How waves form and travel
Why does a bell hanging on a post ring when the other end of a long rope is flicked once, although nobody touches the bell?
Prior knowledge
- The period $T$ is the time of one full vibration, and the frequency $f = \tfrac{1}{T}$ tells the number of vibrations per secondchapter Describing oscillation: period, frequency and amplitude
- The amplitude $A$ is the greatest displacement from the equilibrium positionchapter Describing oscillation: period, frequency and amplitude
- A body in simple harmonic motion moves back and forth on both sides of its equilibrium positionchapter The mass–spring oscillator and the pendulum
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 4.
Think about it
A rope ten metres long has been stretched across a yard, with one end tied to a post. A small bell hangs on the post at the height of the rope. When the free end of the rope is flicked once quickly up and down, a hump appears in the rope and travels towards the post. After a moment the bell rings, although nobody touched it.
What travelled along the rope from the hand to the post and made the bell ring?
Optional Revision: period and frequency
Predict
Make your predictions before investigating 2 questions
Prediction 1: how does a point of the rope move?
Prediction 2: more rapid shaking
Investigate
Press Start and watch how the wave is formed at the left end of the rope. At the same time follow the marked point $P$ and its graph.
Simulation
How to use
Press Start: the handle at the left end starts to move, and the wave travels to the right. The upper picture is a snapshot of the wave, that is, the whole rope or spring at one moment; the readings of the ruler are in metres. The lower graph shows the displacement $y$ of the marked point $P$ as a function of time over the last $4\,\text{s}$. In the Wave menu choose a transverse wave in a rope or a longitudinal wave in a coil spring, in the Source menu a continuous wave or a single pulse, and in the Tension menu a high or a low tension. The slider changes the frequency $f$ of the source. When you change a value, the wave starts again from the beginning; Reset does the same. In a continuous wave the distance between two crests (or compressions) is marked with $\lambda$. Speed changes only the pace of the animation, not the speed $v$ of the wave.
Assumptions of the model
Every point of the rope or spring repeats the movement of the source with a delay that is the distance of the point from the source divided by the speed of the wave; ahead of the front the medium is at rest. The speed of the wave depends only on the tension: at a high tension $v = 2.0\,\text{m/s}$, at a low one $1.0\,\text{m/s}$. The amplitude is $5.0\,\text{cm}$. In the picture of the rope the displacement is drawn enlarged, in the picture of the spring to the scale of the ruler; in a longitudinal wave the displacement $y$ is parallel to the spring, and the shaded area shows the compressions, where the coils are closer together than usual. The point $P$ is at $x = 1.00\,\text{m}$. The medium continues to the right of the picture so far that no reflected wave returns into the picture, and the wave is not damped. The numbers are values of a model, not measurements.
How the calculation works
The displacement of the point $x$ at the moment $t$ is the displacement of the source at the moment $t - \tfrac{x}{v}$:
$$y = A \sin\left(2\pi f \left(t - \frac{x}{v}\right)\right)$$
when $t - \tfrac{x}{v} \ge 0$, otherwise $y = 0$. In a pulse the source moves for only half a period $\tfrac{T}{2}$, so the pulse is a single crest whose length is $\tfrac{\lambda}{2}$. In the readings $T = \tfrac{1}{f}$, $\lambda = \tfrac{v}{f}$ and the distance travelled by the wave is $x = vt$. The greatest displacement of the point $P$ transversely and longitudinally is the greatest $|y|$ that has occurred since the start.
Experiment 1: does the point $P$ travel with the wave?
Test prediction 1. The frequency and the tension stay constant. Press Start and let the wave travel for at least three seconds, so that it has passed the point $P$ and reached the right edge of the picture. Record the measurement. Then change the kind of wave to longitudinal, start again and record again.
Initial state: wave: transverse, source: continuous wave, tension: high tension, frequency $f$ = $1\,\text{Hz}$. Kept constant: source, tension, frequency $f$. Varied: wave.
| wave | travelled $(\text{m})$ | transverse $(\text{cm})$ | longitudinal $(\text{cm})$ |
|---|
Experiment 2: only the frequency changes
Test prediction 2. The rope and its tension stay the same; change only the frequency of the source. Start the wave, wait until at least two crests are visible in the rope, and record the measurement at least at the frequencies $1.0\,\text{Hz}$ and $2.0\,\text{Hz}$, and if you like also at $0.5\,\text{Hz}$ (then $\lambda = 4.0\,\text{m}$ and two crests fit in the picture only at times, so the reading of the wavelength is enough). Check the wavelength also with the ruler.
Initial state: wave: transverse, source: continuous wave, tension: high tension, frequency $f$ = $1\,\text{Hz}$. Kept constant: wave, source, tension. Varied: frequency $f$.
| frequency $f$ | $\lambda$ $(\text{m})$ | $v$ $(\text{m/s})$ |
|---|
Experiment 3: only the medium changes
Now the frequency stays at $1.0\,\text{Hz}$, and you change the tension of the rope. Record the measurement at a high and at a low tension. What changes and what stays the same?
