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13.2Transverse and longitudinal waves

Transverse waves

How do the particles of a medium move when there is a wave in the medium? We will now see the connection between simple harmonic motion (SHM) and wave motion: as a wave propagates through a medium, the particles of the medium perform SHM with a frequency that is the same as that of the wave.

In transverse waves the particles of the medium perform simple harmonic oscillations in a direction at right angles to the direction of energy transfer.

KEY POINT

A wave is transverse a wave where the displacement of the particles of the medium is at right angles to the direction of energy transfer if the displacement of the particles of the medium is at right angles to the direction of energy transfer.

The diagram in Figure 13.6 shows three particles in the medium (the rope). The wave moves from left to right. The wave is transverse which means that the particles are moving along the double headed arrows as the wave progresses.

Figure 13.6: Particles in the medium move strictly at right angles to the direction of energy transfer.

To understand precisely how the particles in the medium move as the wave propagates through the medium consider Figure 13.7. Think of each graph as a photograph of the rope. Photographs are taken every 0.5 ms. Something is shaking the left end of the rope, creating the waves. Concentrate on just one particle in the medium, point Q. At t = 0, it happens to be at a crest. As we expect, this point will perform oscillations along a line at right angles to the direction of energy transfer. So at 0.5 ms the particle is a bit lower, but always along the vertical dotted line. After 3 ms the particle is back at its initial position so this must be the period of this wave.

Figure 13.7: Snapshots of a rope every 0.5 ms. Particles of the rope move along vertical lines. What moves forward is the position of crests (and troughs).

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In Figure 13.7, look also at the crest at x = 7.5 cm at t = 0. The wavelength is 10 cm (obtained from any one of the graphs). As time passes, the position of the crest shifts to the right. After one period, it has moved to position x = 17.5 cm. The distance covered by the crest is 10 cm and equals the wavelength as it should. The speed of the wave is then v=λ T=0.10 30×10−3=33 m s−1.

A very simple way to determine where a particle will be when a wave moves through the medium is the following: the blue curve in Figure 13.8 shows the wave at some instant. Four points on the wave have been marked (dark blue circles). Where will these points be an instant later? We know they will move along the vertical dotted lines. Now draw a copy of the blue wave, make it dashed and place it on the diagram but shifted a bit forward (because the wave moves to the right). This will tell us the new positions of the particles (light blue circles).

Figure 13.8: How medium particles move when a transverse wave propagates.

A snapshot of the wave shows the displacement of the rope along its length at the moment the picture was taken. In the same way a graph of the displacement of the wave as a function of position, i.e. distance from the left end of the rope, gives the displacement at each point on the rope at a specific point in time (Figure 13.9).

Figure 13.9: A graph of displacement versus position tells us the displacement of any point on the rope at a specific moment in time.

We get two important pieces of information from a displacement–distance graph. The first is the amplitude the maximum displacement from equilibrium of a wave or oscillation of the wave (that is, the largest displacement). The second is the wavelength.

Exam tip

It is a common mistake to think that the amplitude is the crest to trough distance.

For the wave of Figure 13.9 the amplitude is 4.0 cm. The wavelength is 0.40 m. This graph also tells us that at the point on the string that is 0.10 m from the rope’s left end the displacement is zero at that specific instant of time. At that same instant of time at a point 1.0 m from the left end the displacement is −4.0 cm, etc. Thus, a graph of displacement versus position is like a photograph or a snapshot of the rope taken at a particular time. If we take a second photograph of the string some time later, the rope will look different because the wave has moved in the meantime.

There is a second type of graph that we may use to describe waves. This is a graph of displacement versus time: we imagine looking at one specific point on the rope and observe how the displacement of that point varies with time. So, for example, Figure 13.10 shows the variation with time of a specific point on a rope as the wave of Figure 13.9 travels down the length of the rope.

Figure 13.10: The same wave as in Figure 13.9 now showing the variation of the displacement of a specific point with time.

We get two important pieces of information from a displacement–time graph: the first is the amplitude of the wave and the second is the period.

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Exam tip

To find the period from Figure 13.10, it is safer to count three loops so that three periods are 10 ms. Also, make sure to check carefully the units on the axes—in this example time is in ms, not seconds.

We already know the amplitude: it is 4.0 cm. The period is 3.33 ms. So the frequency is 1 3.33×10−3=300 Hz. The speed of this wave is therefore v = f λ = 0.40 × 300 = 120 m s–1.

(Now suppose that we are told that Figure 13.10 shows the displacement of the point at x = 0.10 m (call it P). And suppose that the graph of Figure 13.9 is a snapshot of the wave at 0.0 ms. Which way is the wave travelling? Go to Figure 13.9 and find x = 0.10 m: if you shift the wave slightly to the left (i.e. if the wave moves left) P gets a negative displacement. If you shift it to the right (i.e. if the wave moves right) P gets a positive displacement. Which is correct? Refer to Figure 13.10: immediately after t = 0.0 ms the displacement becomes negative. So the wave is travelling to the left.)

