Physics · Secondary School · Waves

Waves and Oscillations

The swing of a pendulum, a note on a guitar, an echo off a cliff and the ambulance siren that changes pitch as it goes past all depend on a handful of ideas about oscillations and waves. We start with simple harmonic motion, the simplest back-and-forth there is, follow an oscillation as it spreads through a medium and becomes a wave, and finish with sound. In the calculations we take \(g = 10\ \text{m/s}^2\).

1Simple harmonic motion
2Simple pendulum
3What a wave is
4Reflection and refraction
5Interference
6Standing waves
7Sound
✓Challenges

STEP 1

A back-and-forth that repeats itself

When we pull a block attached to a spring and let it go, it moves back and forth around its equilibrium position, always at the same rhythm. If we neglect friction, the spring force is \(F = -k\,x\), proportional to the displacement and always pointing back towards equilibrium, and the motion it produces is called simple harmonic motion (SHM).

The position follows a cosine, \(x(t) = A\cos(\omega t + \varphi_0)\). The amplitude \(A\) is the largest displacement, \(\omega\) is the angular frequency, in rad/s, and the initial phase \(\varphi_0\) tells us at which point of the back-and-forth the clock started counting.

This cosine has an origin we already know from the Circular Motion lesson. A point going round a circle of radius \(A\) in uniform circular motion, with angular velocity \(\omega\), casts a shadow on the horizontal diameter that moves back and forth exactly like the block, which is why the angular frequency of SHM uses the same letter and the same unit as angular velocity.

\(x(t) = A\cos(\omega t + \varphi_0)\)\(\omega = \sqrt{\dfrac{k}{m}} = \dfrac{2\pi}{T}\)\(T = 2\pi\sqrt{\dfrac{m}{k}}\)\(v_{\text{max}} = \omega A\)\(a_{\text{max}} = \omega^2 A\)\(E = \tfrac{1}{2}k\,x^2 + \tfrac{1}{2}m\,v^2 = \tfrac{1}{2}k\,A^2\)\(k\) in N/m, \(m\) in kg, \(\omega\) in rad/s and \(f = 1/T\) in hertz. We assume an ideal spring, massless and obeying Hooke's law, and no friction at all. With a real spring the amplitude tends to die away little by little.

The period depends only on the mass and on the stiffness of the spring, \(T = 2\pi\sqrt{m/k}\). A larger mass takes longer to get moving, and a stiffer spring pulls harder. The amplitude does not appear in the formula, because a block released further out also moves faster and takes the same time to come back.

As the block moves, the energy keeps changing form. At the ends the block stops for an instant and the spring holds all the energy, \(\tfrac{1}{2}k\,A^2\), while at the centre the spring is relaxed and the energy is all kinetic. Without friction, the sum of the two stays constant.

Let's discuss

Release the block and compare the shadow of the point going round the circle with the block itself. At what moment of the turn does the block pass the centre at its highest speed?
Make the mass four times larger without touching the spring. Does the period double, as the square root in the formula suggests?
Change only the amplitude and follow the period on the graph. Our intuition tends to say that a bigger swing takes longer, so does the period seem to change?
Pause when the kinetic energy bar is at its maximum and see where the block is. And where is it when the spring energy is at its maximum?

Block mass

Spring constant k

Amplitude A

Angular frequency ω

Period T

Frequency f

Position x

Velocity v

Total energy

Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A 0.5 kg block is attached to a spring of constant \(k = 50\ \text{N/m}\) and oscillates on a frictionless table. What is the angular frequency \(\omega\), and what is the period? Use \(\pi \approx 3.14\).

Show solution
The angular frequency depends only on the mass and the spring, \(\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{50}{0.5}}\) \(= \sqrt{100} = 10\ \text{rad/s}\).
The period is the time the point on the reference circle takes to go once round, \(T = \dfrac{2\pi}{\omega}\).
\(T = \dfrac{2 \cdot 3.14}{10} \approx 0.63\ \text{s}\)
basic
The position of a body in SHM is given, in SI units, by \(x(t) = 0.2\cos(4t)\). What are the amplitude, the angular frequency, the period and the frequency? Use \(\pi \approx 3\).