Initial state: wave: transverse, source: continuous wave, tension: high, frequency $f$ = $1\,\text{Hz}$. Kept constant: wave, source, frequency $f$. Varied: tension.
| tension | $f$ $(\text{Hz})$ | $v$ $(\text{m/s})$ | $\lambda$ $(\text{m})$ |
|---|
Conclusion: what travels in a wave and what determines its speed?
Write on the basis of your tables: How far did the wave travel in experiment 1, and how much and in which direction did the point $P$ move in the transverse and in the longitudinal wave? Calculate the product $f\lambda$ for each row of experiment 2 and compare it with the speed $v$. What changed the speed of the wave in experiment 3, and what happened to the wavelength then?
Model answer
In experiment 1 the wave travelled about $6\,\text{m}$ in three seconds, but the greatest displacement of the point $P$ was only $5.0\,\text{cm}$, the amplitude of the wave. In the transverse wave $P$ moved up and down perpendicular to the rope, in the longitudinal wave back and forth along the spring, but in neither did it travel with the wave. So a wave carries energy, not matter. In experiment 2 the wavelength was $2.0\,\text{m}$ at the frequency $1.0\,\text{Hz}$ and $1.0\,\text{m}$ at $2.0\,\text{Hz}$; the product $f\lambda$ was $2.0\,\text{m/s}$ in every row, the same as the speed $v$. In experiment 3 the lower tension reduced the speed to $1.0\,\text{m/s}$, and the wavelength shortened to $1.0\,\text{m}$ while the frequency stayed the same. The medium determines the speed, the source the frequency, and $v = f\lambda$. The numbers are values of a model.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
A wave is a travelling disturbance
When the end of a rope is flicked upwards, the hand pulls the nearest point of the rope along with it. The taut rope in turn pulls the next point, and that pulls the next one, so every point starts to move a little later than the one before. In this way the displacement from the equilibrium position, the disturbance, travels along the rope away from the hand. Wave motion is the propagation of a disturbance in a medium. The particles of the medium vibrate only around their equilibrium positions and do not travel with the wave. In a gas the molecules also move randomly because of thermal motion, but the wave does not carry them along. So a wave carries energy away from the source, but not matter. This is why the bell at the other end of the rope can ring although no point of the rope travels all the way from the hand to the bell.
A mechanical wave needs a medium, such as a rope, a spring, water or air. One flick makes a pulse, a single hump that travels along the rope. When the hand is moved steadily up and down, a periodic wave forms, in which crests and troughs repeat. Every point of the rope then vibrates with the same frequency $f$ as the hand. Period $T$, frequency $f$ and amplitude $A$ mean the same as in the chapter Describing oscillation: period, frequency and amplitude. When the wave reaches the end of the rope, it is reflected back; reflection is studied in the chapter A wave meets an obstacle or a gap.
Transverse and longitudinal waves
Waves are divided into two kinds according to the direction in which the particles of the medium vibrate. In a transverse wave the particles vibrate perpendicular to the direction of travel of the wave: the points of the rope move up and down while the wave travels horizontally. In a longitudinal wave the particles vibrate parallel to the direction of travel. When the end of a long coil spring is pushed back and forth, compressions form in the spring, where the coils are close together, and rarefactions, where they are far apart. The compressions and rarefactions travel along, but each coil only swings back and forth around its own place. Sound in air, for example, is a longitudinal wave.
The fundamental equation of wave motion
The wavelength $\lambda$ is the distance between two successive crests or two successive compressions. It is the shortest distance between two points that move in exactly the same way. When the source makes one full vibration, that is, during one period $T$, the wave travels exactly one wavelength. The four moments in the figure show how each quarter of a period takes the wave a quarter of a wavelength further.
So the speed of the wave is the wavelength divided by the period. Since the frequency is $f = \tfrac{1}{T}$, we get the fundamental equation of wave motion
$$v = \frac{\lambda}{T} = f\lambda$$
where $v$ is the speed of propagation of the wave ($\text{m/s}$), $\lambda$ the wavelength ($\text{m}$), $T$ the period ($\text{s}$) and $f$ the frequency ($\text{Hz}$). The units agree, because $1\,\text{Hz} = 1\,\text{s}^{-1}$, so $1\,\text{Hz} \cdot 1\,\text{m} = 1\,\text{m/s}$. The speed of a pulse can also be measured directly: a pulse travels at a steady speed in the same medium, so $v = \tfrac{s}{t}$, where $s$ is the distance and $t$ the time it takes.
The speed depends on the medium, the frequency on the source
The speed of a wave is determined by the medium. In a taut rope a wave travels faster than in a slack one, and sound travels in air at $20\,{}^\circ\text{C}$ at about $343\,\text{m/s}$ but in water at about $1500\,\text{m/s}$. The frequency is determined by the source: it is the frequency of the vibration of the hand, the end of the spring or the loudspeaker. In a rope and in air the speed of the wave practically does not depend on the frequency or on a small amplitude. When the same rope is shaken more rapidly, the speed stays the same and the wavelength gets shorter: doubling the frequency halves the wavelength. When a wave passes from one medium to another, the frequency stays the same, but the speed and the wavelength change.
