AS & A Level Physics · Area 1 · AS Level Content

Waves

A wave transfers energy from one place to another without transferring matter, and a progressive wave is described by displacement, amplitude, wavelength, period, frequency and phase difference. This topic covers v = f lambda, reading frequency and amplitude from a cathode-ray oscilloscope, intensity, transverse and longitudinal waves, the Doppler effect for sound, the spectrum and polarisation.

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Syllabus coverage

  • Describe what is meant by wave motion as illustrated by vibration in ropes, springs and ripple tanks
  • Understand and use the terms displacement, amplitude, phase difference, period, frequency, wavelength and speed
  • Understand the use of the time-base and y-gain of a cathode-ray oscilloscope (CRO) to determine frequency and amplitude
  • Derive, using the definitions of speed, frequency and wavelength, the wave equation v = f lambda, and recall and use it
  • Understand that energy is transferred by a progressive wave
  • Recall and use intensity = power / area, and use the relation intensity is proportional to (amplitude)^2 for a progressive wave
  • Compare transverse and longitudinal waves, and analyse and interpret graphical representations of both
  • Understand that when a source of sound waves moves relative to a stationary observer the observed frequency is different from the source frequency, and use the expression f_o = f_s v / (v +/- v_s) for a source moving relative to a stationary observer
  • State that all electromagnetic waves are transverse waves that travel with the same speed c in free space, recall the approximate range of wavelengths in free space of the principal regions of the electromagnetic spectrum from radio waves to gamma rays, and recall that wavelengths in the range 400-700 nm in free space are visible to the human eye
  • Understand that polarisation is a phenomenon associated with transverse waves, and recall and use Malus's law (I = I0 cos^2 theta) to calculate the intensity of a plane-polarised electromagnetic wave after transmission through a polarising filter or a series of polarising filters

Part of the AS & A Level Physics syllabus 9702.

The properties of a progressive wave, and the oscilloscope traceA progressive wave transfers energy without transferring matter: every particle oscillates about its ownfixed position.Displacement against distance00.100.200.300.40distance x / mdisplacementy / mamplitudewavelengthcentre linetroughcrestOscilloscope trace: y-gain and time-base2.5 divisionsT = 4.0 divisionstime-base 5.0 ms per divisiony-gain 2.0 V per divisionT = 4.0 x 5.0 = 20 ms, so f = 50 Hzamplitude A = 2.5 x 2.0 = 5.0 VIntensity is proportional to the square of the amplitude, so twice the amplitude carries four timesthe intensity.The y-gain turns a height in divisions into a voltage and the time-base turns a width in divisionsinto a time.The amplitude is from the centre line to a crest: a peak-to-peak height of 6.0 divisions at 2.0 Vper division is 6.0 V.
The properties of a progressive transverse wave shown on one displacement against distance graph: the undisturbed centre line, the amplitude measured from that line to a crest, the wavelength measured between successive crests, and two points separated by a quarter of a wavelength to illustrate a phase difference of 90 degrees.

Wave motion, and the words you need to describe it

A wave is a way of transferring energy from one place to another without transferring matter. Shake the end of a rope and a pulse runs along it, but the rope itself does not travel: each small section of rope oscillates about its own fixed position and passes the energy on to the next section. The same is true of a compression travelling along a spring and of water waves in a ripple tank, where a motor-driven dipper makes a ripple pattern that moves outwards while the water itself only bobs up and down. Every wave needs something to oscillate, and the energy travels through that material, or through the electric and magnetic fields in the case of light.

The seven quantities in the syllabus describe the same wave from different angles. Displacement is the distance of an oscillating point from its undisturbed position, and it is positive on one side of that position and negative on the other. Amplitude is the maximum displacement, so it is measured from the undisturbed position to a crest, never from trough to crest. Wavelength is the distance between two successive points that are in phase, such as two successive crests or two successive troughs, and it is also the distance the wave travels in one period.

The remaining three terms describe the timing. Period is the time taken for one complete oscillation of a point, or for one wavelength to pass a fixed point. Frequency is the number of complete oscillations passing a fixed point each second, measured in hertz (Hz), so frequency and period are related by f = 1 / T. Phase difference is how far one oscillation leads or lags behind another at the same instant, measured in degrees or radians: one whole wavelength apart is 360 degrees (or 2 pi radians), half a wavelength apart is 180 degrees, and a quarter of a wavelength is 90 degrees.

Two points do not have to be a whole number of wavelengths apart to be in phase. The crest of one wave and the crest of the next are in phase because they are one wavelength apart, and so are points 270 degrees apart in the other sense, because phase difference is really the fraction of a cycle by which one oscillation is ahead of the other.