Show solution
Comparing with \(x = A\cos(\omega t + \varphi_0)\), we read off \(A = 0.2\ \text{m}\) and \(\omega = 4\ \text{rad/s}\), with \(\varphi_0 = 0\).
The period is \(T = \dfrac{2\pi}{\omega} = \dfrac{2 \cdot 3}{4}\) \(= 1.5\ \text{s}\).
\(A = 0.2\ \text{m}\), \(\omega = 4\ \text{rad/s}\), \(T = 1.5\ \text{s}\) and \(f = \dfrac{1}{1.5} \approx 0.67\ \text{Hz}\).
basic
In SHM, at which points of the motion is the speed greatest, and at which points is the acceleration greatest?

Show solution
At the ends the body stops for an instant to turn back, and the spring is at its most stretched or most compressed, with the largest force, \(|F| = k\,A\).
At the centre the spring is relaxed, the force is zero and the body goes past with all its energy in kinetic form.
The speed is greatest at the centre, \(v_{\text{max}} = \omega A\), and the acceleration is greatest at the ends, \(a_{\text{max}} = \omega^2 A\).
basic
A block on a spring makes 20 complete oscillations in 10 s. What is the period, and what is the frequency?

Show solution
The period is the time of one oscillation, \(T = \dfrac{10}{20} = 0.5\ \text{s}\).
\(T = 0.5\ \text{s}\) and \(f = \dfrac{1}{T} = 2\ \text{Hz}\)
intermediate
A 0.2 kg block oscillates on a 20 N/m spring. We replace it with a 0.8 kg block on the same spring. By what factor does the period increase? Use \(\pi \approx 3.14\) and work out both periods.

Show solution
With the first block, \(T_1 = 2\pi\sqrt{\dfrac{0.2}{20}}\) \(= 2\pi \cdot 0.1 \approx 0.63\ \text{s}\).
With the second, \(T_2 = 2\pi\sqrt{\dfrac{0.8}{20}}\) \(= 2\pi \cdot 0.2 \approx 1.26\ \text{s}\).
The period doubles, because the mass became 4 times larger and \(\sqrt{4} = 2\).
intermediate
A 0.5 kg block, attached to a 200 N/m spring, is pulled 10 cm and released. Neglecting friction, what is the mechanical energy of the system, and what is the speed of the block as it passes the centre?

Show solution
At the end all the energy is in the spring, \(E = \dfrac{1}{2}k\,A^2\) \(= \dfrac{1}{2} \cdot 200 \cdot 0.1^2 = 1\ \text{J}\).
At the centre it is all kinetic, \(\dfrac{1}{2} \cdot 0.5 \cdot v^2 = 1\), and so \(v^2 = 4\).
Checking another way, \(\omega = \sqrt{200/0.5} = 20\ \text{rad/s}\) and \(\omega A = 20 \cdot 0.1\) give the same value.
\(E = 1\ \text{J}\) and \(v = 2\ \text{m/s}\)
intermediate
A point moves in uniform circular motion on a circle of radius 5 cm, making 2 turns per second, and a torch casts its shadow on a wall. What motion does the shadow make, and with what amplitude, frequency and maximum speed? Use \(\pi \approx 3.14\).

Show solution
As we saw in the Circular Motion lesson, the projection of uniform circular motion onto a diameter is SHM with the same frequency and an amplitude equal to the radius, \(A = 5\ \text{cm}\) and \(f = 2\ \text{Hz}\).
The angular frequency is the angular velocity of the point, \(\omega = 2\pi f = 4\pi \approx 12.6\ \text{rad/s}\).
The shadow moves fastest when the point passes the top or the bottom of the circle, and at that moment it has the same speed as the point.
SHM with \(A = 5\ \text{cm}\), \(f = 2\ \text{Hz}\) and \(v_{\text{max}} = \omega A = 12.6 \cdot 0.05\) \(\approx 0.63\ \text{m/s}\)
intermediate
In SHM of amplitude \(A\), what fraction of the mechanical energy is kinetic when the body passes through \(x = A/2\)?

Show solution
The elastic energy at that point is \(\dfrac{1}{2}k\left(\dfrac{A}{2}\right)^2 = \dfrac{1}{4} \cdot \dfrac{1}{2}k\,A^2\), a quarter of the total energy.
Without friction, the rest of the total energy is in kinetic form.
Three quarters, or 75%, of the energy is kinetic at \(x = A/2\).
challenge
We hang a 0.4 kg body from a vertical spring, and the spring stretches 10 cm before reaching equilibrium. Then we pull the body down a little and let go. What is the period of the oscillation? Use \(g = 10\ \text{m/s}^2\) and \(\pi \approx 3.14\).