A snapshot and a graph of one point
A wave can be described by two graphs that look like similar sine waves. A snapshot, the $(x, y)$ graph, shows the shape of the whole rope at one moment: the horizontal axis shows the position $x$, and the distance between two crests is the wavelength $\lambda$. A graph of one point, the $(t, y)$ graph, shows the displacement $y$ of one point of the rope as a function of time: the horizontal axis shows the time $t$, and the distance between two peaks is the period $T$. The amplitude $A$ can be read from either graph. So always read first which quantity is on the horizontal axis.
Common misconception
Many people think that a wave travels faster the more rapidly the hand is shaken. In the same rope, however, the speed stays the same. When the frequency increases from $1.0\,\text{Hz}$ to $2.0\,\text{Hz}$, the wavelength shortens from $2.0\,\text{m}$ to $1.0\,\text{m}$, and the product $f\lambda$ is still $2.0\,\text{m/s}$. The speed can be changed only by changing the medium, for example by tightening the rope.
Another common mistake is to read the wavelength from the graph of one point, the $(t, y)$ graph. There the distance between peaks is the period $T$ in seconds, not the wavelength. A third mistake is to think that the wave carries the matter of the rope along: the points of the rope only vibrate in place, and it is energy that moves forward.
Summary
- Wave motion is the propagation of a disturbance in a medium. The particles of the medium vibrate around their equilibrium positions, and the wave carries energy but not matter.
- In a transverse wave the particles vibrate perpendicular to the direction of travel (rope), in a longitudinal wave parallel to it (coil spring, sound in air).
- One flick makes a pulse, steady shaking makes a periodic wave. The wavelength $\lambda$ is the distance between two successive crests.
- The fundamental equation of wave motion is $v = \tfrac{\lambda}{T} = f\lambda$. The speed depends on the medium, the frequency on the source.
- The wavelength is read from a snapshot, the $(x, y)$ graph, and the period from the graph of one point, the $(t, y)$ graph.
Optional Return to your predictions
Compare your predictions with the values of the model in the experiments. Then return to the question at the beginning: what travelled along the rope from the hand to the post, and why could the bell ring although no point of the rope moved to it?
Would you change your prediction?
Apply
Worked example
Problem
A long rope is stretched taut across a yard. When the end of the rope is flicked once, the pulse travels a distance of $6.0\,\text{m}$ in $2.5\,\text{s}$. Then the same rope is shaken steadily at a frequency of $3.0\,\text{Hz}$, and the tension of the rope does not change. What is the wavelength of the wave that forms?
Step 1/7
Given and find. The distance travelled by the pulse is $s = 6.0\,\text{m}$ and the time taken is $t = 2.5\,\text{s}$; the frequency of the periodic wave is $f = 3.0\,\text{Hz}$. We are asked for the wavelength $\lambda$.
Step 2/7
Principle. The speed of a wave depends on the medium, here the rope and its tension, not on whether a pulse or a periodic wave is made in the rope. So the periodic wave travels at the same speed $v$ as the pulse, and the pulse travels steadily, so $v = \tfrac{s}{t}$. The wavelength then follows from the fundamental equation $v = f\lambda$.
Step 3/7
Formula solved.
$$\begin{aligned} v &= \frac{s}{t} \quad\text{and}\quad v = f\lambda \\ \lambda &= \frac{v}{f} = \frac{s}{f\,t} \end{aligned}$$
Step 4/7
Unit conversions. Not needed: the distance is in metres and the time in seconds, and $1\,\text{Hz} = 1\,\text{s}^{-1}$, so $\tfrac{\text{m/s}}{\text{Hz}} = \text{m}$.
Step 5/7
Substitution. First we calculate the speed and then the wavelength:
$$\begin{aligned} v &= \frac{s}{t} \\ &= \frac{6.0\,\text{m}}{2.5\,\text{s}} = 2.4\,\text{m/s} \\[6pt] \lambda &= \frac{v}{f} \\ &= \frac{2.4\,\text{m/s}}{3.0\,\text{Hz}} = 0.80\,\text{m} \end{aligned}$$
Step 6/7
Result. The wavelength is $0.80\,\text{m}$. The answer is given with two significant figures like the initial values $6.0\,\text{m}$, $2.5\,\text{s}$ and $3.0\,\text{Hz}$.
Step 7/7
Reasonableness. The period is $T = \tfrac{1}{f} = 0.333\,\text{s}$, and in that time the wave travels $2.4\,\text{m/s} \cdot 0.333\,\text{s} = 0.80\,\text{m}$, exactly one wavelength, as it should.
Optional Find the error
Optional Apply to a new situation
A long coil spring has been stretched on a table. Its end is pushed back and forth along the spring steadily $2.5$ times a second, and compressions form in the spring with $0.60\,\text{m}$ between their centres.
How long does it take for one compression to travel to the other end of the spring, $6.0\,\text{m}$ away? Answer in seconds.
Hint 1: a guiding question
Which number is the frequency and which the wavelength? Which quantity is needed to get a time from a distance?
Hint 2: what is the connection
The speed of the wave follows from the fundamental equation $v = f\lambda$, and the compression travels at a steady speed, so $t = \tfrac{s}{v}$.
Hint 3: an intermediate step
so $t = \tfrac{6.0\,\text{m}}{1.5\,\text{m/s}}$.