  • Displacement: how far a point has moved from its undisturbed position, in metres.
  • Amplitude: the maximum displacement, in metres.
  • Wavelength: the distance between two successive points in phase, in metres.
  • Period: the time for one complete oscillation, in seconds.
  • Frequency: the number of oscillations per second, f = 1 / T, in hertz (Hz).
  • Phase difference: the fraction of a cycle by which one oscillation leads another, in degrees or radians.
Amplitude is measured from the centre line to a crest, not from trough to crest, which is twice the amplitude.

Deriving and using v = f lambda

The wave equation follows from the definitions you already have, and an exam question may ask you to derive it rather than recall it. Speed is distance travelled divided by time taken. During one period T, a wavefront moves forward by exactly one wavelength, lambda, so the speed of the wave is v = lambda / T. Frequency is the number of oscillations per second, f = 1 / T, which rearranges to 1 / T = f. Substituting this into v = lambda / T gives v = f x lambda, where v is in metres per second, f is in hertz and lambda is in metres.

A worked example. A ripple tank dipper vibrates at 8.0 Hz and the waves it makes are 4.5 cm apart. Convert the wavelength to metres first: 4.5 cm = 0.045 m. Then v = f lambda = 8.0 x 0.045 = 0.36 m s-1. For a sound wave of frequency 440 Hz travelling through air at 340 m s-1, the wavelength is lambda = v / f = 340 / 440 = 0.77 m. Notice that the two rearrangements are used in opposite directions: in the first the wavelength is known and the speed is wanted, in the second the speed is known and the wavelength is wanted.

The frequency of a wave is fixed by the source and does not change as the wave travels: a 440 Hz tuning fork sends out 440 oscillations every second whether the sound is in air or in water. The speed, however, depends on the medium, because the medium decides how quickly one oscillating point can pass the disturbance to the next. Since f is fixed and v changes, the wavelength must change too, by lambda = v / f, so sound travelling from air into water keeps its frequency but has a longer wavelength because it travels faster.

QuantitySymbolUnitHow to find it
Speedvm s-1v = f lambda, or distance / time
FrequencyfHzf = 1 / T, or f = v / lambda
Wavelengthlambdamlambda = v / f, or the distance between successive crests
PeriodTsT = 1 / f
AmplitudeAmMeasured from the undisturbed position to a crest
Always put the wavelength into metres before using v = f lambda. A wavelength in centimetres is the commonest source of a wrong power of ten.

Reading frequency and amplitude from a cathode-ray oscilloscope

A cathode-ray oscilloscope (CRO) draws the signal as a graph of voltage against time, and its two controls set the scales of that graph. The time-base sets the horizontal scale, in milliseconds per division or microseconds per division, so it tells you how much time each square on the screen represents. The y-gain sets the vertical scale, in volts per division, so it tells you how much voltage each square represents. With a periodic signal the trace is a steady repeating pattern, and measuring that pattern gives the period and the amplitude of the wave.

To find the frequency, measure the horizontal distance for one complete cycle in divisions and multiply by the time-base setting to get the period; then use f = 1 / T. To find the amplitude, measure the vertical peak-to-peak height in divisions, multiply by the y-gain setting to get the peak-to-peak voltage, and halve it. A trace spanning 4.0 divisions with a time-base of 5.0 ms per division has a period of 4.0 x 5.0 = 20 ms = 0.020 s, so f = 1 / 0.020 = 50 Hz. If the same trace has a peak-to-peak height of 6.0 divisions with a y-gain of 2.0 V per division, the peak-to-peak voltage is 6.0 x 2.0 = 12 V, so the amplitude is 12 / 2 = 6.0 V.

  • Time-base: the number of milliseconds (or microseconds) each division represents, read from the horizontal scale.
  • Y-gain: the number of volts each division represents, read from the vertical scale.
  • Period = (number of divisions for one cycle) x (time-base setting).
  • Frequency = 1 / period.
  • Amplitude = (peak-to-peak height in divisions) x (y-gain setting) / 2.
The y-gain and the time-base are scales, not readings. A height of six divisions means nothing until you multiply it by the y-gain setting.

Energy, intensity and amplitude

A progressive wave carries energy away from the source, and intensity is the rate at which that energy arrives at each square metre of surface held at right angles to the direction the wave is travelling: intensity = power / area, measured in W m-2. Intensity is not the same as power. A large loudspeaker emits more power than a small one, but the intensity at your ear also depends on how far away it is, because the same power is spread over a larger area.

For a progressive wave the intensity is proportional to the square of the amplitude, I is proportional to A^2. Doubling the amplitude therefore multiplies the intensity by four, and multiplying the amplitude by three multiplies the intensity by nine. The reason is that the energy carried by a wave depends on how far each oscillating particle is displaced and how fast it is moving, and both of those quantities increase in proportion to the amplitude, so the energy depends on the amplitude twice over.

Worked example. A wave of amplitude 2.0 mm has an intensity of 0.50 W m-2. A wave of the same frequency but amplitude 4.0 mm has an amplitude twice as large, so its intensity is 2^2 = 4 times greater: 4 x 0.50 = 2.0 W m-2. If instead you wanted to double the intensity, you would have to increase the amplitude by a factor of 2^0.5, that is 1.41, because (1.41)^2 is approximately 2.

