Waves, sound and light: Identify, describe and apply principles of waves

# Unit 3: Wave calculations

Linda Pretorius

### Unit outcomes

By the end of this unit you will be able to:

- Calculate frequency, period, wave speed and wavelength of a transverse wave.
- Describe the effect of the medium on wave speed.

## What you should know

Before you start this unit, make sure you can:

- Describe what a wave is, as covered in Subject outcome 3.1, Unit 1.
- Define frequency, period and wavelength of a transverse wave, as covered in Subject outcome 3.1, Unit 1.
- Describe the relationship between speed, distance and time, as covered in Subject outcome 2.1, Unit 1.

## Introduction

In this unit^{[1]} you will apply your knowledge about the basic properties of waves to understand how a transverse wave moves. You will also learn how to calculate the properties of a wave to describe its behaviour.

# Calculating the properties of a wave

You know by now that energy travels in wave form. The speed at which a certain wave moves therefore tells us the speed at which a certain type of energy, such as light or sound, travels. Because speed is defined as the change in distance over time, we can use the speed of a wave to determine how far away the source of the energy is from a reference point. Calculating the speed of a wave has many useful applications in real life, such as in astronomy, navigation, medical imaging and even traffic law enforcement.

### Note

You may also see the term ‘wave velocity’ used to refer to wave speed (*v*).

## The wave equation

Speed is defined as distance covered in a certain time. We can also apply this to wave behaviour. You already know from your work in Unit 1 that the period of a wave tells us how long it takes for one complete wave to move past a fixed point.

Applying the definition of speed to wave behaviour, we can say that the speed of a wave is the distance travelled (the wavelength) in one period:

[latex]\scriptsize \Large v=\displaystyle \frac{{\text{distance travelled}}}{{\text{time taken}}}=\displaystyle \frac{\lambda }{T}[/latex]

But you also know that period (*T*) is inversely related to the frequency (*f*) of a wave: [latex]\scriptsize f=\displaystyle \frac{1}{T}[/latex].

In other words, for a wave with a period of [latex]\scriptsize 0.2[/latex] seconds, [latex]\scriptsize \displaystyle \frac{1}{{0.2}}=5[/latex], so five complete wavelengths will move past a fixed point per second. The wave therefore has a frequency of [latex]\scriptsize \displaystyle \frac{1}{T}=\displaystyle \frac{1}{{0.2}}=5\ \text{Hz}[/latex].

Combining these concepts, we can also write:

[latex]\scriptsize \begin{align*}v & =\displaystyle \frac{{\text{distance travelled}}}{{\text{time taken}}}\\ & =\displaystyle \frac{\lambda }{T}\\ & =\lambda \cdot \displaystyle \frac{1}{T}\\ & =\lambda \cdot f\end{align*}[/latex]

We call this equation the :

[latex]\scriptsize v=\lambda \cdot f[/latex]

where:

[latex]\scriptsize v[/latex] = speed of the wave in metres per second (m.s^{–1})

[latex]\scriptsize \lambda[/latex]= wavelength in metres (m)

[latex]\scriptsize f[/latex] = frequency of the wave (Hz)

### Take note!

The unit of frequency – hertz – is defined as ‘per second: [latex]\scriptsize {{\text{s}}^{{-1}}}[/latex].

By manipulating the wave equation, we can calculate the characteristics of a given wave.

### Example 3.1

- When a particular string is vibrated at a frequency of [latex]\scriptsize 10\text{ Hz}[/latex], a transverse wave of wavelength [latex]\scriptsize 0.25\text{ m}[/latex] is produced. Determine the speed of the wave as it travels along the string.
- A cork on the surface of a swimming pool bobs up and down once every second on some ripples. The ripples have a wavelength of [latex]\scriptsize 20\ \text{cm}[/latex]. If the cork is [latex]\scriptsize 2\text{ m}[/latex] from the edge of the pool, how long does it take a ripple passing the cork to reach the edge?

