Waves and polarisation

Transverse waves

When you think of waves, the first image in your mind is likely to be a transverse wave like a water wave. In a transverse wave the oscillations or vibrations are perpendicular to the direction of energy transfer (Figure 3).

As the wave moves from left to right, the oscillations are at 90° to the direction of the wave's movement - up and down, or side to side, or at any orientation as long as it is in a plane that is perpendicular to the direction of energy transfer.

Transverse waves have peaks and troughs where the oscillating

particles are at a maximum displacement from their equilibrium position (Figure 4).

Examples of transverse waves include:

⚫ waves on the surface of water

⚫ any electromagnetic wave- radio waves, microwaves, infrared,

visible light, ultraviolet, X-rays, and gamma rays

⚫ waves on stretched strings

⚫ S-waves produced in earthquakes.

transfer of energy

particles in their equumption (no w

energy tran

▲ Figure 3 A transverse wave travelling along a horizontal slinky spring

90 in the direction of energy trans

A Figure 4 For a transverse wave, the particles of the medium vibrate in a plane or

to the firection of energy transfer

Longitudinal waves

In longitudinal waves the oscillations are parallel to the direction of energy transfer.

transfer of energy

rarefaction

compression

particles in their equilibrium positions (no wave)

energy transfer

Figure 5 A longitudinal wave looks very different to a transverse wave

Examples of longitudinal waves include:

sound waves

P-waves produced in earthquakes.

Longitudinal waves are often called compression waves. When they travel through a medium they create a series of compressions and rarefactions.

each particle vibrates parallel to the direction of energy transfer

Figure 6 In a longitudinal wave particles vibrate parallel to the action of energy transfer

The electromagnetic spectrum

The different types of EM wave are classified by wavelength. The full range of EM waves is called the electromagnetic spectrum and ranges from radio waves, with the longest wavelength, to gamma rays, with the shortest. Some radio waves have wavelengths longer than a million metres, whilst high-frequency gamma waves have wavelengths of just 10-16m (less than the diameter of an atomic nucleus).

As you can see from Figure 3, the wavelength ranges of X-rays and gamma rays overlap. Unlike other parts of the spectrum, these EM waves are not classified by their wavelength, but by their origin. X-rays are emitted by fast-moving electrons, whereas gamma rays come from the unstable atomic nuclei.

An EM wave is an example of a transverse wave, but it is a little more complex than a ripple on a pond. EM waves can be thought of as electric and magnetic fields oscillating at right angles to each other (Figure 2).

Reducing wavelength (values in m)

106

10-1

10-3

7x10-7 4×10-7

10-8

X-rays

Radio waves

View along direction of wave

Microwaves

Infrared

Visible

Ultraviolet

Oscillating electric field

Oscillating magnetic field

Direction of wave

Figure 2 An electromagnetic wave does not need a medium to be able to transfer energy

10-10 10-13

<10-16

Gamma rays

The possible range of wavelengths of X-rays and gamma rays overlap

Figure 3

700 nm

400 nm

The electromagnetic spectrum ranges from radio waves down to gamma rays

Properties of EM waves

Q.5 ab

Waveleng of em wa

Like all waves, EM waves can be reflected, refracted, and diffracted. As EM waves are transverse waves they can also be plane polarised (see Topic 11.7, Polarisation of electromagnetic waves).

All EM waves travel at the same speed through a vacuum (c), 3.00 x 10 ms. This is a very close approximation to their speed through air. Therefore, we can modify the wave equation for EM waves to

C=fh

Polarising filters

Polarised light

Light is polarised when it passes through a polarising filter. The filter only allows through electric field oscillations in one plane because the filter absorbs energy from oscillations in all other planes. The metal grille shown in Figure 5.16 is a good model for a polarising light filter, such as that shown in Figure 5.17a. Long molecules of quinine iodosulphate are lined up, and electrons in these molecules affect the light in the same way as the microwaves are affected by the grille. Polarised light is less intense than unpolarised light because only half the energy is transmitted through the filter. If a second polarising filter is held at right angles to the original filter, all oscillations are blocked and no light is transmitted. This is called crossing the polarisers.