Show solution
At equilibrium the spring force balances the weight, \(k \cdot 0.1 = 0.4 \cdot 10\), and so \(k = 40\ \text{N/m}\).
On a vertical spring the weight only shifts the equilibrium position, and about that position the body performs SHM with the same period as on a horizontal spring.
\(T = 2\pi\sqrt{\dfrac{0.4}{40}}\) \(= 2\pi \cdot 0.1 \approx 0.63\ \text{s}\)
challenge
A 1 kg block on a 100 N/m spring, on a frictionless table, is released 20 cm from equilibrium. What is its speed as it passes 12 cm from equilibrium?

Show solution
The total energy is that of the spring at the end, \(E = \dfrac{1}{2} \cdot 100 \cdot 0.2^2 = 2\ \text{J}\).
At 12 cm the spring holds \(\dfrac{1}{2} \cdot 100 \cdot 0.12^2 = 0.72\ \text{J}\), and the rest is kinetic, \(2 - 0.72 = 1.28\ \text{J}\).
So \(\dfrac{1}{2} \cdot 1 \cdot v^2 = 1.28\), that is, \(v^2 = 2.56\).
\(v = 1.6\ \text{m/s}\)

STEP 2

A clock made of string and a weight

A simple pendulum is a small mass hanging from a light string and swinging to and fro around the vertical. For small angles, the part of the weight that pulls the mass back is almost proportional to the displacement, and the swing becomes SHM.

From this comes \(T = 2\pi\sqrt{L/g}\), which holds if we assume small oscillations, up to about 15°, a massless string and no air resistance. The mass does not appear, because a heavier body feels a larger force and is also harder to accelerate, in the same proportion, just as in free fall.

The amplitude hardly matters. The story goes that Galileo noticed this while watching a chandelier swing in Pisa cathedral, and it was this property that Huygens used, in 1656, to build the first pendulum clock. For large angles the period grows a little, by about 7% at 60°.

The formula also works the other way round. Measuring \(L\) and \(T\) gives \(g = 4\pi^2 L/T^2\), and a carefully built pendulum can measure the local gravity with good precision.

\(T = 2\pi\sqrt{\dfrac{L}{g}}\)\(g = \dfrac{4\pi^2 L}{T^2}\)\(f = \dfrac{1}{T}\)Valid for small oscillations, up to about 15°. In the simulation the Earth has its measured \(g\), 9.8 m/s², the Moon 1.6 and Mars 3.7; in the calculations we use \(g = 10\ \text{m/s}^2\). The simulation solves the full motion, without the small-angle approximation, and ignores the air.

Let's discuss

With an amplitude of 10°, time 10 oscillations and compare the measured period with the one from the formula.
Take the same pendulum to the Moon. How many times slower does it seem to get, and where can we find that number?
Raise the amplitude to 70° and time it again. Does the difference from the formula still look negligible?
Which length gives a period of 2 s on Earth? Hunt for it with the slider and check it against the formula.

Length L

Amplitude

T from formula

Exact T

Stopwatch

Measured T

Calculated g

Angle now

Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
What is the period of a 2.5 m simple pendulum swinging through a small angle? Use \(g = 10\ \text{m/s}^2\) and \(\pi \approx 3.14\).

Show solution
For small oscillations, \(T = 2\pi\sqrt{\dfrac{L}{g}} = 2\pi\sqrt{\dfrac{2.5}{10}}\) \(= 2\pi \cdot 0.5\).
\(T = \pi \approx 3.14\ \text{s}\)
basic
On a lab pendulum, we swap the 100 g ball for a 500 g one of the same size, without changing the string. What happens to the period?

Show solution
The mass does not appear in \(T = 2\pi\sqrt{L/g}\). A heavier ball feels a larger force and is harder to accelerate, in the same proportion.
The period does not change.
basic
A pendulum completes 30 oscillations in 60 s. What is the period, and what is the frequency?

Show solution
The period is the time of one oscillation, \(T = \dfrac{60}{30} = 2\ \text{s}\).
\(T = 2\ \text{s}\) and \(f = \dfrac{1}{2} = 0.5\ \text{Hz}\)
basic
On the Moon, the acceleration due to gravity is about 6 times smaller than on Earth. If we take the same pendulum there, does it swing faster or slower, and by what factor does the period change?

Show solution
With \(g\) in the denominator, inside the square root, a \(g\) six times smaller makes the period \(\sqrt{6}\) times larger.
The pendulum gets slower, with a period about \(\sqrt{6} \approx 2.4\) times longer.
intermediate
How long must a pendulum be for its period to be 2 s? Use \(g = 10\ \text{m/s}^2\) and \(\pi^2 \approx 10\).