When a wave spreads out from a small source in three dimensions, the same power passes through a spherical surface of ever greater area, so the intensity falls with the square of the distance from the source: doubling the distance quarters the intensity. That is why a siren is much quieter at 200 m than at 100 m, and it is the same geometrical spreading that makes amplitude fall with distance.

Change in amplitudeChange in intensityNumerical example
Doubled (x 2)x 40.50 W m-2 becomes 2.0 W m-2
Tripled (x 3)x 90.50 W m-2 becomes 4.5 W m-2
Halved (x 0.5)x 0.250.50 W m-2 becomes 0.125 W m-2
Multiplied by 2^0.5 (x 1.41)x 20.50 W m-2 becomes 1.0 W m-2
Intensity is proportional to (amplitude)^2. Double the amplitude and the intensity goes up by a factor of four, not two.

Transverse and longitudinal waves

Waves are classified by the direction in which the oscillations point compared with the direction in which the energy travels. In a transverse wave the oscillations are perpendicular to the direction of travel of the wave: shake a rope up and down and the crests travel away from your hand while the rope moves across the line of travel. In a longitudinal wave the oscillations are parallel to the direction of travel: push and pull the end of a spring and compressions travel along it, with the coils moving backwards and forwards along the same line that the energy follows.

Sound in air is longitudinal. The vibrating source creates regions where the air is squeezed and its pressure and density are slightly above the surrounding value, called compressions, and regions where the air is stretched and the pressure is slightly below it, called rarefactions, and these travel away from the source. A microphone records the pressure changes, and the signal is usually displayed as a graph, but that graph is a representation of the wave, not a picture of the air molecules. A longitudinal wave can also be drawn as a row of dots that are closer together at a compression and further apart at a rarefaction.

Electromagnetic waves, waves on a rope or a string and water surface waves are all transverse. Sound in air or water, compression waves in a spring and the ultrasound used in medical scanners are longitudinal. Only transverse waves can be polarised, because polarisation means choosing a single plane for oscillations that are already perpendicular to the direction of travel; there is no equivalent choice for oscillations that lie along the direction of travel, which is why a beam of sound cannot be polarised.

PropertyTransverse waveLongitudinal wave
Direction of oscillationPerpendicular to the direction of travelParallel to the direction of travel
Pattern in the mediumCrests and troughsCompressions and rarefactions
ExampleVisible light, waves on a rope, water surface wavesSound in air, compression waves in a spring
Can it be polarised?Yes, because the oscillations have a plane to be restricted toNo, because the oscillations lie along the direction of travel
Ask one question of any wave: do the oscillations point along the direction of travel, or across it? Along means longitudinal; across means transverse.

The Doppler effect for sound

When a source of sound moves relative to a stationary observer, the observer hears a frequency that is different from the frequency the source is emitting. A source approaching the observer is heard at a higher frequency than the source frequency, and a source moving away is heard at a lower frequency. The source itself has not changed: only 500 compressions leave the siren each second, but they arrive at the observer more often when the siren is closing in and less often when it is drawing away.

The reason is the crowding of the wavefronts. As the source moves towards the observer, each new wavefront is emitted from a point closer to the observer than the one before, so the wavefronts are squeezed together in front of the source and the distance between them, which is the observed wavelength, is shorter than the wavelength you would measure with the source at rest. The speed of sound in the air is unchanged, so f = v / lambda with a smaller lambda means a higher observed frequency. Behind a receding source the wavefronts are stretched further apart, and the observed frequency is lower.

The observed frequency is given by f_o = f_s v / (v +/- v_s), where f_s is the source frequency, f_o is the frequency heard by the stationary observer, v is the speed of sound in the air and v_s is the speed of the source. Use the minus sign when the source is approaching, because the source is moving into its own wavefronts, and the plus sign when the source is receding. The equation only applies to a source moving directly towards or directly away from the observer, and to a stationary observer: the syllabus requires the source to move, not the observer and the source together.

Worked example. A siren emits 500 Hz and the speed of sound is 340 m s-1. Approaching at 30 m s-1, f_o = 500 x 340 / (340 - 30) = 500 x 340 / 310 = 548 Hz. After passing, f_o = 500 x 340 / (340 + 30) = 500 x 340 / 370 = 459 Hz. The observer hears a sudden drop of about 90 Hz as the vehicle goes past, which is why an ambulance seems to change note at the moment it passes you.

Motion of the sourceSign used in the denominatorObserved frequency compared with the source frequency
Moving straight towards a stationary observerv - v_sHigher than f_s
Moving straight away from a stationary observerv + v_sLower than f_s
Stationary, or moving across the line of sightv_s = 0Equal to f_s at the instant of passing
The source frequency is unchanged; what changes is how often the wavefronts arrive. Approaching means a shorter observed wavelength and a higher observed frequency.