Unpolarised light

Unpolarised ray

Outdoor television and FM radio aerials must be correctly aligned for the best reception. Transmitters generate plane-polarised electromagnetic waves, which are picked up most effectively by a receiver with the same alignment. It is possible to reduce interference between nearby transmitters if one of the two transmitters is aligned vertically, and the other is aligned horizontally as shown in Figure 5.19.

Horizontally mounted aerial

Vertically mounted aerial.

Figure 5.19 A horizontally mounted serial absorbs horizontally polarised signals efficiently.

If a wave has a frequency of 5.0 Hz, each second there are 5 complete oscillations. If each wave has a wavelength of 2.0m, then the wave has travelled 10m from the source in that time. Therefore, its speed must be 10ms. So, for a frequency f, the wave would have travelled a distance of fx λ per second, that is, the wave speed v.

We can also see from the definitions that the period of oscillation and the frequency of a wave are reciprocals of each other. If a wave has a frequency of 2.0 Hz, there are two complete wave cycles each second; therefore, the period for each wave must be 0.50s or Therefore, 2.0 we have a second important equation relating the frequency f of a wave in Hz to its period T in s.

Worked example: Findin musical note

Phase difference

Particles along a wave that move in phase move in the same direction with the same speed. The particles have the same displacement from their mean position. Particles in phase are separated by a whole number, n, of wavelengths, nλ.

Particles along a wave that move out of phase are at different points in their cycle at a particular time.

Particles along a wave that move in antiphase move in opposite directions at

The same speed. The particles have opposite displacements from their mean position. Particles moving in antiphase are separated by a distance of a whole number, n, of wavelengths plus an extra half wavelength, +슬

Displacement/cm

A wave repeats its cycle at regular time intervals. The phase of a wave describes the fraction of a cycle completed compared to the start of the cycle. Particles in parts of a wave that are moving at the same speed in the same direction are in phase. They are out of phase if they are at different points in their cycle at a particular time. Particles in parts of a wave that move in opposite directions and at exactly the same speed are moving in antiphase, or completely out of phase.

Direction of wave travel

In phase: EM anti phase: EI

Out of phase: E and M. With all other particles.

10

B

K undisturbed position of slinky

Distance

-20-

Along slinky /metres

Figure 5.6 The displacement of a slinky along its length. The arrows on the graph show the direction of motion of the slinky, the wave is travelling from right to left.

One wavelength is the shortest distance between two points that are moving in phase. Points that are a whole number of wavelengths apart move in phase; points that are half a wavelength or 1.5 wavelengths or 2.5 wavelengths, etc. Move exactly out of phase (or in antiphase) with

Each other.

Figure 5.6 shows a transverse wave travelling on a slinky. In this diagram,

Points E and M are in phase as they move in the same direction at the same speed and are at the same point in their cycle. These three points

Are out of phase with all other points points E and I are moving in antiphase, because they are moving in opposite directions with the same speed.

Phase difference is measured as a fraction

Of the wave cycle between two points along a wave, separated by a distance x. =2 rad

One complete rotation involves turning through 2x radians, so 2x (or 6.28) radians is equivalent to 360°

Any phase difference can be measured as an angle in degrees or radians. The motion of many waves can be described using a sine function. Since a sine function repeats itself over an angle of 360° or 2 radians, we say the phase change of a wave during one complete cycle is 360° or 2x radians. You can read more about radian and degree measure in Chapter 15.

The maths of phase differences

One complete rotation of a circle involves turning through an angle of 360° You can also measure this angle in radians. One complete rotation involves turning through 2 radians, so 2x radians is equivalent to 360°.

The motion of the wave in Figure 5.6 is sinusoidal with a time period of T. When time t, equals T, one cycle has been completed, so the value of the angle in the sine function must be 2x radians. This happens when t = T for

The angle, 2at

T

# represents the fraction of an

20

Wave properties: Reflection Refraction Diffraction

Reflection

Reflection occurs when a wave changes direction at a boundary between two different media, remaining in the original medium.

A simple example is light reflecting off a mirrored surface (Figure 1). The light waves remain in the original medium (the air). We often sar represent the direction taken by a wave as a ray (like those in Figure 1). The ray shows the direction of energy transfer and so the path taken by the wave.

The law of reflection applies whenever waves are reflected. It states that the angle of incidence is equal to the angle of reflection.