*Solutions*

- We can use the wave equation to calculate the speed of the wave: [latex]\scriptsize v=\lambda \cdot f[/latex].

We know that the wavelength is [latex]\scriptsize 0.25\text{ m}[/latex] and the frequency is [latex]\scriptsize 10\text{ Hz}[/latex]. The quantities are already given in SI units, and therefore we can substitute them directly into the equation:

[latex]\scriptsize \begin{align*}v & =\lambda \cdot f\\ & =(0.25\text{ m)(}10\text{ Hz)}\\ & =2.5\text{ m}\cdot {{\text{s}}^{{-1}}}\end{align*}[/latex]

The wave travels at [latex]\scriptsize 2.5\text{ m}\cdot {{\text{s}}^{{-1}}}[/latex] along the string. - From the information given, we know that the frequency of the wave is [latex]\scriptsize 1\ \text{Hz}[/latex].

We are required to determine the time it takes for a ripple to travel between the cork and the edge of the pool. The wavelength is not in SI units and should be converted.

.

Let the distance between the cork and the edge of the pool be*D*.

.

We know that speed (*v*) is distance (*D*) over time (*t*), so we can write: [latex]\scriptsize v=\displaystyle \frac{D}{t}[/latex].

We can then manipulate the equation to make time the subject of the formula: [latex]\scriptsize t=\displaystyle \frac{D}{v}[/latex].

.

We know that [latex]\scriptsize v=\lambda \cdot f[/latex], so we can write: [latex]\scriptsize t=\displaystyle \frac{D}{{\lambda \cdot f}}[/latex].

.

Remember to convert the wavelength to SI units before substituting the values into the equation: [latex]\scriptsize 20\text{ cm = 0}\text{.2 }\text{m}[/latex]

.

Therefore:

[latex]\scriptsize \begin{align*}t & =\displaystyle \frac{D}{{\lambda \cdot f}}\\ & =\displaystyle \frac{{(2\ \text{m)}}}{{(0.2\text{ m)}(1\text{ Hz)}}}\\ & =\displaystyle \frac{{(2\ \text{m)}}}{{(0.2\text{ m)}(1\text{ }{{\text{s}}^{{-1}}}\text{)}}}\\ & =10\text{ s}\end{align*}[/latex]

.

It therefore takes the ripple [latex]\scriptsize 10[/latex] seconds to reach the edge of the pool.

### Exercise 3.1

*Questions 1, 3 and 4 were sourced from or based on questions in Siyavula Physical Science Gr 10 Learner’s Book, pp. 152–154, released under a CC-BY licence.*

- A transverse wave has a frequency of [latex]\scriptsize 15\text{ Hz}[/latex]. The horizontal distance from a crest to the nearest trough is measured to be [latex]\scriptsize 2.5\text{ cm}[/latex]. Find the:
- period of the wave
- speed of the wave.

- Calculate the frequency of a wave that is travelling at a speed of [latex]\scriptsize 6.0\text{ m}\text{.}{{\text{s}}^{{-1}}}[/latex] and has a wavelength of [latex]\scriptsize 1.8\text{ m}[/latex].
- Microwave ovens produce radiation with a frequency of [latex]\scriptsize 2\ 540\text{ MHz}[/latex] [latex]\scriptsize \text {(} 1\text{ MHz = 1}{{\text{0}}^{6}}\text{ Hz)}[/latex] and a wavelength of [latex]\scriptsize 0.122\ \text{m}[/latex]. What is the wave speed of the radiation?
- Tom is fishing from a pier and notices that four waves pass by in 8 s and he estimates the distance between two successive crests is [latex]\scriptsize 4\text{ m}[/latex]. The timing starts with the first crest and ends with the fourth. Calculate the speed of the wave.

The full solutions are at the end of the unit.

## The effect of the medium on wave speed

The speed of a wave depends on the properties of the medium in which it is travelling. This means that a wave of a fixed frequency will travel at different speeds through different mediums.

