Interactive textbook: Sound and light

What are the similarities between sound and light?

Someone with no knowledge of physics would probably answer that there aren't any similarities. You see light with your eyes and you hear sound with your ears. Sound is produced by vibrating objects but light is (generally) produced by hot ones. Light travels in straight lines whereas sound goes around corners.

It is true that the natures of sound and light are completely different. Sound is changes in air pressure and light is electromagnetic. However when we start to take controlled measurements, we find that they behave in the same way.

Both sound and light reflect, refract, diffract and interfere which makes them similar too to water waves. They all have wave-like properties. The consequence? We can use the mathematical models to predict the behaviour of water waves and, in turn, to predict the behaviour of sound and light.

Sound

Sound is produced by bodies vibrating in air. The body could actually be vibrating in a liquid or solid but, unless we are swimming under water, all the sounds we hear come to us through the air (a gas). We know about the behaviour of air molecules so let's use our kinetic model of a gas to see how a vibrating object can transfer energy through the air.

It's not easy to see but you should be able to make out a compression travelling from left to right. This is a sound wave! The increase in pressure arriving at the right pushes out the piston, thereby transferring energy.

The wavelength of the sound wave is the distance between two compressions.

Remember that the centre of the compressions and rarefactions are the places where the displacement is minimum so the displacement-position graph looks like this:

A common misconception is that the air molecules vibrate but they don't. To vibrate there would have to be a force acting on them, but our gas model is based on the assumption that there is no force between them. Except when colliding, the molecules move with random motion. When we talk about displacement we mean the average displacement of the molecules.

The speed of sound at room temperature is 340 m s-1

The average speen of the air molecule is approximately:

The speed of sound must be less than the speed of the molecules

Frequency

The frequency of a sound wave is perceived as pitch - high frequency is high pitch. We can see this if we plot the displacement-time graph for the cone of a loudspeaker while we listen to the sound.

We could have plotted the output of a microphone but a microphone actually detects changes in pressure and not displacement.

The musical scale was not created by a physicist but by musicians, it goes like this:

• C  130.82 Hz
• C#  138.59 Hz
• D  146.83 Hz
• D#  155.56 Hz
• E  164.81 Hz
• F  174.61 Hz
• F#  185 Hz
• G  196 Hz
• G#  207.65 Hz
• A  220 Hz
• A#  233.08 Hz
• B  246.94 Hz

The human ear can detect sounds from $$20$$ to $$20,000\text{ Hz}$$. Press the button below for a sound of $$30,000\text{ Hz}$$.

We'll never know if it works!

Is a sound of 1cm wavelength audible?

f = v/λ = 34000 Hz

What is the ratio speed of 200 Hz sound / speed of 400 Hz sound? This is ultrasound

The speed is the same for all frequencies.

Amplitude

Using the previous simulation you can also see that the amplitude is related to loudness. How loud a note sounds is dependant on your ears; two notes of the same amplitude will not sound equally loud.

Loudness is measured in decibels, an example of a logarithmic scale. A sound with loudness of $$70\text{ dB}$$ is twice as loud as a sound of $$60\text{ dB}$$. The decibel scale is calculated from the ratio of the quietest sound, so a sound of $$10\text{ dB}$$ is twice as loud as the quietest sound (a mosquito $$5\text{ m}$$ away).

Velocity

The velocity of a sound wave at room temperature is $$340\text{ ms}^{-1}$$. The energy in a sound wave is transferred by molecules hitting each other so it stands to reason that it is dependent on temperature. The velocity of sound at $$0\text{ }℃$$ is therefore less, $$330\text{ ms}^{-1}$$. The speed of the wave could not possibly be faster than the speed of the air molecules (and in fact is about half).

Wavelength

You don't have to remember the values of everything but you should have some idea of the sort of size. A sound of $$300\text{ Hz}$$ has a wavelength of about $$1\text{ m}$$.

Wave properties of sound

Reflection

Reflection of sound is called an echo. A room with no reflection is called an anechoic chamber. To do experiments with sound you need to go into an anechoic chamber or a large open space because reflections off the walls of a laboratory could affect your results.