Show the solution
Given
frequency $f = 2.5\,\text{Hz}$ (pushes per second), distance between compressions, that is, the wavelength $\lambda = 0.60\,\text{m}$, distance $s = 6.0\,\text{m}$
Find
the time $t$ in which a compression travels the distance $s$, in seconds
Principle
Compressions are parts of a longitudinal wave, and they travel at the speed of the wave. The distance between two successive compressions is the wavelength, and the frequency is that of the source, that is, the pushes of the hand, so the speed follows from the fundamental equation. In the same spring the speed is constant, so the time is the distance divided by the speed.
Formula
$v = f\lambda$ and $s = vt$, so
$$t = \frac{s}{v} = \frac{s}{f\lambda}$$
Conversions
Not needed: $1\,\text{Hz} = 1\,\text{s}^{-1}$, so $f\lambda$ is in m/s and the distance is in metres.
Substitution
Answer
$4.0\,\text{s}$
Two significant figures like the initial values $2.5$, $0.60\,\text{m}$ and $6.0\,\text{m}$.
Reasonableness
The spring has room for $6.0\,\text{m} : 0.60\,\text{m} = 10$ wavelengths, and the wave travels one wavelength during each period
: $10 \cdot 0.40\,\text{s} = 4.0\,\text{s}$. The speed $1.5\,\text{m/s}$ is a walking pace, which suits a slack table spring.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- What does a wave carry away from the source, and what does it not carry?
Answer
A wave carries energy but not matter. The particles of the medium vibrate around their equilibrium positions, and only the disturbance, that is, the phase of the vibration, moves forward.
- The same rope is shaken three times as rapidly as before, and the tension of the rope does not change. How does the wavelength change?
Answer
The wavelength is a third of the former. The speed $v = f\lambda$ depends on the rope and does not change, so when $f$ triples, $\lambda = \tfrac{v}{f}$ decreases to a third.
1. What stays the same?
2. Read a snapshot
The figure shows a snapshot of a rope at one moment. The end of the rope is shaken at a frequency of $1.8\,\text{Hz}$.
What is the speed of the wave? Answer in metres per second.
Hint 1: a guiding question
On the horizontal axis of the snapshot there is position. Which distance in the figure is the wavelength?
Hint 2: what is the connection
The wavelength is the distance between two successive crests. The crests are at the grid lines.
Hint 3: an intermediate step
The crests are at $0.5\,\text{m}$ and $2.0\,\text{m}$, so
Show the solution
Given
a snapshot of the rope (figure), frequency $f = 1.8\,\text{Hz}$
Find
the speed of the wave $v$ in metres per second
Principle
A snapshot shows the whole rope at one moment, so the distance between two successive crests is the wavelength. During one period the wave travels one wavelength, so the fundamental equation of wave motion gives the speed.
Reading the graph
The crests, at a height of $4\,\text{cm}$, are at the grid lines at $x = 0.5\,\text{m}$, $2.0\,\text{m}$ and $3.5\,\text{m}$. The distance between successive crests is $2.0\,\text{m} - 0.5\,\text{m} = 1.5\,\text{m}$, and the same distance repeats between the following crests.
Formula
$v = f\lambda$
Conversions
Not needed: $1\,\text{Hz} \cdot 1\,\text{m} = 1\,\text{m/s}$. The centimetres of the vertical axis are not needed, because the amplitude does not affect the speed.
Substitution
Answer
$2.7\,\text{m/s}$
Two significant figures like the frequency $1.8\,\text{Hz}$ and the wavelength $1.5\,\text{m}$ read from the figure.
Reasonableness
The period is
and $\tfrac{1.5\,\text{m}}{0.56\,\text{s}} \approx 2.7\,\text{m/s}$ gives the same. A few metres per second is a typical speed in a taut rope.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
3. Sound as a wave
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Wave motion
- An oscillation travelling from place to place so that energy is transferred but the matter does not travel with it.
- Medium
- A gas, liquid or solid in which a mechanical wave travels through the interactions of its particles.
- Pulse
- A single disturbance travelling through a medium.
- Wavelength
- The distance between two neighbouring points in the same phase of oscillation.
- Wave speed
- The speed at which a disturbance travels through a medium; it depends on the properties of the medium.
- Fundamental equation of wave motion
- The link between the wave speed, the frequency and the wavelength.$$v = \lambda f$$
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 10-1Core
The end of a long coil spring is shaken sideways, and a transverse wave with a frequency of $4.0\,\text{Hz}$ and a wavelength of $0.75\,\text{m}$ travels in the spring.
What is the speed of the wave?
Answer
$3.0\,\text{m/s}$
Show the solution
Given
frequency $f = 4.0\,\text{Hz}$, wavelength $\lambda = 0.75\,\text{m}$
Find
the speed of the wave $v$
Principle
During one period the wave travels one wavelength, so the speed is the wavelength times the frequency. The fundamental equation holds for all periodic waves, also for a transverse wave travelling in a spring.
Formula
$v = f\lambda$
Conversions
Not needed: $1\,\text{Hz} = 1\,\text{s}^{-1}$, so $\text{Hz} \cdot \text{m} = \text{m/s}$.