- The denser the medium, the faster a wave travels. Waves therefore travel faster through solids than through liquids, and faster through liquids than through gases.
- The higher the tension of a medium (in other words, the less elastic it is), the faster the wave will travel, because each section of the medium is in tighter contact with the adjacent section. A wave will therefore travel faster through a taut string than a loose one.
- Waves slow down when they move from deeper water into shallower water.

It is important to remember that once a wave has been generated, its frequency cannot change. From the wave equation [latex]\scriptsize v=\lambda \cdot f[/latex], it follows then that if the wave’s speed changes, its wavelength must also change. Because speed is directly proportional to wavelength ([latex]\scriptsize v\propto \lambda[/latex]) according to the wave equation, a decrease in speed will lead to a decrease in wavelength; similarly, an increase in speed will lead to an increase in wavelength. This is the reason why a produces shorter and higher waves near the shore.

## Summary

In this unit you have learnt the following:

- The period of a wave is the inverse of its frequency: [latex]\scriptsize T=\displaystyle \frac{1}{f}[/latex].
- Wave speed is calculated according to the wave equation: [latex]\scriptsize v=\lambda \cdot f[/latex], where
*v*is the speed, [latex]\scriptsize \lambda[/latex] is the wavelength and*f*is the frequency. - When working with the wave equation, all quantities have to be expressed in SI units:
- Frequency must be expressed in hertz (Hz)
- Wavelength must be expressed in metres (m).
- Wave speed must be expressed in metres per second (m.s
^{–1}).

- The speed of a wave depends on the properties of the medium:
- The denser a medium, the faster a wave will travel.
- The more tension in a medium, the faster a wave will travel.
- The shallower the depth of the medium, the slower a wave will travel.

- The frequency of a wave does not change when it moves from one medium to another; only its speed and wavelength change.

# Unit 3: Assessment

#### Suggested time to complete: 40 minutes

*Questions 1–4 were sourced from or based on questions from OpenStax College released under a Creative Commons Attribution Licence 4.0 licence. Questions 5–6 were sourced from or based on questions in Siyavula Physical Science Gr 10 Learner’s Book, pp. 155–156 released under a CC-BY licence.*

- When is the wavelength directly proportional to the period of a wave?
- when the velocity of the wave is halved
- when the velocity of the wave is constant
- when the velocity of the wave is doubled
- when the velocity of the wave is tripled

- If a seagull sitting in water bobs up and down once every [latex]\scriptsize 2[/latex] seconds and the distance between two crests of the water wave is [latex]\scriptsize 3 \text{ m}[/latex], what is the speed of the wave?
- [latex]\scriptsize 1.5 \text{ m}\text{.}{{\text{s}}^{{-1}}}[/latex]
- [latex]\scriptsize 3 \text{ m}\text{.}{{\text{s}}^{{-1}}}[/latex]
- [latex]\scriptsize 6 \text{ m}\text{.}{{\text{s}}^{{-1}}}[/latex]
- [latex]\scriptsize 12 \text{ m}\text{.}{{\text{s}}^{{-1}}}[/latex]

- A boat in the trough of a wave takes [latex]\scriptsize 3[/latex] seconds to reach the highest point of the wave. The speed of the wave is [latex]\scriptsize 5 \text{ m}\text{.}{{\text{s}}^{{-1}}}[/latex]. What is its wavelength?
- [latex]\scriptsize \text{0}\text{.83 m}[/latex]
- [latex]\scriptsize \text{15 m}[/latex]
- [latex]\scriptsize \text{30 m}[/latex]
- [latex]\scriptsize \text{180 m}[/latex]

- A woman creates two waves every second by shaking a slinky spring up and down.
- What is the period of each wave?
- If each wave travels [latex]\scriptsize 0.9[/latex] metres after one complete wave cycle, what is the speed of wave propagation?

- A wave travels along a string at a speed of [latex]\scriptsize 1.5\text{ m}\text{.}{{\text{s}}^{{-1}}}[/latex]. If the frequency of the source of the wave is [latex]\scriptsize 7.5\text{ Hz}[/latex], calculate:
- the wavelength of the wave
- the period of the wave.