The electromagnetic spectrum

All electromagnetic waves are transverse waves, in which oscillating electric and magnetic fields travel together, and in free space they all travel at the same speed, c = 3.00 x 10^8 m s-1, whatever their frequency. In a material the speed is lower and depends on the material, but the frequency is set by the source and does not change. Because v = f lambda in free space becomes c = f lambda, a high-frequency region of the spectrum always has a short wavelength and a low-frequency region always has a long wavelength.

The spectrum is usually split into seven regions, from the longest wavelength to the shortest: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. The boundaries between neighbouring regions are conventions rather than sharp walls, and the two shortest regions overlap: the real difference between X-rays and gamma rays is where they come from, since X-rays are produced by fast electrons hitting a metal target in an X-ray tube while gamma rays are produced by changes inside a nucleus.

Visible light occupies only a tiny window: wavelengths from about 700 nm (red) to about 400 nm (violet), which is 7 x 10-7 m to 4 x 10-7 m, and it is the only region the human eye detects. Red light has the longest wavelength of the visible colours and violet the shortest.

RegionApproximate wavelength range in free space / mOne real use
Radio wavesLonger than about 0.1 mRadio and television broadcasting
Microwaves0.1 m to 1 x 10-3 mSatellite links and mobile phone signals; a microwave oven uses 2.45 GHz
Infrared1 x 10-3 m to 7 x 10-7 mRemote controls and thermal imaging cameras
Visible7 x 10-7 m to 4 x 10-7 m (700 nm to 400 nm)Sight, photography and optical fibre communications
Ultraviolet4 x 10-7 m to 1 x 10-8 mSterilising water and equipment, and producing fluorescence
X-rays1 x 10-8 m to 1 x 10-13 mImaging bones and teeth in hospital radiography
Gamma raysShorter than about 1 x 10-10 mSterilising medical equipment and treating cancer
Learn the order from the long-wavelength end: radio, microwave, infrared, visible, ultraviolet, X-rays, gamma. Wavelength gets shorter and frequency gets higher as you move along that list.
The electromagnetic spectrum, and polarisationAll electromagnetic waves are transverse and travel at the same speed in free space, c = 3.00 x 10^8 m s-1. Only thewavelength and the frequency change from one region to the next.radiowavelengths infree spacemicrowavewavelengths infree spaceinfraredwavelengths infree spacevisiblewavelengths infree spaceultravioletwavelengths infree spaceX-rayswavelengths infree spacegammawavelengths infree spacewavelength in free space / mfrequency in free space / Hzlonger than0.1 mbelow 3 x 10^9Hz0.1 m to1 x 10-3 m3 x 10^9 to3 x 10^11 Hz1 x 10-3 m to7 x 10-7 m3 x 10^11 to4.3 x 10^14 Hz7 x 10-7 m to4 x 10-7 m4.3 to 7.5x 10^14 Hz4 x 10-7 m to1 x 10-8 m7.5 x 10^14to 3 x 10^16 Hz1 x 10-8 m to1 x 10-13 m3 x 10^16 to3 x 10^21 Hzshorter than1 x 10-13 mabove 3 x 10^21Hzwavelength decreases, frequency increasesc = f x lambda for every regionThe regions are not drawn to scale: the boundaries between them are conventions, and X-rays and gamma rays overlap.The visible band is only 700 nm (red) to 400 nm (violet). X-rays come from fast electrons hitting a metal target; gamma rayscome from a nucleus.Polarisation: a transverse wave passing through a filterunpolarised light: the electric fieldoscillates in every planepolarising filtertransmission axis verticalplane-polarised light: one plane onlyhalf the intensity of the unpolarised beam
The regions of the electromagnetic spectrum arranged in a single band shared by all electromagnetic waves, showing the approximate wavelength range in free space of each region from radio waves at the long-wavelength end to gamma rays at the short-wavelength end, with the narrow visible band from 700 nm to 400 nm marked.

Polarisation and Malus's law

Polarisation is the restriction of the oscillations of a transverse wave to a single plane. Light from a lamp or from the Sun is unpolarised: its electric field oscillates in every plane containing the direction of travel. A polarising filter only transmits the component of the oscillation parallel to its transmission axis, so light leaving a filter is plane-polarised and cannot pass through a second filter whose axis is at right angles to the first. Because polarisation requires oscillations that are perpendicular to the direction of travel, it is direct evidence that light is a transverse wave, and it is impossible for a longitudinal wave such as sound.

Unpolarised light passing through a single filter always comes out at half the original intensity, whatever the orientation of the filter, because on average half of the oscillations lie closer to the transmission axis than to the perpendicular direction. When unpolarised light meets a filter, the angle between the axis of the filter and the direction of polarisation is therefore meaningless; the quantity that matters is the angle between the plane of polarisation of the incoming light and the transmission axis of the filter.