Substitution
$v$$= 4.0\,\text{s}^{-1} \cdot 0.75\,\text{m}$$= 3.0\,\text{m/s}$Answer
$3.0\,\text{m/s}$
Two significant figures like the frequency $4.0\,\text{Hz}$ and the wavelength $0.75\,\text{m}$.
Reasonableness
The period is
$T$$= \dfrac{1}{4.0\,\text{Hz}}$$= 0.25\,\text{s}$and $0.75\,\text{m} : 0.25\,\text{s} = 3.0\,\text{m/s}$ gives the same. A few metres per second is a typical speed in a stretched spring: the wave can be seen travelling at a pace the eye can follow.
- Exercise 10-2Core
A loudspeaker plays a note whose frequency is $440\,\text{Hz}$. The sound travels in the air of the room at a speed of $343\,\text{m/s}$.
How long is the wavelength of the sound?
Answer
$0.780\,\text{m}$
Show the solution
Given
frequency $f = 440\,\text{Hz}$, speed of sound in air $v = 343\,\text{m/s}$
Find
the wavelength $\lambda$
Principle
Sound is a longitudinal wave travelling in air, and the fundamental equation of wave motion holds for it as well: during one period the sound travels one wavelength. The frequency is the frequency of the vibration of the loudspeaker cone.
Formula
$v = f\lambda$, so $\lambda = \dfrac{v}{f}$
Conversions
Not needed:
$\tfrac{\text{m/s}}{\text{Hz}}$$= \tfrac{\text{m/s}}{1/\text{s}}$$= \text{m}$Substitution
$\lambda$$= \dfrac{343\,\text{m/s}}{440\,\text{s}^{-1}}$$= 0.7795\,\text{m}$Answer
$0.780\,\text{m}$
Three significant figures like the speed $343\,\text{m/s}$ and the frequency $440\,\text{Hz}$.
Reasonableness
An estimate with round numbers: $\tfrac{350\,\text{m/s}}{450\,\text{Hz}} \approx 0.8\,\text{m}$. The wavelengths of audible sound in air range from about two centimetres to over ten metres, so a wavelength of a little under a metre suits a note in the middle of the range.
- Exercise 10-3
In a taut rope a wave travels at a speed of $6.0\,\text{m/s}$, and the distance between the crests of the wave is $1.5\,\text{m}$.
At what frequency is the end of the rope shaken?
Answer
$4.0\,\text{Hz}$
Show the solution
Given
speed of the wave $v = 6.0\,\text{m/s}$, distance between crests, that is, the wavelength $\lambda = 1.5\,\text{m}$
Find
the frequency $f$
Principle
The distance between two successive crests is the wavelength. Every point of the rope vibrates with the same frequency as the hand, so the frequency of the wave is the frequency of the shaking, and it follows from the fundamental equation.
Formula
$v = f\lambda$, so $f = \dfrac{v}{\lambda}$
Conversions
Not needed:
$\tfrac{\text{m/s}}{\text{m}}$$= \text{s}^{-1}$$= \text{Hz}$Substitution
$f$$= \dfrac{6.0\,\text{m/s}}{1.5\,\text{m}}$$= 4.0\,\text{s}^{-1}$$= 4.0\,\text{Hz}$Answer
$4.0\,\text{Hz}$
Two significant figures like the initial values $6.0\,\text{m/s}$ and $1.5\,\text{m}$.
Reasonableness
Four shakes per second is fast but possible by hand. A check: in a second the wave travels $6.0\,\text{m}$, and $6.0\,\text{m} : 1.5\,\text{m} = 4$ wavelengths fit in that distance, that is, four crests formed in a second.
- Exercise 10-4
The wave machine of an indoor wave pool makes waves whose period is $2.5\,\text{s}$ and wavelength $4.0\,\text{m}$. At what speed do the waves travel?
Answer
$1.6\,\text{m/s}$
Show the solution
Given
period $T = 2.5\,\text{s}$, wavelength $\lambda = 4.0\,\text{m}$
Find
the speed of the waves $v$
Principle
During one period the wave travels one wavelength. So the speed is the wavelength divided by the period; this form of the fundamental equation fits because the problem gives the period and not the frequency.
Formula
$v = \dfrac{\lambda}{T}$
Conversions
Not needed: the wavelength is in metres and the period in seconds.
Substitution
$v$$= \dfrac{4.0\,\text{m}}{2.5\,\text{s}}$$= 1.6\,\text{m/s}$Answer
$1.6\,\text{m/s}$
Two significant figures like the period $2.5\,\text{s}$ and the wavelength $4.0\,\text{m}$.
Reasonableness
The same follows through the frequency:
$f$$= \tfrac{1}{2.5\,\text{s}}$$= 0.40\,\text{Hz}$$v$$= 0.40\,\text{Hz} \cdot 4.0\,\text{m}$$= 1.6\,\text{m/s}$A pace a little faster than walking suits the gentle waves of a wave pool.
- Exercise 10-5
In a rope a wave travels at a speed of $4.8\,\text{m/s}$, and its wavelength is $1.2\,\text{m}$. What is the period of the wave?