- Ocean waves crash against a seawall around the harbour. Eight complete waves hit the seawall in [latex]\scriptsize 5[/latex] seconds. The distance between successive troughs is [latex]\scriptsize 9\text{ m}[/latex]. The height of the waveform trough to crest is [latex]\scriptsize \text{1}\text{.5 m}[/latex].

Note that you have to draw on your prior knowledge about wave characteristics in this question.- How many complete waves are indicated in the sketch?
- Write down the letters that indicate any TWO points that:
- are in phase
- are out of phase
- represent ONE wavelength.

- Calculate the amplitude of the wave.
- Show that the period of the wave is [latex]\scriptsize 0.625\text{ s}[/latex].
- Calculate the velocity of the waves.

- Wind gusts create ripples on the ocean that have a wavelength of [latex]\scriptsize \text{0}\text{.5 m}[/latex] and propagate at [latex]\scriptsize \text{2}\text{.00 m}\cdot {{\text{s}}^{{-1}}}[/latex]. What is their frequency?
- What is the wavelength of an earthquake that shakes you with a frequency of [latex]\scriptsize \text{10}\text{.0 Hz}[/latex] and gets to another city [latex]\scriptsize \text{84}\text{.0 km}[/latex] away in [latex]\scriptsize \text{12}\text{.0 s}[/latex]?
- Your ear can differentiate sounds that arrive at the ear in just [latex]\scriptsize \text{1}\text{.00 ms}[/latex]. What is the minimum distance between two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is [latex]\scriptsize \text{340 m}\cdot {{\text{s}}^{{-1}}}[/latex]?
- A [latex]\scriptsize \text{660 Hz}[/latex] source emits a wave of [latex]\scriptsize \text{30 cm}[/latex]. How much time is needed for the wave to travel [latex]\scriptsize \text{594 m}[/latex]?

The full solutions are at the end of the unit.

# Unit 3: Solutions

### Exercise 3.1

- .
- .

[latex]\scriptsize \begin{align*}T & =\displaystyle \frac{1}{f}\\ & =\displaystyle \frac{1}{{15}}\\ & =0.067\text{ s}\end{align*}[/latex] - The distance from a crest to the next trough is [latex]\scriptsize 2.5\text{ cm}[/latex], which is half a wavelength. A full wavelength is therefore [latex]\scriptsize 5\text{ cm}[/latex]. This must be converted to metres to be used in the wave equation: [latex]\scriptsize 5\,\text{cm = 0}\text{.05}\ \text{m}[/latex].

[latex]\scriptsize \begin{align*}v & =\lambda \cdot f\\ & =(0.05\ \text{m)(15}\ \text{Hz)}\\ & = \text{ 0}\text{.75}\ \text{m}\cdot {{\text{s}}^{{-1}}}\end{align*}[/latex]

- .
- .

[latex]\scriptsize \begin{align*}v=\lambda \cdot f\\ \therefore f & =\displaystyle \frac{v}{\lambda }\\ & =\displaystyle \frac{{6.0\,\text{m}\cdot {{\text{s}}^{{-1}}}}}{{1.8\text{ m}}}\\ & =3.33\ \text{Hz}\end{align*}[/latex] - Frequency should first be converted to the SI unit, hertz:

[latex]\scriptsize f=2\ 540\text{ MHz = }2\ 540\times {{10}^{6}}\text{ Hz = 2}\text{.540}\times \text{1}{{\text{0}}^{9}}\text{ Hz}[/latex]

[latex]\scriptsize \begin{align*}v & =\lambda \cdot f\\ & =(0.122\ \text{m)(2}\text{.540}\times \text{1}{{\text{0}}^{9}}\ \text{Hz)}\\ & = \text{ 3}\text{.09 }\times \text{ 1}{{\text{0}}^{8}}\text{ m}\cdot {{\text{s}}^{{-1}}}\end{align*}[/latex] - The description of [latex]\scriptsize 4[/latex] waves passing in [latex]\scriptsize 8[/latex] seconds translates to a frequency of [latex]\scriptsize 0.5[/latex] waves per second. Therefore, the frequency is [latex]\scriptsize 0.5\text{ }\text{Hz}[/latex].