For light that is already plane-polarised, the transmitted intensity is given by Malus's law, I = I0 cos^2 theta, where I0 is the intensity of the light arriving at the filter, I is the intensity leaving it and theta is the angle between the plane of polarisation of the incoming light and the transmission axis of the filter. Worked example: polarised light of intensity 400 W m-2 falls on a filter with its plane of polarisation at 60 degrees to the transmission axis. Then I = 400 x cos^2 60 = 400 x (0.50)^2 = 400 x 0.25 = 100 W m-2.

A series of filters is handled one at a time, using the light that leaves each filter as the input to the next. If a second filter is placed after the first with its axis 30 degrees away, the light reaching it is the 100 W m-2 polarised in the direction of the first axis, so I = 100 x cos^2 30 = 100 x 0.75 = 75 W m-2. The angle that matters is always between the polarisation direction of the light and the axis of the next filter, never between the first and last filter alone.

  • Polarisation can only happen to transverse waves, so it is evidence that light is transverse.
  • A polarising filter transmits only oscillations parallel to its transmission axis.
  • Unpolarised light loses half its intensity in one filter, at any orientation.
  • For plane-polarised light, I = I0 cos^2 theta.
  • For several filters, apply Malus's law once for each filter in turn.
  • Malus's law uses cos^2, not cos: at 60 degrees the transmitted intensity is 25 percent, not 50 percent.
In I = I0 cos^2 theta, theta is measured from the plane of polarisation of the incoming light to the transmission axis of the filter it is about to enter.

Key terms you must be able to define

Progressive wave
A wave that transfers energy from one place to another without transferring matter, in which energy is carried away from the source and each particle of the medium oscillates about a fixed position.
Displacement
The distance of an oscillating point from its undisturbed position, measured in metres, taken as positive in one direction and negative in the other.
Amplitude
The maximum displacement of a point on a wave from its undisturbed position, measured in metres; it is half the peak-to-peak value and the intensity of the wave is proportional to its square.
Wavelength
The distance between two successive points on a wave that are in phase, such as two successive crests or two successive troughs; it is also the distance travelled by the wave in one period, and it is measured in metres.
Period
The time taken for one complete oscillation of a point on the wave, or for one wavelength to pass a fixed point, measured in seconds; it is the reciprocal of the frequency.
Frequency
The number of complete oscillations passing a fixed point each second, measured in hertz (Hz); for a progressive wave f = 1 / T and the frequency is fixed by the source and does not change as the wave travels.
Phase difference
The fraction of a cycle by which one oscillation leads or lags another, expressed in degrees or radians; one wavelength of separation corresponds to 360 degrees or 2 pi radians, and half a wavelength to 180 degrees.
Intensity
The power transmitted by a wave per unit area perpendicular to the direction of travel, measured in W m-2; for a progressive wave the intensity is proportional to the square of the amplitude.
Observed frequency (Doppler effect)
The frequency heard by a stationary observer when a source of sound moves relative to the observer; it is given by f_o = f_s v / (v +/- v_s), and it is higher than the source frequency when the source approaches and lower when it recedes.

Applied to the real world

Cambridge sets 9702 in unfamiliar contexts - a fairground ride, a power station, a hospital scanner, a satellite, an astronomical observatory. Use these as models for how to apply the physics, then write your own.

In the laboratory

A signal generator is connected to a cathode-ray oscilloscope and the trace is read, not guessed. With the time-base at 5.0 ms per division and one cycle spanning 4.0 divisions, the period is 20 ms and the frequency is 50 Hz; with the y-gain at 2.0 V per division and a peak-to-peak height of 6.0 divisions, the amplitude is 6.0 V. Changing the time-base alters how many cycles fit on the screen but never changes the frequency of the signal itself.

In medicine

Doppler ultrasound measures blood flow. A probe sends ultrasound of a few MHz into an artery and detects the tiny frequency shift of the sound scattered back from moving red blood cells; blood flowing towards the probe returns a slightly higher frequency and blood flowing away returns a slightly lower one. Remember that the usual equation applies to a moving source, and here it is the blood that is moving, so the reasoning is the same but the source and observer exchange roles.

On the road

A police speed gun transmits microwaves at about 24 GHz (a wavelength of roughly 12 mm) and receives the reflection from the vehicle. The frequency of the reflected wave is shifted by the motion of the vehicle, and the electronics convert that shift into a speed in km per hour. The same physics explains why an ambulance siren drops in pitch by about 90 Hz as it passes a stationary listener, even though the siren itself is emitting a steady 500 Hz.

In astronomy

The light from a galaxy contains dark absorption lines at wavelengths that its atoms emit in a laboratory on Earth. Almost every distant galaxy shows those lines shifted towards longer wavelengths, by an amount proportional to the speed at which the galaxy is receding; this is the Doppler effect applied to light, and it is the evidence for the expansion of the Universe. Radio astronomers use the same idea to map the motion of hydrogen clouds from the 21 cm hydrogen line.

