Physics
Double Award
Double Award
Cambridge IGCSE
TOPIC 3: WAVES
Have you ever watched a loudspeaker as music is playing? The speaker cone moves backwards and forwards creating the sound. In fact, all sound waves are created by vibrations, and the wave then travels through air, water and any other solid, liquid or gas. However, sound waves cannot travel in a vacuum, because they need molecules or atoms to pass on the vibrations.
When a sound is made, for example by a loudspeaker, air molecules are pushed forward by the speaker cone, and this makes a pressure wave in the air. This is a longitudinal wave, as discussed in section 3.1. In regions where the molecules are closer together, there is a high-pressure region (called a 'compression') and where they are more spaced out, the pressure is lower. (A 'rarefaction'). This is shown in figure 1:
Figure 1. A sound wave moving through air
The particles move backwards and forwards, left to right, whilst the sound pressure wave moves to the right in this diagram. This means that sound is a longitudinal wave, as the particles vibrate parallel to the direction of the wave motion. (See section 3.1).
Here's an excellent animation showing air particles creating the wave motion:
Figure 2. A sound wave animation (L. Perera)
Note that when we hear a reflected sound, we usually call it an echo. Echos are used by dolphins to navigate by listening for the sound reflection from pulses.
Humans can hear a wide range of frequencies. The lowest sounds that we can detect are about 20 Hz. The highest sounds for a healthy young adult are about 20 000 Hz or 20 kHz. As we get older, our range of hearing decreases, particularly with the high frequency sounds.
This YouTube clip sweeps through the whole human hearing range. Put some headphones on or play this through some good loudspeakers. Can you hear to the end of the clip right up to 20 000 Hz?
YouTube - The range of human hearing - 20 Hz to 20 000 Hz -test
(Sonic Electronix)
If you could not hear the end of this clip, it may be that your sound system cannot play sounds that high, rather than your hearing is defective! However, listening to loud music for a long time can eventually damage your hearing and reduce your range and sensitivity to sounds. Try playing this clip through your laptop or phone speaker to find out how limited the range is - this is why music does not sound so great through the speaker in a phone or laptop.
When listening to music, if we hear a very high musical note, we say it has a high pitch. This means that the loudspeaker or headphones that made it are vibrating extremely quickly - at a very high frequency.
A loud sound is made when the amplitude of the vibrations is very large, and conversely quiet sound waves have a small amplitude. It is important to understand this so that when people describe sounds using every-day terms like pitch and loudness, we can relate the description to physical measurements.
Sound waves are invisible. We cannot see them moving through the air. However, we can convert sounds to an electrical signal using a microphone. The electrical signal can then be displayed on a device called an oscilloscope. An oscilloscope looks similar to a heart rate monitor used in hospitals - a small dot moves left to right across the screen, and any change in the signal moves the dot upwards or downwards. To display sound waves, the dot moves so quickly that it looks like a continuous line to us. We will learn more about using the oscilloscope later in this section.
The 'height' of a wave from the centre line is called the amplitude, and this gives us a measure of the loudness of the sound. If the waves are close together, then the wavelength is short and the frequency is high - it will sound high pitched.
Figure 3. Sound waves displayed on an oscilloscope
Questions:
1. The diagram here shows a wave displayed a computer screen.
Add a second wave on the diagram to show a wave with a higher pitch and louder than the original sound.
The wave drawn should be:
An example is shown here as a dotted line:
To find the speed of any object, we need to find the time it takes to cover a specific distance. We can use this principle with sound. For this course, you will need to be able to describe an experiment to measure the speed of sound. There are many methods available, but this one is simple to perform if you have a nearby building next to open ground:
You will need:
Figure 4. Measuring the speed of sound
To perform this experiment, simply bang the two wooden blocks together at the same time as a partner starts a stop watch. You will hear an echo from the wall - stop the stop watch as soon as you hear the echo. (An echo is a reflection of the sound from the wall).
Do this several times to get an average - it can be difficult to measure this accurately. If you have the distance to the wall, and the time it took for the echo to return to you, then we can use:
| speed = | distance |
| time |
This will give us the speed of sound. Remember that the echo travels there and back, so you will need to double the distance to the wall to find the distance travelled by the sound wave.
Alternative Method:
You could also try this without using an echo. One person ('A') with a stop watch stands a long distance away, typically 300-500 m from person ('B') who is going to bang wooden blocks together as described above. This time, person A starts the stop watch when they see the blocks being banged together, and stops the stop watch when they hear the sound less than a second later. Light travels much faster than sound, almost instantaneously in this experiment, so the delay is due to the slower speed of sound. If the distance is known as well as the time taken for the sound to travel from B to A, the speed can be calculated.
The expected value for the speed of sound is about 330 to 350 m/s, (although you do not need to learn this) and the exact value depends on various factors such as air pressure and temperature.
Questions:
2. A wall is 80 m away from some observers. A loud bang is produced by a drum, and an echo is heard 0.48 s later.
Using this data, calculate the speed of sound.
The sound wave travels there and back, so the distance is 160 m. The time taken is 0.48 s, so:
| speed = | distance |
| time |
| speed = | 160 |
| 0.48 |
speed of sound = 330 m/s ( to 2 sig. figs.)
Have you noticed that it is still possible to hear sounds when swimming under water? in fact sound waves are transmitted efficiently in liquids and solids as the particles are closer together. This also means that the speed of sound in liquids and solids is generally much faster than in air, with solids having the fastest speeds:
Speed of sound data:
As you may well be aware, some animals like dogs and cats can hear much higher frequencies than humans, well above 20 000 Hz. Elephants and some other large animals can hear sounds below 20 Hz. All sounds above human hearing (above 20 kHz) are called ultrasound waves. These can have many uses from treating muscle injuries to sensing a heartbeat in a developing foetus.
This principle is also used in 'sonar', used by ships and submarines to detect objects underwater. Ultrasound waves can also be used in this way to test materials for defects, without needing to break them open ('non-destructive' testing).
*Note: For this course, you should be able to describe ultrasound as a sound with a frequency higher than 20 kHz, but you will not be asked questions about applications. They are included here for context only.
Questions:
3. A sound wave travels at a speed of 1500 m/s in sea water. A pulse of sound is emitted by an ocean liner, and an echo from the bottom of the ocean is detected 6 seconds later.
Calculate the depth of the ocean.
We know that:
| speed of wave = | distance |
| time |
Rearranging the equation gives:
Distance = speed x time.
So distance travelled = 1500 x 6,
distance = 9000 m.
However this is not the final answer, as the wave travelled from the liner to the ocean floor and back again. Therefore the ocean is only 4500 m below (4.5 km).
4. An ultrasound wave is used in a medical scan. The wave travels at 1600 m/s through a certain body tissue and has a frequency of 30 kHz.
Calculate the wavelength of the ultrasound wave.
30 kHz is equal to 30 000 Hz. We also know that v = ƒ λ, and rearranging this gives:
| λ = | v |
| ƒ |
| λ = | 1600 |
| 30 000 |
So λ = 0.0533 m (to 3 sig. figs). This is 5.33 cm.