The distance from crest to crest is estimated to be [latex]\scriptsize 4\text{ m}[/latex], which represents the wavelength.

[latex]\scriptsize \begin{align*}v & =\lambda \cdot f\\ & =(4\ \text{m)(0}\text{.5}\ \text{Hz)}\\ & = \text{ 2 m}\cdot {{\text{s}}^{{-1}}}\end{align*}[/latex]

### Unit 3: Assessment

- A

We know that period is the inverse of frequency. We can therefore rewrite the wave equation as [latex]\scriptsize v=\lambda \cdot \displaystyle \frac{1}{T}[/latex]. Manipulating the equation to make period (*T*) the subject gives [latex]\scriptsize T=\lambda \cdot \displaystyle \frac{1}{v}[/latex]. This tells us that for period (*T*) to be directly proportional to wavelength ([latex]\scriptsize \lambda[/latex]), velocity (*v*) has to be reduced. The only option that describes a reduction in velocity is A. - A

The frequency is [latex]\scriptsize 0.5\text{ Hz}[/latex] and the wavelength is given as [latex]\scriptsize 3\text{ m}[/latex]. To calculate the speed of the wave, we use the wave equation:

[latex]\scriptsize \begin{align*}v & =\lambda \cdot f\\& =(3\ \text{m)(}0.5\ \text{Hz)}\\ & = \text{ 1}\text{.5 m}\cdot {{\text{s}}^{{-1}}}\end{align*}[/latex] - C

The distance from trough to crest represents half a wavelength. If it takes the boat [latex]\scriptsize 3\text{ s}[/latex] to complete half a wavelength, it will take [latex]\scriptsize 6\text{ s}[/latex] to complete a full wavelength. The period of the wave is therefore [latex]\scriptsize 6\text{ s}[/latex]. This translates to a frequency of [latex]\scriptsize f=\displaystyle \frac{1}{T}=\displaystyle \frac{1}{{6\text{ s}}}=0.167\ \text{Hz}[/latex].

According to the wave equation then:

[latex]\scriptsize \begin{align*}v=\lambda \cdot f\\\therefore \lambda & =\displaystyle \frac{v}{f}\\ & =\displaystyle \frac{{5\text{ m}\cdot {{\text{s}}^{{-1}}}}}{{0.167\text{ Hz}}}\\ & = \text{ 29}\text{.9 m}\\ & \approx 30\text{ m}\cdot {{\text{s}}^{{-1}}}\end{align*}[/latex] - .
- The frequency is given as [latex]\scriptsize 2\text{ Hz}[/latex]. Therefore [latex]\scriptsize T=\displaystyle \frac{1}{f}=\displaystyle \frac{1}{2}=0.5\text{ s}[/latex].
- .

[latex]\scriptsize \begin{align*}v & =\lambda \cdot f\\ & =(0.9\ \text{m)(2}\ \text{Hz)}\\ & = \text{ 1}\text{.8 m}\cdot {{\text{s}}^{{-1}}}\end{align*}[/latex]

- .
- .

[latex]\scriptsize \begin{align*}v=\lambda \cdot f\\\therefore \lambda & =\displaystyle \frac{v}{f}\\ & =\displaystyle \frac{{\text{1}\text{.5 m}\cdot {{\text{s}}^{{-1}}}}}{{7.5\text{ Hz}}}\\ & = \text{ 0}\text{.2 m}\end{align*}[/latex] - .