Answer
$0.25\,\text{s}$
Show the solution
Given
speed of the wave $v = 4.8\,\text{m/s}$, wavelength $\lambda = 1.2\,\text{m}$
Find
the period $T$
Principle
The period is the time in which the wave travels one wavelength, so it follows from the form $v = \tfrac{\lambda}{T}$ of the fundamental equation. The period is also the time of one full vibration of every point of the rope.
Formula
$v = \dfrac{\lambda}{T}$, so $T = \dfrac{\lambda}{v}$
Conversions
Not needed: $\tfrac{\text{m}}{\text{m/s}} = \text{s}$.
Substitution
$T$$= \dfrac{1.2\,\text{m}}{4.8\,\text{m/s}}$$= 0.25\,\text{s}$Answer
$0.25\,\text{s}$
Two significant figures like the speed $4.8\,\text{m/s}$ and the wavelength $1.2\,\text{m}$.
Reasonableness
The frequency is
$f$$= \dfrac{1}{0.25\,\text{s}}$$= 4.0\,\text{Hz}$and $4.0\,\text{Hz} \cdot 1.2\,\text{m} = 4.8\,\text{m/s}$, as given. The wavelength is a quarter of the distance the wave travels in a second, so the period must be a quarter of a second.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 10-6Core
The graph shows the displacement of one point of a taut rope as a function of time when the end of the rope is shaken steadily. From a snapshot of the rope it has been measured that the distance between successive crests is $1.0\,\text{m}$. Read the necessary value from the graph.
What is the speed of the wave?
Answer
$2.5\,\text{m/s}$
Show the solution
Given
the $(t, y)$ graph of one point of the rope (figure), distance between crests in the snapshot, that is, the wavelength $\lambda = 1.0\,\text{m}$
Find
the speed of the wave $v$
Principle
On the horizontal axis of the graph of one point there is time, so the distance between two similar points is the period $T$. The wavelength comes from the snapshot. During one period the wave travels one wavelength.
Reading the graph
The curve crosses the time axis going upwards at the moments $0$, $0.4\,\text{s}$, $0.8\,\text{s}$, … , $2.0\,\text{s}$. So in the interval $2.0\,\text{s}$ there are $5$ full vibrations, and
$T$$= 2.0\,\text{s} : 5$$= 0.40\,\text{s}$Formula
$v = \dfrac{\lambda}{T}$
Conversions
Not needed: the wavelength is in metres and the period in seconds. The displacement in centimetres is not needed, because the amplitude does not affect the speed.
Substitution
$v$$= \dfrac{1.0\,\text{m}}{0.40\,\text{s}}$$= 2.5\,\text{m/s}$Answer
$2.5\,\text{m/s}$
Two significant figures like the wavelength $1.0\,\text{m}$ and the period $0.40\,\text{s}$ read from the graph.
Reasonableness
The frequency is
$f$$= \tfrac{1}{0.40\,\text{s}}$$= 2.5\,\text{Hz}$and
$f\lambda$$= 2.5\,\text{Hz} \cdot 1.0\,\text{m}$$= 2.5\,\text{m/s}$gives the same. The distance between peaks, $0.40\,\text{s}$, is not the wavelength, because it is a time.
- Exercise 10-7Core
A student shakes the end of a long skipping rope steadily and counts that the hand makes $18$ full swings in $12.0$ seconds. The distance between the crests of the rope is $1.40\,\text{m}$.
What is the speed of the wave in the rope?
Answer
$2.10\,\text{m/s}$
Show the solution
Given
$N = 18$ full swings in the time $t = 12.0\,\text{s}$, distance between crests, that is, the wavelength $\lambda = 1.40\,\text{m}$
Find
the speed of the wave $v$
Principle
Every point of the rope vibrates with the frequency of the hand, so the frequency of the wave is the number of swings of the hand per second. The speed then follows from the fundamental equation, because during one period the wave travels one wavelength.
Formula
$f = \dfrac{N}{t}$ and $v = f\lambda$, so $v = \dfrac{N\lambda}{t}$
Conversions
Not needed: $\text{s}^{-1} = \text{Hz}$ and $\text{Hz} \cdot \text{m} = \text{m/s}$.
Substitution
$f$$= \dfrac{18}{12.0\,\text{s}}$$= 1.5\,\text{Hz}$$v$$= 1.5\,\text{s}^{-1} \cdot 1.40\,\text{m}$$= 2.10\,\text{m/s}$Answer
$2.10\,\text{m/s}$
Three significant figures like the time $12.0\,\text{s}$ and the wavelength $1.40\,\text{m}$; the number of swings $18$ is a counted number, so it does not limit the significant figures.
Reasonableness
The period is
$T$$= \tfrac{12.0\,\text{s}}{18}$$= 0.667\,\text{s}$and $\tfrac{1.40\,\text{m}}{0.667\,\text{s}} = 2.10\,\text{m/s}$. The unit also agrees: $\text{s}^{-1} \cdot \text{m} = \text{m/s}$, and a few metres per second is a typical speed for a skipping rope.