CHECK YOURSELF 2

Figure 13.11 shows, at a particular instant, a transverse wave travelling towards the left. A point in the medium is marked.

Figure 13.11: For check yourself question 2.

What is the direction of velocity of the marked particle at this instant?

Worked Example 13.4

A transverse wave of amplitude 2.0 cm, wavelength 0.40 m and frequency 350 Hz propagates on a string.

aWhat is the average velocity of a point on the string during one period?

bWhat is the average speed of a point on the string during one period?

cWhat is the ratio wave speed speed in (b)?

Answer

aDuring one period any point on the string will have a displacement of zero so the average velocity will be zero.

bDuring one period any point on the string will move a distance equal to 4 times the amplitude, i.e. 8.0 cm. The period is T=1 350=0.002857 s. Hence the average speed is v¯=0.080 0.002857=28 m s−1.

cThe wave speed is v = 350 × 0.40 = 140 m s−1, hence the ratio is 140 28=5.0.

Longitudinal waves

Imagine that you push the left end of a slinky in and out as in Figure 13.12. The coils of the slinky move in a direction that is parallel to that of the wave. As the hand moves to the right it forces coils to move forward, causing a compression (coils crowd together). As the hand moves to the left, coils right in front of it also move left, causing an expansion or rarefaction (coils move apart).

KEY POINT

A wave is longitudinal a wave where the displacement is parallel to the direction of energy transfer if the displacement is parallel to the direction of energy transfer.

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Figure 13.12: A longitudinal wave in which the medium moves parallel to the direction of energy transfer.

Sound waves are longitudinal waves that can travel in gases and liquids as well as in solids. A sound wave consists of a series of compressions and rarefactions in the medium in which it is travelling. Figure 13.13 shows the compressions and rarefactions produced in the air by a loudspeaker.

Figure 13.13: A vibrating loudspeaker cone produces compressions and rarefactions. These compressions and expansions move through air as a wave called sound.

As with transverse waves, we can plot a graph of displacement versus distance along the wave. The graph in Figure 13.14 shows the displacement y of the molecules against distance x at some instant of time. Positive displacement means motion to the right and negative displacement motion to the left. The red arrows represent the displacement of the molecules. Molecules at x = 0, 2.0 cm and 4.0 cm have not moved (y = 0); those between x = 0 and 2.0 cm and between x = 4.0 cm and 5.0 cm have moved to the right (y > 0); and those between x = 2.0 cm and 4.0 cm have moved to the left (y < 0).

Below the graph in Figure 13.14 is a row of the medium molecules shown in their equilibrium positions; these are the positions of the molecules when no wave travels in the medium. The row below shows their positions when the wave is present.

The molecule at x = 2.0 cm is, therefore, at the centre of a compression a point where the density of the medium is highest (a region of higher than normal density), while that at x = 4.0 cm is at the centre of a rarefaction a point where the density of the medium is lowest (a region of lower than normal density).

Exam tip

You cannot tell whether a wave is transverse or longitudinal by looking at displacement–distance or displacement–time graphs. The graphs look the same for both.

Figure 13.14: Molecules to the left of that at x = 2.0 cm move to the right, while the neighbours to the right move left. This means that the region at x = 2.0 cm is the centre of a compression. At x = 4.0 cm we have the centre of a rarefaction.

Since a compression is a region where molecules crowd together, the pressure and density of the medium in a compression is higher than normal. To give an idea of the differences in pressure involved, a sound of frequency 1000 Hz can be heard by the human ear when the pressure of air at the eardrum exceeds atmospheric pressure by just 20 μPa. (Normal atmospheric pressure is 105 Pa.) The amplitude of oscillations for air molecules under these conditions is about 10−11 m, or a tenth of the diameter of the hydrogen atom! In a rarefaction the reverse is true, with the molecules moving farther apart so that the density and pressure are a bit less than normal.

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CHECK YOURSELF 3

Figure 13.15 shows, at a particular instant, a longitudinal wave travelling towards the left. A point in the medium is marked.

Figure 13.15: For check yourself question 3.

What is the direction of velocity of the marked particle at this instant?

Worked Example 13.5

The solid line in Figure 13.16 shows a wave travelling to the right and the dashed line shows the wave 5.0 ms later (the period of the wave is longer than 5.0 ms).

Figure 13.16: For worked example 13.5.

Determine the speed and frequency of the wave.

Answer

In 5.0 ms crests moved forward 0.40 m and so the speed is v=0.40 5.0×10−3=80 m s−1.

The wavelength is 1.5 m and so the frequency is f=80 1.5≈53 Hz.