A sound in a room:

The ratio of the amplitude of sound reflected to the amplitude of sound passing into a medium is related to the difference between the density of the two materials. When sound meets a wall, most is reflected. This was an important factor in the evolution of the human ear. Life started under water, like a fish. Fish ears are full of liquid so sound passes easily from the water to the ear but when fish are out of water they can't hear.

The fish might not notice this as their gills are rendered useless...!

To combat the fact that little sound enters the ear, land-based creatures evolved a complex series of small bones: the hammer, anvil and stirrup to amplify the sound.

Loud music  can damage an unborn baby's ears?

Most of the sound is reflected off the mothers tummy.

Refraction

Refraction of sound isn't an everyday experience but if you are ever sitting in a boat on a still evening you might notice that sounds seem amplified. This is due to refraction. On a still evening the cold air settles near the water with the warmer air on top. Sound waves travelling up through the air are refracted so much by the different layers that they are directed back down again. What you hear is the sound directly from the source plus the sound refracted down.

This can be demonstrated using Paul Falstad's 'ripple tank' simulation.

Here you can see the different layers. The wave speed increases with increasing height. The wall in this simulation is there so that the sound travelling directly to the receiver doesn't interfere with the refracted wave. If the wall was removed sound would arrive at the receiver directly and from above, this is why sounds are louder when sitting in boats on still evenings.

Diffraction

We have seen (or maybe heard!) that sound reflects and refracts. These effects often hide others such as diffraction. Imagine your teacher is talking in the classroom with the door open. If you go down the corridor you will still hear your teacher (although admittedly this is because of reflection off the walls not diffraction through the door). In an open space you may notice how sound diffracts around objects, something that animators have to be aware of when adding sound to computer games.

The diffraction of waves around an object can also be simulated with the ripple tank.

Will a 300 Hz sound be diffracted significantly as it passes through an open door?

The wavelength and door size are both of the order of 1 m.

Interference

If a note is played on the stereo speakers of a computer there will be quiet and loud areas around the room where the sound waves are interfering destructively and constructively. This won't be a simple pattern like we got in the ripple tank; reflections from the walls will mess it up. Some noise-cancelling headphones work by producing a sound that is the same as the noise from the outside but phase shifted by $$π$$ to interfere destructively with the original noise. You may be wondering where the energy is going (and if you weren't, you are now!).

When waves interfere there are regions of low amplitude and regions of high amplitude. The energy is in the high amplitude areas.

When 2 sounds interfere constructively the resultant sound is twice as loud.

The amplitude is double but loudness is based on a logarithmic scale.

Standing waves

Closed pipes

If a sound wave travels along a closed pipe it will reflect off the end. The reflected wave superposes with the incident wave forming a standing wave.

The simulation below shows the possible standing waves (harmonics) in a closed pipe. Use the slider to change between harmonics, noting the position of the nodes.

The oscillating lines represent layers of air and not individual molecules. You can see that, at the closed end, the pressure changes a lot but the layer of air doesn't move. This is a displacement node and a pressure antinode. In this course we always represent standing waves using the displacement vs position graph, but if you were to do an experiment moving a microphone along the pipe you would find that the sound is loudest at the closed end.

The 1st harmonic is a quarter wavelength:

$$\lambda=4L$$

$$v=f_1\lambda$$

$$f_1={v\over \lambda}={v\over 4L}$$

The 3rd harmonic (note there is no 2nd harmonic!) is three quarters of a wavelength:

$$\lambda={4\over3}L$$

$$v=f_3\lambda$$

$$f_3={v\over \lambda}={3v\over 4L}=3f_1$$

The 5th harmonic is one and a quarter wavelengths:

$$\lambda={4\over 5}L$$

$$v=f_5\lambda$$

$$f_5={v\over \lambda}={5v\over 4L}=5f_1$$

Unlike the standing wave in a string, a standing wave in a closed pipe has only odd harmonics.

Infrasound is low frequency sound, it can be damaging to your health.

A machine is producing sound of 10 Hz at the entrance of a 25.5 m long closed tunnel, where is the safest place to stand?

v = 340 m s-1

wavelength = 34 m

25.5 = 3λ/4 so the 3rd harmonic will be formed

safest place is at a displacement antinode, this is a pressure node so there will be no sound.

There is also a pressure node at the open end but you wil be right next to the machine.