When waves are reflected their wavelength and frequency do not change. This can be seen by reflecting water waves using a ripple tank. In Figure 2 we have represented the wave as a series of wavefronts. Each wavefront is a line joining points of the wave which are in phase. They can be thought of as the peak of each ripple. By definition (see Topic 11.2, Wave properties) the distance between wavefronts is equal to the wavelength of the wave.

Like plane (straight) waves, circular waves – like ripples from dropping a stone into a pond – can be reflected too (Figure 3).

Refraction

Refraction occurs when a wave changes direction as it changes speed when it passes from one medium to another. You will look at refraction of light in more detail in Topic 11.8. Refractive index.

Whenever a wave refracts there is always some reflection off the surface (partial reflection).

If the wave slows down it will refract towards the normal, if it speeds up it refracts away from the normal. Sound waves normally speed up when they enter a denser medium, whereas electromagnetic waves. Like light, normally slow down. This results in the waves refracting in different directions.

Unlike reflection, refraction does have an effect on the wavelength of the wave, but not its frequency. If the wave slows down its wavelength decreases and the frequency remains unchanged, and vice versa.

Normal

Normal

Incident ray

Partial flection

Refracted

If the wave slows down it refracts toward the normal

If the wave speeds up it retracts away from the normal

Incident

The normal

▲ Figure 4 Achange in wave speed causes the wave to refract towards or away from

Reflected

Angle of

Reflection

Mincred surface

Normal

Angle of incidence

Incident ray

A Figure 1 When waves are reflected

They always obey the law of reflection (note that all angles are measured to the normal

Incident wave

Reflected wave

A Figure 2 Plane waves con be mode by bobbing a ruler up and down in a ripple tank-when they reflect off a surface their frequency and wavelength remain unchanged

Wave source

A Figure 3 When circular waves reflect off surfaces their wavelength and frequency remain the same, just like plane waves

3

Boundary

7 reflection

Partial

A Figure 5: When woves refract their wavelength changes: if the wave slows down, as shown here, the wavelength decreases so the wavefronts get closer together

Diffraction

Diffraction is a property unique to waves. When waves pass through a gap or travel around an obstacle, they spread out.

All waves can be diffracted. The speed, wavelength, and frequency of a wave do not change when diffraction occurs.

How much a wave diffracts depends on the relative sizes of the wavelength and the gap or obstacle. Diffraction effects are most significant when the size of the gap or obstacle is about the same as the wavelength of the wave. This is why sound diffracts when it passes through a doorway, allowing you to hear conversations around the corner. The wavelength of the sound is similar to the size of the gap. However, light has a much smaller wavelength, so it does not diffract through such a large gap. In order to observe the diffraction of light we need a much smaller gap.

A Figure 2 The size of the gap compared with the wavelength of the wave offects how much diffraction takes place, and the spacing between the wavefronts shows that there is no change in wavelength

Diffraction occurs when waves spread out after passing through a gap or round an obstacle. The effect can be seen in a ripple tank when straight waves are directed at a gap, as shown in Figure 4.

⚫ The narrower the gap, the more the waves spread out. ⚫ The longer the wavelength, the more the waves spread out.

Snell’s Law and Refractive Index

However, the speed of light in a vacuum is almost exactly the same as the speed of light in air. This leads to two useful and accurate approximations:

⚫ the refractive index of air » 1 the refractive index of a material

Speed of light in air speed of light in the material

The refractive index of window glass is about 1.5, which means that the speed of light in air is about 1.5 times faster than it is in glass. The refractive index of water is 1.33, so light travels 1.33 times faster in air than it does in water

When light travels from one material to another (other than air we can define the relative refractive index between them as follows:

Where n₂ = refractive index between materials 1 and 2

4 is the speed of light in material 1

Is the speed of light in material 2.

However, if we divide the top and bottom of the equation above by the

Speed of light in air, c, we get:

(c)

Since-and-

It follows that:

(1/m)

The advantage of this formula is that we only need to know one refractive index for a material, and we can calculate a new refractive index when light passes between any pairs of materials other than air.

This last example leads to an important result, which shows that the refractive index travelling from water to glass is the reciprocal of the refractive index when light travels from glass to water.