- Exercise 10-8
The sound of a loudspeaker at the edge of a pool travels in air at a speed of $343\,\text{m/s}$, and its wavelength in air is $0.750\,\text{m}$. The sound continues into water, where it travels at a speed of $1480\,\text{m/s}$.
How long is the wavelength of the sound in water?
Answer
$3.24\,\text{m}$
Show the solution
Given
in air $v_\text{a} = 343\,\text{m/s}$ and $\lambda_\text{a} = 0.750\,\text{m}$, in water $v_\text{w} = 1480\,\text{m/s}$
Find
the wavelength in water $\lambda_\text{w}$
Principle
When a wave passes from one medium to another, the frequency stays the same, because the water is set moving by the vibration of the air at the boundary in step with it. The speed changes with the medium, so the wavelength changes. In both media $v = f\lambda$ holds.
Formula
$f$$= \dfrac{v_\text{a}}{\lambda_\text{a}}$$\lambda_\text{w}$$= \dfrac{v_\text{w}}{f}$so
$\lambda_\text{w}$$= \dfrac{v_\text{w}\lambda_\text{a}}{v_\text{a}}$Conversions
Not needed: the speeds are in the same unit, so their ratio has no unit, and the wavelength is in metres.
Substitution
$f$$= \dfrac{343\,\text{m/s}}{0.750\,\text{m}}$$= 457.3\,\text{Hz}$$\lambda_\text{w}$$= \dfrac{1480\,\text{m/s}}{457.3\,\text{s}^{-1}}$$= 3.236\,\text{m}$Answer
$3.24\,\text{m}$
Three significant figures like the initial values $343\,\text{m/s}$, $0.750\,\text{m}$ and $1480\,\text{m/s}$.
Reasonableness
Sound travels in water about $\tfrac{1480}{343} \approx 4.3$ times faster than in air, so at the same frequency the wavelength is about $4.3$ times as long: $4.3 \cdot 0.750\,\text{m} \approx 3.2\,\text{m}$.
- Exercise 10-9
The longitudinal wave of an earthquake travels in the bedrock. A seismometer measures the frequency of the wave as $0.50\,\text{Hz}$, and the wavelength is $12\,\text{km}$.
What is the speed of the wave in metres per second?
Answer
$6.0 \cdot 10^{3}\,\text{m/s}$
Show the solution
Given
frequency $f = 0.50\,\text{Hz}$, wavelength $\lambda = 12\,\text{km}$
Find
the speed of the wave $v$ in metres per second
Principle
The longitudinal wave of an earthquake is a mechanical wave in which the rock vibrates parallel to the direction of travel. The fundamental equation of wave motion holds for it as well, because during one period the wave travels one wavelength.
Formula
$v = f\lambda$
Conversions
$\lambda$$= 12\,\text{km}$$= 12 \cdot 1000\,\text{m}$$= 12\,000\,\text{m}$Substitution
$v$$= 0.50\,\text{s}^{-1} \cdot 12\,000\,\text{m}$$= 6000\,\text{m/s}$Answer
$6.0 \cdot 10^{3}\,\text{m/s}$
That is, $6000\,\text{m/s}$, two significant figures like the frequency $0.50\,\text{Hz}$ and the wavelength $12\,\text{km}$.
Reasonableness
The period is
$T$$= \tfrac{1}{0.50\,\text{Hz}}$$= 2.0\,\text{s}$and $12\,\text{km}$ in two seconds is $6\,\text{km/s}$. Sound travels in solid rock much faster than in air ($343\,\text{m/s}$), so kilometres per second is a believable speed.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 10-10
Two photographs are taken of a rope at an interval of $0.15\,\text{s}$. In the first picture a crest is at $x = 0.60\,\text{m}$ and in the second at $x = 1.05\,\text{m}$, and between the pictures the crest has not had time to travel one wavelength. In both pictures the distance between successive crests is $1.20\,\text{m}$.
At what frequency is the end of the rope shaken? Also justify why the displacement of the crest is not a displacement of the matter of the rope.
Answer
$2.5\,\text{Hz}$
Show the solution
Given
a crest at $x_1 = 0.60\,\text{m}$ and $\Delta t = 0.15\,\text{s}$ later at $x_2 = 1.05\,\text{m}$; the crest has not travelled one wavelength; distance between crests $\lambda = 1.20\,\text{m}$
Find
the frequency of the shaking $f$
Principle
A crest is a part of the wave, so it travels at the speed of the wave, $v = \tfrac{\Delta x}{\Delta t}$. Since the crest has travelled less than a wavelength, the photographs show the same crest and not already the next one. The displacement of the crest is not a displacement of the matter of the rope: the points of the rope only move up and down, and the crest is always at the point that is at its highest position at that moment. The frequency follows from the fundamental equation.
Formula
$v = \dfrac{x_2 - x_1}{\Delta t}$ and $v = f\lambda$, so
$f$$= \dfrac{x_2 - x_1}{\lambda\,\Delta t}$Conversions
Not needed: all lengths are in metres and the time in seconds.