[latex]\scriptsize \begin{align*}T & =\displaystyle \frac{1}{f}\\ & =\displaystyle \frac{1}{{7.5}}\\ & =0.13\text{ s}\end{align*}[/latex]

- .
- .
- Three complete waves are shown.
- .
- Any of the pairs: B and D; D and F; E and G; C and E; or A and H.
- Some examples are: A and B; A and D; B and C; A and C; C and D; D and E; E and F; F and G; or E and H.
- Any of the pairs: B and D; C and E; D and F; or E and G.

- Amplitude is defined as the deviation from the rest position. Therefore amplitude is [latex]\scriptsize \displaystyle \frac{{1.5\text{ m}}}{2}=0.75\text{ m}[/latex].
- If [latex]\scriptsize 8[/latex] waves hit the wall in [latex]\scriptsize 5[/latex] seconds, it means [latex]\scriptsize 1.6[/latex] waves hit the wall every second. Therefore:

[latex]\scriptsize T=\displaystyle \frac{1}{f}=\displaystyle \frac{1}{{1.6}}=0.625\text{ s}[/latex] - .

[latex]\scriptsize \begin{align*}v & =\lambda \cdot f\\ & =(9\ \text{m)(1}\text{.6}\ \text{Hz)}\\ & = \text{ 14}\text{.4 m}\cdot {{\text{s}}^{{-1}}}\end{align*}[/latex]

- .

[latex]\scriptsize \begin{align*}v=\lambda \cdot f\\\therefore f & =\displaystyle \frac{v}{\lambda }\\ & =\displaystyle \frac{{\text{2 m}\cdot {{\text{s}}^{{-1}}}}}{{\text{0}\text{.05 m}}}\\ & = \text{ 40}\text{ Hz}\end{align*}[/latex] - It takes the wave [latex]\scriptsize 12.0\text{ s}[/latex] to travel [latex]\scriptsize 84.0\text{ km}[/latex]. The speed of the wave is therefore: [latex]\scriptsize 7\text{ km}\cdot {{\text{s}}^{{-1}}}\text{ = 7}\ \text{000}\ \text{m}\cdot {{\text{s}}^{{-1}}}[/latex].

[latex]\scriptsize \begin{align*}v=\lambda \cdot f\\\therefore \lambda & =\displaystyle \frac{v}{f}\\ & =\displaystyle \frac{{\text{7}\ \text{000 m}\cdot {{\text{s}}^{{-1}}}}}{{10.0\text{ Hz}}}\\ & = \text{ 700 m}\end{align*}[/latex] - We know that speed is defined as distance over time. If we define the distance here as D, we can write [latex]\scriptsize v=\displaystyle \frac{D}{t}[/latex].

[latex]\scriptsize \begin{align*}\therefore D & =vt\\ & =(340)(0.001\text{ s})\\ & =0.34\text{ m}\end{align*}[/latex]

The minimum distance of the two speakers should therefore be [latex]\scriptsize 0.34\text{ m}[/latex]. - The speed of the wave described here is [latex]\scriptsize v=\lambda \cdot f=(0.3\text{ m})(660\text{ Hz})=198\text{ m}\cdot {{\text{s}}^{{-1}}}[/latex].

To cover a distance of [latex]\scriptsize 594\ \text{m}[/latex], we use the relationship: [latex]\scriptsize v=\displaystyle \frac{D}{t}[/latex].

[latex]\scriptsize \therefore t=\displaystyle \frac{D}{v}=\displaystyle \frac{{594\text{ m}}}{{198\ \text{m}\cdot {{\text{s}}^{{-1}}}}}=3\text{ s}[/latex]

### Media Attributions

- Img01_Radio telescope © Pierluigi D'Amelio is licensed under a CC0 (Creative Commons Zero) license
- Img02_Assmnt_Q6 © Siyavula is licensed under a CC BY (Attribution) license

- Parts of the text in this unit were sourced from Siyavula Physical Science Gr 10 Learner’s Book, p. 139–152, released under a CC-BY licence. ↵

equation used to relate the frequency and wavelength of a wave to its speed

a high-energy wave caused by an earthquake at sea, which increases in amplitude as it approaches the coast