In general:

1 2

Law of refraction

For many hundreds of years scientists have studied refraction and have been able to predict the angles of refraction inside transparent materials. Willebrord Snellius was the first person to realise that the following ratio is always constant for all materials:

Sin 8 sin 0, = constant

Where 0 is the angle of incidence and e, is the angle of refraction. This is now known as Snell’s law

At a later date it was understood that this constant ratio is the refractive index between the two materials that the light passes between.

More generally, Snell’s law of refraction is stated as

Sin 0 sin 02

Where

12 = refractive index between materials 1 and 2 0,= angle of incidence in material 1 0,= angle of refraction in material 2

Or, since

It follows that:

N2 sin e 12 = n₁ sin 0₂ n₁ sin 0,= n, sin 0₂ and

Where

N = refractive index of material 1

N2= refractive index of material 2

Comparing glass to air refraction with air to glass refraction

In Figure 3, notice that the angle of refraction of the light ray emerging from the block is the same as the angle of incidence of the light ray entering the block. This is because the two sides of the block at which refraction occurs are parallel to each other.

• If i, and r, are the angles of incidence and refraction at the point where the light ray enters the block, then the refractive index of the glass n = sini sinh

At the point where the light ray leaves the block, i₂ = r, and r₂ = i₁, SO sini1 sinh, n

Refraction of a light ray by a triangular prism

Figure 4 shows the path of a monochromatic light ray through a triangular prism. The light ray refracts towards the normal where it enters the glass prism then refracts away from the normal where it leaves the prism.

Optical fibre is a thin glass (or plastic) fibre that transmits light.

Optical fibres

An optical fibre is a thin glass (or plastic) fibre that transmits light or infrared radiation. These waves travel through the glass but are trapp inside by repeated total internal reflection. This is possible if each reflection has an angle of incidence larger than the critical angle. The critical angle depends on the ratio between the refractive index of the optical fibre and its cladding (coating).

Step index optical fibre is an optical fibre with a uniform refractive index in the core and a smaller uniform refractive index for the cladding

A step index optical fibre has a central core with a uniform refractive index, while the cladding has a different, smaller, refractive index. By choosing a suitable material for the core and cladding, only certain wavelengths of light or infrared radiation can travel through the fibre by total internal reflection.

Low refractive index cladding

High refractive index core

Material and modal dispersion

A spectrum seen using a prism is caused by dispersion. Different colours of light travelling through the glass slow down by different amounts The refractive index varies with wavelength. A similar effect happens inside optical fibres. As a signal travels within an optical fibre, it disperses. Two types of dispersion occur in step index optical fibres:

⚫ material dispersion occurs because the refractive index of the optical fibre varies with frequency. Different wavelengths of light in the signal travel at different speeds. This causes a sharp pulse to spread into a broader signal. Therefore, the duration of each pulse increases. This is called pulse broadening. Pulse broadening is a problem because it limits the maximum frequency of pulses and therefore the bandwidth available.

Violet light travels move sloudy then red in

Solution: use monochromatic light

Figure 5.26 Optical fibre.

Material dispersion is the spreading of a signal caused by the variation of refractive index with wavelength.

Pulse broadening occurs when the duration of a pulse increases as a result of dispersion in an optical fibre.

Modal dispersion is the spreading of a signal caused by rays taking slightly different paths in the fibre.

Figure 5.27 Material dispersion: different frequencies have a different refractive index.

Modal dispersion occurs when rays inside an optical fibre take slightly different paths. Rays taking longer paths take longer to travel through the fibre, so the duration of the pulse increases and the pulse broadens. Modal dispersion is significant in multimode fibres, because these fibres are broad enough to allow rays to take different paths. For communications, monomode fibres are used. These have a very narrow core, so that light is very nearly confined to one single path along the axis of the cable. Solution: we necrow core (manomode)

Travels lurther

Travels less far

Figure 5.28 Modal dispersion: rays can take more than one path in the fibre.

Path difference and Phase difference

Interference patterns contain a series of maxima and minima. At a maximum the waves interfere constructively, at a minimum they interfere destructively. For example, a pair of loudspeakers emitting coherent sound waves produces an interference pattern with regions that are louder (maxima) and others that are quieter (minima) than the original waves. The same effect can be seen in Figure 2 with water waves. In places the water waves have increased amplitude (maxima) and in others the amplitude appears to be zero (minima).