Substitution
$v$$= \dfrac{1.05\,\text{m} - 0.60\,\text{m}}{0.15\,\text{s}}$$= \dfrac{0.45\,\text{m}}{0.15\,\text{s}}$$= 3.0\,\text{m/s}$$f$$= \dfrac{3.0\,\text{m/s}}{1.20\,\text{m}}$$= 2.5\,\text{Hz}$Answer
$2.5\,\text{Hz}$
Two significant figures, because the time $0.15\,\text{s}$ and the displacement $0.45\,\text{m}$ are known to two figures.
Reasonableness
The period is
$T$$= \tfrac{1}{2.5\,\text{Hz}}$$= 0.40\,\text{s}$so the interval between the photographs, $0.15\,\text{s}$, is less than a period, and the crest travelled $0.45\,\text{m}$, less than the wavelength $1.20\,\text{m}$, as assumed.
- Exercise 10-11
One end of a long railway rail is struck with a hammer. A listener at the other end of the rail hears two bangs: first the sound that came along the rail and $0.50\,\text{s}$ later the sound that came through the air. Sound travels in steel at a speed of $5100\,\text{m/s}$ and in air at $340\,\text{m/s}$.
How long is the rail? Justify why the same blow of the hammer produces two sounds that arrive at different times.
Answer
$1.8 \cdot 10^{2}\,\text{m}$
Show the solution
Given
time difference $\Delta t = 0.50\,\text{s}$, speed of sound in steel $v_\text{s} = 5100\,\text{m/s}$ and in air $v_\text{a} = 340\,\text{m/s}$
Find
the length of the rail $d$
Principle
The blow of the hammer produces sound both in the rail and in the air. The speed of a wave depends on the medium, so the two sounds travel the same distance $d$ at different speeds and arrive at different times: the sound that travelled in the steel first. Each travels at a steady speed, so the travel time is $\tfrac{d}{v}$, and the time difference is the difference of the travel times.
Formula
$\Delta t$$= \dfrac{d}{v_\text{a}} - \dfrac{d}{v_\text{s}}$so
$d$$= \dfrac{\Delta t}{\dfrac{1}{v_\text{a}} - \dfrac{1}{v_\text{s}}}$Conversions
Not needed: the speeds are in metres per second and the time in seconds.
Substitution
$d$$= \dfrac{0.50\,\text{s}}{\dfrac{1}{340\,\text{m/s}} - \dfrac{1}{5100\,\text{m/s}}}$$= \dfrac{0.50\,\text{s}}{0.002\,745\,\text{s/m}}$$= 182.1\,\text{m}$Answer
$1.8 \cdot 10^{2}\,\text{m}$
That is, about $180\,\text{m}$, two significant figures, because the time difference $0.50\,\text{s}$ is known to two figures.
Justification
The same blow produces sound both in the rail and in the air, and sound travels much faster in steel than in air. Both sounds travel the same distance, but the one that travelled in the steel arrives first, so the listener hears two bangs.
Reasonableness
In air the distance takes $\tfrac{182\,\text{m}}{340\,\text{m/s}} = 0.536\,\text{s}$ and in steel $\tfrac{182\,\text{m}}{5100\,\text{m/s}} = 0.036\,\text{s}$; the difference is $0.50\,\text{s}$. The sound in the steel is so fast that the length is only slightly more than the sound in air would travel in half a second ($170\,\text{m}$).
- Exercise 10-12
A wave goes round a stadium stand: every spectator stands up $0.050\,\text{s}$ later than the one sitting next to them and then sits down again. The distance between adjacent seats is $0.55\,\text{m}$, and the length of the row going round the stand is $440\,\text{m}$.
How long does it take for the wave to go once round the stand? Justify why the wave travels although no spectator changes places.
Answer
$40\,\text{s}$
Show the solution
Given
delay relative to the neighbour $\Delta t = 0.050\,\text{s}$, distance between seats $\Delta x = 0.55\,\text{m}$, length of the row $s = 440\,\text{m}$
Find
the time $t$ in which the wave goes once round the stand
Principle
The wave in a stadium stand is a model of wave motion: every spectator repeats the movement of the neighbour after a small delay, so the standing up, the disturbance, travels along the row although the spectators stay in their places and only stand up and sit down. The disturbance travels one seat spacing in the time $\Delta t$, so its speed is constant and the time of one round is the distance divided by the speed.
Formula
$v = \dfrac{\Delta x}{\Delta t}$ and $s = vt$, so
$t$$= \dfrac{s}{v}$$= \dfrac{s\,\Delta t}{\Delta x}$Conversions
Not needed: the lengths are in metres and the time in seconds.
Substitution
$v$$= \dfrac{0.55\,\text{m}}{0.050\,\text{s}}$$= 11\,\text{m/s}$$t$$= \dfrac{440\,\text{m}}{11\,\text{m/s}}$$= 40\,\text{s}$Answer
$40\,\text{s}$
Two significant figures like the delay $0.050\,\text{s}$ and the seat spacing $0.55\,\text{m}$.
Reasonableness
The row has $440\,\text{m} : 0.55\,\text{m} = 800$ spectators, and each adds a delay of $0.050\,\text{s}$: $800 \cdot 0.050\,\text{s} = 40\,\text{s}$. About $11\,\text{m/s}$ is close to the speed of the wave observed in stadiums, and an individual spectator moves only about a metre up and down.