Open pipes

When a sound wave meets the end of an open pipe it, too, reflects. This may seem a bit odd but remember the string with a loose end; an open pipe is not an infinitely long pipe.

An open pipe has displacement antinodes at each end.

The 1st harmonic is half a wavelength:

$$\lambda=2L$$

$$v=f_1\lambda$$

$$f_1={v\over \lambda}={v\over 2L}$$

The 2nd harmonic is one wavelength:

$$\lambda=L$$

$$v=f_2\lambda$$

$$f_2={v\over\lambda}={v\over L}=2f_1$$

The 3rd harmonic is one and a half wavelengths:

$$\lambda={2\over 3}L$$

$$v=f_3\lambda$$

$$f_3={v\over \lambda}={3v\over2L}=3f_1$$

An open pipe has all the harmonics.

Here is not a physics teacher demonstrating on a trumpet.

A and B are two simple pipe based instruments based musical instruments.

A produces a note of 300 Hz when blown softly and 900 Hz when blown strongly.

B produces a note of 600 Hz when blown softly and 1200 Hz when blown strongly.

When blown harder A gives 3rd harmonic B 2nd, so A is closed, B open

approx. v = 300 ms-1

When blown softly both give 1st harmonic

A is closed

λ = 1 m, LA = λ/4 = 0.25 m

B is open

λ = 0.5 m 1st harmonic LB = λ/2 = 0.25 m

Measuring the speed of sound

This practical is listed in the Subject Guide so you are expected to know about one method. This one uses a drinking straw. If you blow across the top of a drinking straw you can produce a sound. This sound can be analysed using a program such as Audacity, which will produce a frequency spectrum. If you're lucky, it will look like the one below.

You can see the regularly spaced peaks corresponding to the harmonics. Each peak is a multiple of $$f_1$$ so the difference between each peak is $$f_1$$. In this case it is about $$500\text{ Hz}$$.

Notice that $$f_1$$ is actually missing (or very small). This is due to the force of the air being blown across the pipe being too great to excite the lowest harmonic. We know that for $$f_1$$ the wavelength is $$2L$$ so we can calculate the wavelength. This is repeated for at least five different lengths.

The equation relating frequency and wavelength is:

$$f={v\over\lambda}$$

So a graph of $$f$$ vs $$1\over λ$$ will be a straight line with gradient $$v$$.

This graph has a negative y intercept, this could be due to:

The antinodes are beyond the ends but considering the size of the error bars, it could also be due to random errors.

Light

Electromagnetic spectrum

When you have completed the Electricity and Magnetism topics you will understand more about this but for now it is enough to simply state that light is an electromagnetic (EM) wave. Unlike water and sound waves, electromagnetic waves require no medium - they can travel through a vacuum.

The properties of EM waves depend on their wavelength. Radio waves have the longest wavelength and gamma waves the shortest. We can display all the different wavelengths on a chart called a spectrum.

You are probably already familiar with the spectrum of visible light (from a rainbow).

This is produced by separating the wavelengths using a prism, which refracts the different wavelengths by different amounts. In this image, blue bends best.

As with sound waves the wave properties of light are related to the way our eyes perceive it. Wavelength is analagous to colour and amplitude to brightness. The velocity of light in a vacuum (and air) is $$3 \times 10^8\text{ ms}^{-1}$$.

What is the order of magnitude of the frequency of light?

3 x 108/600 x 10-9 = 5 x 1014 Hz

Colour

Colour perception is a combination of intensity, wavelength and the surroundings. The retina is the part of the eye that detects light; this is covered in light-sensitive cells called rods and cones. The cones are also sensitive to wavelength. If you don't have enough cones then your brain has less information to work with so sometimes gets the colour wrong.

Brightness

The brightness is directly related to the intensity (or power per unit area). This is proportional to the square of the amplitude, so if the amplitude doubles the intensity quadruples. The unit of intensity is $$\text{Wm}^{-2}$$.

The brightness of a star is used to calculate its distance from the Earth by using the fact that the power spreads out in a sphere resulting in the intensity being proportional to the inverse of the square of the distance. This is called an inverse square law.

The luminosity is the total power radiated by a star.

Star A has 2x the luminosity of B but is 2x further away from the Earth.