In engineering

A model made of transparent plastic is placed between crossed polarising filters and loaded. Where the plastic is stressed it becomes slightly birefringent, so the plane of polarisation of the light passing through it is rotated and coloured bands appear against a dark background. Engineers photograph the pattern of bands to find the points of highest stress before a real component is built, using exactly the polarising filters and Malus's law that are examined in this topic.

In sport

Polarising sunglasses cut glare for skiers, sailors and anglers. Light reflected from snow or from a water surface becomes partly polarised in the plane of the surface, and a filter whose transmission axis is set vertically blocks much of that reflected component while still transmitting plenty of light from the rest of the scene. The same principle is used in the liquid crystal displays of a scoreboard, which contain two polarising filters and switch the liquid crystal to brighten or darken each pixel.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What is meant by a progressive wave?
A wave that transfers energy from one place to another without transferring matter, in which every particle of the medium oscillates about its own fixed position and passes energy on to the next particle.
Q2Define the wavelength of a wave.
The distance between two successive points on the wave that are in phase, such as two successive crests, measured in metres; it is also the distance travelled by the wave in one period.
Q3How are period and frequency related?
f = 1 / T, where T is the period in seconds and f is the frequency in hertz. A period of 0.020 s is a frequency of 50 Hz.
Q4What phase difference corresponds to one wavelength, and to half a wavelength?
One wavelength corresponds to 360 degrees or 2 pi radians; half a wavelength corresponds to 180 degrees or pi radians, and a quarter of a wavelength to 90 degrees.
Q5State the wave equation and the units of each quantity in it.
v = f lambda, where v is the speed in m s-1, f is the frequency in Hz and lambda is the wavelength in metres.
Q6Derive v = f lambda in three steps.
Speed is distance divided by time; in one period T a wavefront advances by one wavelength, so v = lambda / T; and f = 1 / T, so v = lambda x f, which is v = f lambda.
Q7What does the time-base of a cathode-ray oscilloscope control, and how is frequency found from the trace?
The time-base sets the time represented by each horizontal division, in ms or microseconds per division. Period = divisions for one complete cycle x time-base setting, then f = 1 / period.
Q8What does the y-gain control, and how is the amplitude of the signal found?
The y-gain sets the voltage represented by each vertical division. Amplitude = peak-to-peak height in divisions x y-gain setting / 2.
Q9Define intensity, and state how it depends on amplitude.
Intensity is the power transmitted per unit area perpendicular to the direction of travel, in W m-2, and for a progressive wave intensity is proportional to the square of the amplitude.
Q10Give two differences between transverse and longitudinal waves, with one example of each.
In a transverse wave the oscillations are perpendicular to the direction of travel (visible light, a wave on a rope); in a longitudinal wave they are parallel to it (sound in air, a compression in a spring). Only transverse waves can be polarised.
Q11Why does a moving source of sound sound higher in pitch as it approaches?
Each wavefront is emitted from a point closer to the observer than the one before, so the wavefronts are crowded together and the observed wavelength is shorter. The speed of sound is unchanged, so f = v / lambda gives a higher observed frequency.
Q12State the expression for the observed frequency when a source of sound moves relative to a stationary observer, and say which sign to use when approaching.
f_o = f_s v / (v +/- v_s). Use the minus sign in the denominator when the source approaches and the plus sign when it recedes. Here v is the speed of sound and v_s is the speed of the source.
Q13What is polarisation, and which waves show it?
Polarisation is the restriction of the oscillations of a transverse wave to a single plane. It is shown by transverse waves such as light, microwaves and waves on a string, and never by a longitudinal wave such as sound.
Q14State Malus's law and give one numerical example.
For plane-polarised light, I = I0 cos^2 theta, where theta is the angle between the plane of polarisation of the incoming light and the transmission axis of the filter. At 60 degrees, I = I0 x (0.50)^2 = 0.25 I0.
0 of 14 marked as known

Exam-style questions

9 questions, 31 marks. Write your answer, then reveal the mark scheme and tick each point you actually made. Answering honestly here is what makes the rest of the site useful.

DeduceAO13 marks
A progressive wave of frequency f and wavelength lambda travels with speed v. Deduce, from the definitions of speed, frequency and wavelength, that v = f lambda.

Mark scheme — tick each point you actually made

Examiner tip: A derivation needs the reasons as well as the algebra. Quote the definitions of speed and frequency, then say that the wave moves one wavelength in one period.
CalculateAO33 marks
A signal is displayed on a cathode-ray oscilloscope. The time-base is set to 5.0 ms per division and one complete cycle of the trace spans 4.0 divisions. The y-gain is set to 2.0 V per division and the trace has a peak-to-peak height of 6.0 divisions. Calculate (a) the period of the signal, (b) the frequency of the signal, (c) the amplitude of the signal.