In both cases these maxima and minima are a result of the two waves having travelled different distances from their sources. This difference in the distance travelled is called the path difference.

Figure 3 shows two sources emitting coherent waves. The wavelength of the progressive wave is 2. If the path difference to a point is zero or a whole number of wavelengths (0, 2, 22, ..., nλ, where n is an integer), then the two waves will always arrive at that point in phase. This produces constructive interference at that point. The resultant wave at this point has maximum amplitude.

If the path difference to a point is an odd number of half wavelengths (λ...... (n+), where n is an integer) the two waves will always arrive at that point in antiphase. This produces destructive interference. The resultant wave at this point has minimum amplitude.

At the central maxima, shown in Figure 4, the path difference is zero and so the phase difference is zero. At the first-order maxima the path difference is one whole wavelength, so the phase difference is 360° or

2л radians. The peaks from the first waves perfectly line up with the peaks of the second waves and so constructive interference occurs.

At the first-order minima the path difference is half a wavelength, a phase difference of 180° or a radians. The peaks from the first waves line up with the troughs of the second waves, resulting in destructive interference.

Young’s Double slit Experiment

Constructive interference (crest on crest)

Young’s double slit experiment-Required practical number 2

Destructive interference (crest on trough)

Constructive

Young’s double slit experiment demonstrates interference between coherent light sources, thus showing the wave nature of light. The experiment uses twe coherent sources of light waves, produced from a single source of hight, which then pass through two very narrow, parallel slits. The light diffracts (spreads) through the slits, producing an interference pattern of fringes on a screen Interference occurs because the waves overlap and superpose in a stable patter

-destructive

Constructive

Light has such a short wavelength it is difficult to see its interference patterns. This is why Young’s double slit experiment works best using a blacked out room and a very bright white light source, or a laser as a sou of intense single-wavelength (monochromatic) light. The shits must be very narrow and less than 1 mm apart. An interference pattern can be seen whe patches of bright light alternate with regions of darkness. These correspon to areas of constructive and destructive interference. These patterns are called fringes. The interference fringes are visible on a screen placed at lea 1m away from the slits. If the screen is further away, the fringe separation increases but their appearance becomes fainter.

Figure 6.6 Interference in a ripple tank

Central axs

Figure 6.7 Young’s double slit experiment.

Light from each slit travels a slightly different route to the screen, creating a path difference as shown in Figure 6.7. Dark fringes occur where there is destructive interference (the path difference between the two slits is (n+1)). Bright fringes occur where there is constructive interference (the path difference between the two slits is nλ).

Fringes are ordered. The order of the central bright fringe is n = 0. The order of the fringes closest to the central fringe is n = 1; the order of the next pair of fringes is n = 2, etc.

Referring to Figure 6.7, the condition for constructive interference (bright fringes) is:

Ssin 0 = nλ

Where s = spacing between the slits

@-angle from the beam towards the screen

N = a whole number

2. = wavelength of light.

The extra distance travelled by the waves leaving S, is ssin 6, and for constructive interference this distance (or path difference) must be a whole number of wavelengths, nλ

Wavelength and colour

In the double slit experiment, the fringe separation depends on the colour of light used. Each colour of light has its own narrow band of wavelengths. The fringe separation is greater for red light than for blue light because red

Light has a longer wavelength than blue light.

White light fringes

For white light, each component colour of white light produces its own fringe pattern, and each pattern is centred on the screen at the same position. As a resul

⚫ the central bright fringe is white because every colour contributes at the centre of the pattern

⚫ the inner bright fringes are tinged with blue on the inner side and red on the outer side. This is because the red fringes are more spaced out than th blue fringes so the two fringe patterns do not overlap exactly

⚫ the outer bright fringes become fainter with increasing distance from the centre. Also, the fringes merge because the colours overlap.

Single slit diffraction

Diffraction of light by a single slit can be demonstrated by directing a parallel beam of light at the slit. The diffracted light forms a pattern that can be observed on a white screen. The pattern shows a central fringe with further fringes either side of the central fringe, as shown in Figure 2. The intensity of the fringes is greatest at the centre of the central fringe. Figure 2 also shows the variation of the intensity of the diffracted light with the distance from the centre of the fringe pattern.