Phase difference

In Chapter 12 we defined the phase difference in the context of simple harmonic motion. There we had two graphs of displacement versus time for the same oscillation. We defined phase difference as

Δ ϕ=2 π Δ t T

where T is the period and Δt is the time separating two consecutive peaks. With waves we will be mainly interested in the phase difference between two identical waves travelling in the same medium or two points on the same wave. Figure 13.17 shows two identical waves travelling in the same medium. The phase difference between two waves is defined as

Δ ϕ=2 π Δ x λ

where Δx is the distance between consecutive peaks of the waves and λ is the wavelength. For the waves in Figure 13.17 we find a phase difference of

Δ ϕ=2 π 1.0 4.0=π 2

Figure 13.17: Two identical waves in the same medium with a phase difference.

The same formula can also be used to find the phase difference between two different points on the same wave. In that case Δx is the distance between the two points. So consider the blue wave in Figure 13.17. The phase difference between the point at x = 0 and the point at x = 2 cm is Δ ϕ=2 π 2.0 4.0=π. We can see that the phase difference increases as we move away from x = 0. By the time we get one wavelength away, at x = 4 cm, the phase difference is Δ ϕ=2 π 4.0 4.0=2 π, i.e. no difference at all since phase difference is an angle. (This holds for travelling waves; we will see a different behaviour of phase difference when we discuss standing waves in Chapter 15.)

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NATURE OF SCIENCE

Careful observations of the vibrations of a plucked violin string led the Swiss mathematician Daniel Bernoulli to find a way to describe the oscillation using mathematics. This was a simple setting—a single string fixed at both ends—and led to a simple solution. Scientists found similar patterns in more complex oscillations and waves in the natural world, but also differences in the way that waves propagated. By looking for trends and discrepancies in their models for different waves produced under different conditions, scientists developed wave equations that apply across many areas of physics.

TEST YOUR UNDERSTANDING

8Figure 13.18 shows three points on a string on which a transverse wave propagates to the right.

Figure 13.18: For question 8.

aWhat do you understand by the term transverse wave?

bIndicate how these three points will move in the next instant of time.

cHow would your answers change if the wave were moving to the left?

9The graph (Figure 13.19) shows, at an instant of time, a longitudinal wave travelling to the right. Displacements to the right are positive.

Figure 13.19: For question 9.

aWhat do you understand by the term longitudinal wave?

bOn the graph, draw arrows to represent the velocity of points in the medium whose equilibrium positions are at x = 1.0 m and x = 3.0 m.

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10Figure 13.20 shows a piece of cork floating on the surface of water when a wave travels through the water to the right. What is the shape of the water surface half a wave period later? What is the position of the cork then?

Figure 13.20: For question 10.

11The solid graph of Figure 13.21 shows the variation with distance of the displacement of a wave travelling to the right. The dashed graph shows the same wave 1.25 ms later.

Figure 13.21: For question 11.

Determine, for this wave:

athe speed

bthe frequency.

12Two waves are emitted from the same point and arrive at an observer 4.2 s apart.The waves have speeds 3.4 km s–1 and 8.0 km s–1.How far away was the emission point?

13Figure 13.22 shows a transverse wave at a particular instant of time. The wave is travelling to the right.

Figure 13.22: For question 13.

At this instant, for the particle in the medium whose equilibrium position is at x = 0.50 m,

aExplain why the speed is a maximum.

bDraw an arrow on the graph to indicate the direction of the velocity.

cState the value of the acceleration.

dLabel one point whose acceleration is positive and maximum.

eWhat is the phase difference between the points at x = 0.5 m and x = 1.5 m?

14Figure 13.23a shows the displacement d of the particles in a medium against position x when a longitudinal wave pulse travels through the medium from left to right with speed 2.0 cm s–1. This is the displacement at t = 0.

Figure 13.23b shows a line of nine molecules separated by 1.0 cm. The positions shown are the equilibrium positions of the molecules when no wave travels in the medium.

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Figure 13.23: For question 14.

aState what is meant by a longitudinal wave pulse.

b iCopy the horizontal axis from Figure 13.23b. Immediately below the copied axis draw another line to show the position of these molecules at t = 0.

iiIndicate on the diagram the position of a compression.

c iRepeat b i to show the position of these molecules at t = 2.0 s.

iiComment on the position of the compression at 2.0 s.

15Figure 13.24a shows the equilibrium positions of molecules in a medium, numbered 1 to 13. At equilibrium they are separated by 1 cm. A wave pulse travels through a medium.

Figure 13.24a also shows the positions of the 13 molecules at t = 0. The equilibrium position of the first molecule is at x = 0.

Figure 13.24: For question 15.

aCopy the axes (Figure 13.24b) and draw the shape of the pulse at t = 0. Take displacements to the right to be positive.

bThe pulse moves to the right with speed 20 cm s–1. On the same axes draw the shape of the pulse at t = 0.1 s.

16Figure 13.25 shows the variation with distance x of the displacement y of air molecules as a sound wave travels to the right through air.

Figure 13.25: For question 16.

Positive displacement means motion to the right. The speed of sound in air is 340 m s–1.

aDetermine the frequency of the sound wave.

bState a distance x at which i a compression and ii a rarefaction occurs.