Mark scheme — tick each point you actually made

Examiner tip: Write the scale down first. The halving in part (c) is a mark in its own right, and part (b) needs the period in seconds, not in milliseconds.
ExplainAO22 marks
A progressive wave of amplitude 2.0 mm has an intensity of 0.50 W m-2. Explain what is meant by the intensity of a wave, and state the intensity of a wave of the same frequency whose amplitude is 4.0 mm.

Mark scheme — tick each point you actually made

Examiner tip: Say 'power per unit area' and give the unit. Adding the area at right angles to the direction of travel shows you understand which area is meant.
CompareAO23 marks
Compare a transverse wave with a longitudinal wave, giving one example of each, and state which of the two can be polarised and why.

Mark scheme — tick each point you actually made

Examiner tip: A comparison needs both waves in the same sentence. Do not give two examples of the same type.
CalculateAO24 marks
In a ripple tank, straight water waves are produced at a frequency of 12 Hz and the distance between successive crests is 8.0 cm. (a) Calculate the speed of the waves. (b) Two points on the water surface are 6.0 cm apart, measured along the direction in which the waves travel. Calculate the phase difference between the oscillations at these two points, in degrees.

Mark scheme — tick each point you actually made

Examiner tip: Convert centimetres to metres before substituting. In part (b) find the separation as a fraction of a wavelength first, then multiply by 360 degrees.
ExplainAO24 marks
An ambulance sounding its siren at a constant frequency drives at a steady speed past a stationary observer. Explain why the observer hears a frequency higher than the source frequency as the ambulance approaches and lower than the source frequency after it has passed.

Mark scheme — tick each point you actually made

Examiner tip: Explain it with wavefronts, not with the source 'sounding different'. The mark comes from the change in the distance between successive wavefronts and from the fact that the speed of sound stays the same.
CalculateAO23 marks
An ambulance siren emits sound of frequency 500 Hz and the speed of sound in the air is 340 m s-1. The ambulance travels at 30 m s-1 in a straight line towards a stationary observer, passes the observer and continues at the same speed. Calculate the frequency heard by the observer (a) while the ambulance is approaching and (b) after it has passed and is moving away.

Mark scheme — tick each point you actually made

Examiner tip: Pick the sign before you calculate and check the result: an approaching source must give an answer larger than 500 Hz and a receding source an answer smaller than 500 Hz. Give both answers in Hz.
DescribeAO14 marks
(a) State the type of wave that every electromagnetic wave is, and state their speed in free space. (b) State the approximate wavelength range in free space of the region of the spectrum that is visible to the human eye, and name the region whose wavelengths are longer than those of microwaves. (c) Give one real use of infrared radiation and one real use of X-rays.

Mark scheme — tick each point you actually made

Examiner tip: Give the visible range in both nanometres and metres and remember that every region of the spectrum is electromagnetic: the regions differ in frequency and wavelength, not in nature or in speed in free space.
CalculateAO25 marks
Plane-polarised light of intensity 400 W m-2 is incident on a polarising filter, with the plane of polarisation of the light at 60 degrees to the transmission axis of the filter. (a) State what is meant by plane-polarised light. (b) Calculate the intensity of the light transmitted by the first filter. (c) A second identical filter is placed after the first, with its transmission axis at 30 degrees to the transmission axis of the first filter. Deduce the intensity of the light leaving the second filter and explain your reasoning.

Mark scheme — tick each point you actually made

Examiner tip: Use cos squared, not cos. For a series of filters, work through them in order, taking the intensity leaving one filter as the intensity entering the next, and measure the angle from the polarisation of the light that actually arrives at each filter.

Exam technique for this topic

  • Convert every length into metres before substituting into v = f lambda: 8.0 cm is 0.080 m and 600 nm is 6.00 x 10-7 m. Nearly every lost mark on this topic comes from a wavelength left in centimetres or nanometres.
  • Frequency is fixed by the source and never changes as a wave travels; if a wave changes speed, the wavelength changes with it. Quote the equation lambda = v / f as your reason rather than saying the wavelength 'changes to match'.
  • For a cathode-ray oscilloscope, write down the scale before you read the screen: period = divisions for one cycle x time-base, and amplitude = peak-to-peak divisions x y-gain / 2. The factor of a half is the mark most often dropped.
  • In a Doppler calculation, choose the sign first and say why: the minus sign with an approaching source because it moves into its own wavefronts, and the plus sign with a receding source. Give the observed frequency in Hz and show that it is higher than f_s when approaching, which is a quick check on your own answer.
  • In a Malus's law question, identify which angle is being asked for: for plane-polarised light it is the angle between the incident plane of polarisation and the filter axis, and for a chain of filters you must work through them one at a time, taking the output of each filter as the input to the next.
  • Learn the electromagnetic regions in order with a rough wavelength for each, and keep them in one band: radio, microwave, infrared, visible, ultraviolet, X-rays, gamma. Saying that the spectrum is 'made of different waves' loses the point that they are all electromagnetic, all transverse and all travelling at c in free space.