3

2

1

1

2

3

Bright fringe

Dark fringe

Intensity

Ли position on screen

0

▲ Figure 2 Single slit diffraction

Note

⚫ The central fringe is twice as wide as each of the outer fringes (measured from minimum to minimum intensity).

• The peak intensity of each fringe decreases with distance from the centre.

. Each of the outer fringes is the same width.

The outer fringes are much less intense than the central fringe.

Single slit diffraction and Young’s fringes

In the double slit experiment in Topic 5.4, light passes through the two slits of the double slit arrangement and produces an interference pattern. However, if the slits are too wide and too far apart, no interference pattern is observed. This is because interference can only occur if the light from the two slits overlaps. For this to be the case:

⚫ each slit must be narrow enough to make the light passing through it diffract sufficiently

⚫the two slits must be close enough so the diffracted waves overlap

On the screen.

In general, for monochromatic light of wavelength A, incident on two slits of aperture width a at slit separation s (from centre to centre).

⚫ the fringe spacing of the interference fringes, w= AD

The width of the central diffraction fringe, W= slit-screen distance. 2AD, where D is the a

Figure 4 shows the intensity variation with distance across the screen in terms of distance from the centre of the pattern. Using the expressions above, you should be able to see that only a few interference fringes will be observed in the central diffraction fringe if the slit separation s is small compared to the slit width a.

Without single slit effect

Intensity

A Figure 4 Intensity distribution for Young’s fringes

0

21.Dis distance from centre

Diffraction gratings

We can see interference patterns using a diffraction grating. A diffraction grating has thousands of slits spaced very closely together. The slits are very narrow so that the light diffracts through a wide angle. The pattern is a result of light overlapping (or superposing) and interfering from a great number of slits.

When light passes straight through the slits, the path difference between each slit is zero, so the light from each slit is in phase and the overlapping

White paper

Diffraction

Grating

Ray box

Filter

3rd order

2nd order

1st order

Zero order

1st order

2nd order

3rd order

A Figure 1 The diffraction grating

Applications of diffraction gratings Diffraction gratings are used to separate light of different wavelengths in great detail (high resolution). This is useful when the diffraction grating is part of a spectrometer and used for investigating atomic spectra in laboratory measurements. Diffraction gratings are also used in telescopes to analyse light from galaxies. CDs and DVDs use diffraction gratings but they reflect light from a grating rather than transmitting it. The light source is a laser installed inside the CD or DVD drive.

Stationary waves only form on a guitar string at specific frequencies. Where the waves are in phase, the displacements add to form a peak or a trough of double the original amplitude. Where the waves are in anti-phase, their displacements cancel out. For stationary waves, the positions of maximum and minimum amplitude remain in the same places, with particles, at antinodes, vibrating rapidly between their positions of maximum displacement. This is why the waves don’t appear to progress along the string.

A stationary wave alternates between the positions shown by the red wave and the blue wave in Figure 6.9.

Nodes and antinodes

Nodes and antinodes are at fixed points along the guitar string. At a node, the amplitudes of the two progressive waves moving in opposite directions always cancel out so the particles do not oscillate at all. At an antinode, the amplitudes add together and the guitar string is displaced between the peak and trough during a cycle. The amplitude of the peak or trough is double the amplitude of the two progressive waves. Nodes are always separated by half a wavelength, as are antinodes. The displacements of particles in positions between the nodes and antinodes vary but do not remain zero or reach the amplitude of an antinode during the cycle according to their positions.

How stationary waves form

Figures 6.10 (a) to (d) show how stationary waves form on a guitar string. Wave 1 and wave 2 are progressive waves travelling in opposite directions along the string with the same frequency and amplitude. The amplitude of the wave 3 is the amplitude of the two waves superimposed.

The diagrams show a sequence of snapshots of the wave at different times during one complete cycle.

Energy transfer

Progressive wave

Stationary wave

No net energy transfer

Energy transferred in the direction of the wave

Wavelength

Minimum distance between two adjacent points oscillating in phase, for example, the distance between two peaks or two compressions

Twice the distance between adjacent nodes (or antinodes) is equal to the wavelength of the progressive waves that created the stationary wave

Phase differences

The phase changes across one complete cycle of the wave

All parts of the wave between a pair of nodes are in phase, and on different sides of a node they are in antiphase

AB in phase •phase diff=0

All parts of A in phase All ports of B in phase A B out of phase

Amplitude

All parts of the wave have the same amplitude (assuming no energy is lost to the surroundings)

Maximum amplitude occurs at the antinode then drops to zero at the node

All partides will experience max Amplitude

All particles experience -1/2 different Amplitudes

A guitar string can support different modes of vibration for stationary waves. The different modes of vibration are called harmonics. Harmonics are numbered: the first harmonic is the mode of vibration with the longest wavelength. The second harmonic is the mode of vibration with the next longest wave.

