude in a real system. Damping limits the response, and large amplitudes may alter behaviour or cause deformation. A swing is an accessible example: push timing matters, because pushing outside the appropriate phase may supply little energy or even remove it. Understanding phase and losses is as important as comparing frequencies.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Example 1

Mass and spring

A 0,50 kg mass oscillates on a spring with k = 200 N/m and amplitude 0,040 m. Calculate period, energy and maximum speed.
  1. ω = √(200/0,50) = 20 rad/s. T = 2π/20 ≈ 0,314 s.
  2. E = kA²/2 = 200 × 0,040²/2 = 0,160 J.
  3. Maximum v = Aω = 0,040 × 20 = 0,80 m/s.
Result
T ≈ 0,314 s; E = 0,160 J; maximum v = 0,80 m/s.
Check the result
At equilibrium, K = 0,50 × 0,80²/2 = 0,160 J, confirming conservation.

Example 2

A string and a source

An ideal string has tension 36 N and μ = 0,010 kg/m. A source oscillates at 30 Hz. Calculate speed and wavelength.
  1. v = √(36/0,010) = √3600 = 60 m/s.
  2. By the propagation relationship, λ = v/f = 60/30 = 2,0 m.
  3. The source period is T = 1/30 ≈ 0,0333 s; in this interval the wave advances 2,0 m.
Result
v = 60 m/s; λ = 2,0 m.
Check the result
The unit inside the square root is (kg·m/s²)/(kg/m) = m²/s². Frequency was not used to calculate v because the model assigns v to the medium.

Example 3

Modes of a fixed string

A string of length 0,80 m has wave speed 40 m/s and fixed ends. Calculate the first three allowed frequencies.
  1. The fundamental mode fits half a wave: λ₁ = 2L = 1,60 m.
  2. f₁ = v/(2L) = 40/1,60 = 25 Hz.
  3. The following modes have f₂ = 2f₁ = 50 Hz and f₃ = 3f₁ = 75 Hz.
Result
25 Hz, 50 Hz and 75 Hz.
Check the result
The corresponding wavelengths are 1,60 m, 0,80 m and approximately 0,533 m; in each case L contains a whole number of half-waves.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Exercise 1

An oscillator completes 24 cycles in 12 s. Determine f and T.
A hint
Count complete cycles and divide by the correct quantities.
See the worked solution
f = 24/12 = 2,0 Hz. T = 12/24 = 0,50 s. The product fT = 1 confirms that period and frequency are inverses.

Exercise 2

Two disturbances superpose with displacements of +3 mm and −2 mm at a point. What is the resultant displacement?
A hint
Superposition adds signed values.
See the worked solution
Resultant y = +3 − 2 = +1 mm. There is partially destructive interference at that point and instant. This result alone does not tell us the total energy of the two waves.

Exercise 3

In an ideal oscillator, amplitude changes from 2 to 6 cm without changing m and k. How does energy change?
A hint
Energy depends on amplitude squared.
See the worked solution
Amplitude triples, so energy is multiplied by 3² = 9. Period remains the same in the linear harmonic model; amplitude and frequency are not automatically proportional.

Exercise 4

A simple pendulum has L = 0,98 m. Estimate its small-angle period with g = 9,8 m/s².
A hint
The ratio L/g has unit s².
See the worked solution
T ≈ 2π√(0,98/9,8) = 2π√0,10 ≈ 1,99 s. It is approximately 2,0 s for a complete cycle, not merely going from one side to the other.

Exercise 5

A 5 Hz wave propagates at 15 m/s. Calculate λ and the distance travelled by the disturbance in 4 s.
A hint
Separate wavelength from the path travelled over several periods.
See the worked solution
λ = 15/5 = 3 m. In 4 s, the disturbance advances d = vt = 60 m. A material point of the medium need not move 60 m: it oscillates locally.

Exercise 6

Why does real resonance not grow indefinitely?
A hint
Consider losses and the limits of the linear model.
See the worked solution
Damping dissipates mechanical energy; the steady response balances supply and losses. At large amplitudes, stiffness and geometry may change or the system may fail. Unlimited growth is predicted by a lossless idealisation under special conditions.

Mistakes worth avoiding

At equilibrium the oscillator is always at rest.
It can pass through at maximum speed.
The wave carries the entire medium with it.
The disturbance and energy propagate; particles oscillate locally in the model.
Frequency and speed are the same property.
Frequency is cycles per time; speed is distance per time.
A node means there is no energy anywhere in the system.
It is a position of zero displacement in a standing pattern.
Resonance requires unlimited amplitude.
Losses and physical limits determine the real response.

The essentials to take with you

  • Oscillation and propagation are different motions.
  • f = 1/T and v = λf connect quantities with different units.
  • A linear restoring force produces simple harmonic motion.
  • Energy alternates between kinetic and potential in the ideal oscillator.
  • Superposition adds signed disturbances.
  • Damping and phase govern resonance.

If you still cannot explain one of these ideas, return to the corresponding section and work through an example again. Recognising the difficulty is part of studying.

See the idea in motion

Now you can change the values, observe the graph and compare the experiment with the results you calculated.

Try it in the lab

References for further study

Text and examples prepared for this learning path. These sources help check the principles and continue studying:

The organisation by school year is a study proposal. Pedagogical review by teachers remains a stage of the project.