Why students lose marks here

A student wroteA student wrote: 'The intensity of a wave is proportional to its amplitude, so a wave with twice the amplitude has twice the intensity.'
Why it loses marksIntensity is proportional to the square of the amplitude, not to the amplitude. Doubling the amplitude multiplies the intensity by 2 x 2 = 4, and the relationship is written as I is proportional to A^2.
The answer that scoresThe intensity of a progressive wave is proportional to the square of its amplitude. A wave of amplitude 2.0 mm carrying an intensity of 0.50 W m-2, when its amplitude is changed to 4.0 mm, has an amplitude twice as large and therefore an intensity of 4 x 0.50 = 2.0 W m-2. Going the other way, to double the intensity you must increase the amplitude by a factor of 2^0.5, which is 1.41.
The properties of a progressive wave, and the oscilloscope traceA progressive wave transfers energy without transferring matter: every particle oscillates about its ownfixed position.Displacement against distance00.100.200.300.40distance x / mdisplacementy / mamplitudewavelengthcentre linetroughcrestOscilloscope trace: y-gain and time-base2.5 divisionsT = 4.0 divisionstime-base 5.0 ms per divisiony-gain 2.0 V per divisionT = 4.0 x 5.0 = 20 ms, so f = 50 Hzamplitude A = 2.5 x 2.0 = 5.0 VIntensity is proportional to the square of the amplitude, so twice the amplitude carries four timesthe intensity.The y-gain turns a height in divisions into a voltage and the time-base turns a width in divisionsinto a time.The amplitude is from the centre line to a crest: a peak-to-peak height of 6.0 divisions at 2.0 Vper division is 6.0 V.
A wave drawn as displacement against distance, with the amplitude shown as the distance from the centre line to a crest and the wavelength as the distance between two successive crests: the intensity carried by such a wave is proportional to the square of the amplitude marked on the diagram.
A student wroteA student wrote: 'The trace on the oscilloscope has a peak-to-peak height of 6.0 divisions and the y-gain is 2.0 V per division, so the amplitude of the signal is 12 V.'
Why it loses marksMultiplying the height by the y-gain gives the peak-to-peak voltage, which is the distance from the bottom of a trough to the top of a crest. Amplitude is measured from the centre line to a crest, so it is half of that value.
The answer that scoresPeak-to-peak voltage = 6.0 x 2.0 = 12 V, and the amplitude is half of that: 12 / 2 = 6.0 V. Write the two steps out separately in the exam, because the halving is worth a mark of its own. The same trap applies to the time-base, where the divisions for one complete cycle, not for a half cycle, give the period.
A student wroteA student wrote: 'The siren emits 500 Hz and the ambulance approaches at 30 m s-1, so f_o = 500 x 340 / (340 + 30) = 459 Hz, which is higher than 500 Hz.'
Why it loses marksThe wrong sign was used, and the answer does not even agree with the student's own statement that a higher frequency is heard. An approaching source moves into its own wavefronts, so the wavefronts are crowded together and the observed wavelength is shorter, which means the denominator must be smaller than v and the observed frequency larger than f_s.
The answer that scoresFor an approaching source use the minus sign: f_o = f_s v / (v - v_s) = 500 x 340 / (340 - 30) = 500 x 340 / 310 = 548 Hz, which is correctly higher than the source frequency. After the ambulance passes, the plus sign applies: f_o = 500 x 340 / (340 + 30) = 500 x 340 / 370 = 459 Hz, which is lower than 500 Hz. Always check the sign of the answer against the direction of motion.
A student wroteA student wrote: 'Plane-polarised light of intensity 400 W m-2 meets a filter at 60 degrees to its plane of polarisation, so I = 400 x cos 60 = 200 W m-2, and a beam of sound could be polarised in the same way.'
Why it loses marksMalus's law uses cos squared, not cos, so the transmitted intensity is 400 x (0.50)^2 = 100 W m-2 and not 200 W m-2. The second part is wrong for a different reason: polarisation means restricting oscillations to one plane, and sound is longitudinal, with its oscillations along the direction of travel, so there is no plane to restrict them to.
The answer that scoresI = I0 cos^2 theta = 400 x cos^2 60 = 400 x (0.50)^2 = 400 x 0.25 = 100 W m-2. Only transverse waves can be polarised, which is why polarising filters work on light, on microwaves from a transmitter and on the transverse waves in a stretched string, but never on sound.

Where this fits in AS & A Level Physics

This page is one topic of Cambridge AS & A Level Physics 9702. AS & A Level Physics is the course home: every area and every topic in Cambridge's own order, the papers you sit and what each one tests, and the exam-practice tools. The full topic list, with each topic linked to its revision page, is on the AS & A Level Physics syllabus page.