The frequency of the vibration is found using

Where f= frequency of the harmonic in Hz

V = speed of wave in ms”¹

A-wavelength of harmonic in m

In Figure 6.11.

⚫ the first harmonic on a string includes one antinode and two nodes. Stringed instruments have nodes at each end of the string as these points are fixed. For a guitar string of length 1, the wavelength of the lowest harmonic is 21; this is because there is one loop only of the stationary wave, which is a half wavelength. Therefore the frequency is:

⚫the second harmonic has three nodes and two antinodes; the wavelength

Is I and frequency is:

--or 24

The third harmonic has four nodes and three antinodes;

21/3 and frequency is:

J3—3 or 3/1

String at maximum displacement

Node

Frequency generator

Vibrator (nearly at a node)

Pulley

N = node A antinode

Weight-

(dotted line shows string half a cycle earlier)

A

String

N

No

A First harmonic

N.E

B Second harmonic

N

String

N4

C Third harmonic

N

String

A Figure 1 Stationary waves on a string

The first harmonic on a string

For waves travelling along a string in tension, the speed of a wave is given by:

Where:

T = tension in the string in N

= mass per unit length of the string in kgm-l

Using Figure 6.11, you can see that the first harmonic for a stationary wave

On a string is half a wavelength. The wavelength of the first harmonic is 21

For a string of length 1.

We can combine the equation above and the wave equation as follows:

Therefore:

Y2

V=f2 =f2l for first harmonic ===f of first harmonic ス

T

This equation gives the frequency of the first harmonic where:

F=frequency in Hz

T = tension in the string in N

Μ= mass per unit length of the string in kgm-1

1= the length of the string length in m

EXAMPLE

A string is stretched between two points 0.45m apart. Al Calculate the frequency of the first harmonic when the tension in the string is 80N and the mass per

80 N

2x0,45m 3.2 10-kgm = 175.7 Hz or 176 Hz to 3 st

Blil Frequency is proportional to, so doubling the length halves the frequency (to 88 Hzl. Iil Frequency is proportional to T. So doubling the tension increases the frequency by a factor of 12 Ito 248 Hz).

Unit length is 3.2gm-1 b) Predict the effect on frequency of: i) doubling the length of the string

Iil doubling the tension in the string.

Answers

1 T 21

Harmonic

Shape

Frequency/Hz

Frequency as a multiple off

8 20

Wavelength of the progressive wave (where Lis the length of the string)

21

L

First harmonic

Is the fundamental

4

40

60

45

100

Stationary waves: Sound waves and musical instruments

Longitudinal waves such as sound waves also form stationary waves. Sound waves are pressure waves with particles vibrating in the direction the wave travels. Stationary wave diagrams for sound waves show the amplitude for particles vibrating longitudinally in the air column. The amplitude is greatest at the open end of pipes, where there is an antinode. The particles cannot vibrate at a closed end and so there is a node. If the pipe has two open ends, the stationary wave has at least two antinodes, at either end of the pipe.

Figure 6.12 Stationary waves for sound waves in open-ended tubes. When there is one open end, there is an antinode at the open end and a node at the closed end. In a pipe that is open at both ends, there is a node in the middle and antinodes at both ends.

Musical instruments include stringed instruments, where stationary waves form on the string when it is plucked or bowed. Stationary waves also form when the air column in organ pipes or wind instruments is forced to vibrate. The harmonic frequencies available to players will vary with different instruments, even if the air column is the same length, depending on whether the instrument has one or both ends open.