Calculate the ratio brightness A/brightness B

bA = 2L/4π(2r)2

bB = L/4πr2

ratio = 0.5

Light we cannot see

In this section we are mostly interested in visible light... but light is just a small part of the whole spectrum.

Note that this version has been plotted with the longest wavelength to the left. Always look at the axis carefully. You should remember the approximate wavelengths of the different regions.

In this animation (made with Prezi) the size of the words are about the same scale as the wavelength:

Wave properties of light

Reflection

The reason we see objects is because they reflect light into our eyes. When light reflects, the angle of reflection equals the angle of incidence... but this is only noticeable when the surface is flat.

Refraction

When light passes from one medium to another of different optical its velocity changes. This results in a change of direction.

We can calculate the angle of refraction using Snell's law: $${\sin i_1\over\sin i_2} = {v_1\over v_2}$$ (the ratio of velocities). However it is more convenient to use the relative refractive index, $$_1n_2$$, of the combination of materials.

In the diagram above medium 1 and medium 2 could be:

1 is less dense than 2

vacuum and air are almost the same so very little refraction.

Absolute refractive index

Absolute refractive index is the ratio of $$\sin i_1\over \sin i_2$$ when light passes from a vacuum into the medium. If light passes from one medium to another then Snell's law can be written as:

$$n_1\sin i_1=n_2\sin i_2$$

• $$n_1$$ is the absolute refractive index of medium 1
• $$n_2$$ is the absolute refractive index of medium 2

Note that we take air to have refractive index $$= 1$$ (same as a vacuum).

You should also do this with real light. See Refractive index.

Total internal reflection

When light travels from glass to air it refracts away from the normal. If the angle of incidence is large enough then the angle of refraction will be $$90°$$. The angle at which this happens is called the critical angle.

If the critical angle is exceeded, no ray is refracted. This is called total internal reflection.

Total internal reflection can be used to reflect light along long optical fibres used in the communications industry.

The velocity of sound in water is approximately 1500 ms-1 in water and 300 ms-1 in air.

The critical  angle for sound travelling from water to air is:

Sound slow down when it travels into the air so will refract towards the normal, there will be no critical angle.

Real optical fibres are thin transparent plastic fibres with an outer layer of lower optical density plastic.

How does this affect the angle of entry into the fibre compared to a simple one plastic fibre surrounded by air?

The refactive index of the coating will be greater than air so the critial angle will be greater, the angle of entry will therefore be smaller.

Single slit diffraction

When light passes through a narrow slit ($$<0.1\text{ mm}$$) it can be observed spreading out. This is diffraction.

We can see how this is related to the diffraction pattern in the ripple tank. The central wedge of high amplitude corresponds to the central maximum.

Laser light is passed through an adjustable narrow slit.

The slit width increases. What effect does this have on the width and intensity of central maximum of the diffraction pattern:

Wider slit means more light but less spread out.

Double-slit interference

If light passes through two narrow slits, interference will take place where the light overlaps.

Here you can see the effect of diffraction at each slit plus interference in the light emerging from each slit. Interference causes the bright dots and diffraction causes them to have different intensity. This can be shown in the following simulation.

Reducing the slit width spreads out the diffraction pattern and reducing separation spreads out the dots.

For the image above, the ratio slit separation / slit width is approximately:

The diffraction pattern is more spread out than the interference fringes, use the simulation to recreate the same pattern.

Polarisation

Light can be polarised by passing it through a special plastic called Polaroid. We can show that this is happening by using two pieces of polaroid. Alternatively you can use one piece and your computer screen. This is because the light from your screen is already polarised (if it is an LED screen). In the animation below you can see what happens as a polarising filter is rotated in polarised light.

Malus' law gives the relationship between the intensity and angle between the polarisers:

$$I=I_0\cos^2\theta$$

• $$I$$ is the transmitted intensity
• $$I_0$$ is the incident intensity

NB: Intensity is proportional to the square of the amplitude.

Light of intensity 8 mW m-2 is passed through two polarisers arranged with their planes of polarisation at 60°.

The intensity of light emerging from the seond polariser is:

The first polarise reduces the light by 1/2, the second by (cos60°)2 = 1/4

Total Score:

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