A

F=fo

Motion of air particles

A

N

N

1=2

A

32 1=2

N

A

N

F=3fo

F=2fo

A

N = node A antinode

1=4

F=fo

N

1=

F=3fo

N

N

A

F=5fo

N

ACTIVITY

Interference patterns from microwaves

Electromagnetic waves are carried by oscillating electric and magnetic fields. For clarity of explanation we will consider only the electric fields in this chapter When two microwaves overlap and superpose, interference can take place. Where the electric fields of the two waves are in phase, there is constructive interference and regions of high intensity. The intensity is low where the electric components are out of phase. The changes in intensity are detected using a probe and this information can be used to calculate the wavelength of the microwaves.

A student sets up a microwave transmitter, microwave detector and aluminum plate placed about 40 cm from the transmitter as shown in Figure 6.16. Microwaves that reflect off the metal plate interfere with the transmitted microwaves. The interference pattern has regions of low intensity where the waves are out of phase, and regions of high intensity where the waves are in phase. She uses a receiving aerial, or probe, to detect nodes.

The student discovers that the distance between successive regions of destructive interference is 1.4cm.

1 By considering the path difference between the wave reaching the receiver directly from the transmitter, and the wave reaching the receiver

Answer

After it has been reflected from the metal plate, deduce the wavelength of the microwaves.

2 The student takes measurements to check the wavelength. Describe how she could reduce

Uncertainties in her measurements.

3 Explain the student’s could explain her observations. Using ideas about stationary waves.

Receiving aerial

Reflecting board

Mircowave transmitter

Figure 6.16 An experiment demonstrating interference of

Microwaves

Interference patterns from microwaves

1 Suppose the probe is placed at a position of minimum intensity. If it is moved towards the plate by a distance of ½, the distance from the transmitter increased by ½, but the distance from the reflecting late decreases by A/2. So, the change in path difference is A and there will again be a minimum. Thus the wavelength is twice the distance between adjacent minima: λ = 2.8 cm.

2 The uncertainty is reduced by measuring over several minimums – eg if the receiver is moved from one minimum to the 7th, a distance of 3 wavelength has been measured. Now the uncertainty has been reduced.

3 Waves travelling directly from the transmitter meet waves travelling from the opposite direction, having been reflected from the metal plate. When two progressive waves meet, stationary waves are set up. The minima are nodes, and the maxima are the antinodes. Because the reflected waves have a smaller amplitude than the waves which travel directly to the receiver, the nodes will not be places of exactly zero intensity.

The Oscilloscope

An oscilloscope consists of a specially made electron tube and associated control circuits. An electron gun at one end of the glass tube emits electrons in a beam towards a fluorescent screen at the other end of the tube, as shown in Figure 1. Light is emitted from the spot on the screen where the beam hits the screen.

Filament

Anode

Tube

Screen

Spot

X₂ behind X

A Figure 1 An oscilloscope tube

How to use an oscilloscope

The position of the spot of light on the screen is affected by the pd across either pair of deflecting plates. With no pd across either set of deflecting plates, the spot on the screen stays in the same position. If a pd is applied across the X-plates, the spot deflects horizontally. A pd across the Y-plates makes it deflect vertically. In both cases, the displacement of the spot is proportional to the applied pd.

To display a waveform:

The X-plates are connected to the oscilloscope’s time base circuit which makes the spot move at constant speed left to right across the screen, then back again much faster. Because the spot moves at constant speed across the screen, the x-scale can be calibrated, usually in milliseconds or microseconds per centimetre.

The pd to be displayed is connected to the Y-plates via the Y-input so the spot moves up and down as it moves left to right across the screen. As it does so, it traces out the waveform on the screen. Because the vertical displacement of the spot is proportional to the pd applied to the Y-plates, the Y-input is calibrated in volts per centimetre (or per division if the grid on the oscilloscope screen is not a centimetre grid). The calibration value is usually referred to as the Y-sensitivity or Y-gain of the oscilloscope.

Figure 2 shows the trace produced when an alternating pd is applied to the Y-input. The screen is marked with a centimetre grid.

1 To measure the peak pd, observe that the waveform height from the bottom to the top of the wave is 3.2cm. The amplitude (ie... peak height) of the wave is therefore 1.6 cm. As the Y-gain is set at

5.0 V cm, the peak pd is therefore 8.0V (= 5.0V cm x 1.6cm). 2 To measure the frequency of the alternating pd, we need to measure the time period 7 (the time for one full cycle) of the waveform. Then we can calculate the frequency fas f= 1/T.