Show solution
Squaring \(T = 2\pi\sqrt{L/g}\) and isolating \(L\), we get \(L = \dfrac{g\,T^2}{4\pi^2}\).
\(L = \dfrac{10 \cdot 2^2}{4 \cdot 10} = 1\ \text{m}\)
intermediate
A 0.4 m pendulum is replaced by a 1.6 m one. Work out both periods and say by what factor the period increased. Use \(g = 10\ \text{m/s}^2\) and \(\pi \approx 3.14\).

Show solution
With 0.4 m, \(T_1 = 2\pi\sqrt{\dfrac{0.4}{10}}\) \(= 2\pi \cdot 0.2 \approx 1.26\ \text{s}\).
With 1.6 m, \(T_2 = 2\pi\sqrt{\dfrac{1.6}{10}}\) \(= 2\pi \cdot 0.4 \approx 2.51\ \text{s}\).
The period doubled, because the length became 4 times larger.
intermediate
Joaquim measures a 0.90 m pendulum and times 10 small oscillations in 19.0 s. What value of \(g\) does he get? Use \(\pi \approx 3.14\).

Show solution
The period is \(T = \dfrac{19.0}{10} = 1.90\ \text{s}\). Timing several oscillations and dividing reduces the weight of the reaction-time error on the stopwatch.
Isolating \(g\) in the pendulum formula, \(g = \dfrac{4\pi^2 L}{T^2}\).
\(g = \dfrac{4 \cdot 3.14^2 \cdot 0.90}{1.90^2}\) \(\approx 9.8\ \text{m/s}^2\)
intermediate
On a hot day, the metal rod of a pendulum clock expands a little. Does the clock start to run fast or slow?

Show solution
With a longer rod, \(L\) increases and so does the period \(T = 2\pi\sqrt{L/g}\).
Each swing takes longer, and the clock counts fewer swings than it should in the same stretch of time.
The clock runs slow.
challenge
A pendulum clock, set on Earth for a period of 2 s, is taken to a planet where the same pendulum has a period of 3 s. What is the gravity on that planet? Use \(g = 10\ \text{m/s}^2\) for the Earth.

Show solution
With the same \(L\), the period is inversely proportional to \(\sqrt{g}\), and so \(\dfrac{g_{\text{planet}}}{g_{\text{Earth}}} = \left(\dfrac{T_{\text{Earth}}}{T_{\text{planet}}}\right)^2\) \(= \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}\).
\(g_{\text{planet}} = 10 \cdot \dfrac{4}{9} \approx 4.4\ \text{m/s}^2\)
challenge
A 0.8 m pendulum is released from 60° to the vertical. Neglecting the air, at what speed does the mass pass the lowest point? Why can we not use the small-oscillation period formula for this swing? Use \(g = 10\ \text{m/s}^2\).

Show solution
The mass drops through a height \(h = L - L\cos 60^\circ\) \(= 0.8 - 0.4 = 0.4\ \text{m}\).
By conservation of energy, \(v = \sqrt{2\,g\,h} = \sqrt{2 \cdot 10 \cdot 0.4} = \sqrt{8}\).
At 60°, the force bringing the mass back is no longer proportional to the angle, and the real period is about 7% longer than the formula gives.
\(v \approx 2.8\ \text{m/s}\), and the angle is too large for the simple formula.

STEP 3

The wave carries the energy and leaves the rope in place

A wave is a disturbance that travels, carrying energy from one place to another without carrying the matter of the medium along with it. When we shake the end of a rope, each piece of it only moves up and down around its own place, and what travels is the shape.

In a transverse wave, like the one on the rope, the medium vibrates at right angles to the direction of travel. In a longitudinal wave, like the one on a stretched spring that we push and pull at one end, or sound in air, the medium vibrates along the direction in which the wave moves, and regions of compression and of stretching form.

Waves on a rope and on water, and sound too, are mechanical waves, which need a material medium. Light, radio waves and X-rays are electromagnetic and travel even through a vacuum, where they all move at \(3 \cdot 10^8\ \text{m/s}\).

\(v = \lambda\,f\)\(v = \dfrac{\lambda}{T}\)\(f = \dfrac{1}{T}\)\(1\ \text{Hz} = 1\ \text{s}^{-1}\)In the simulation the wave travels at 2 m/s on both the rope and the spring, the ruler is in metres and the far end is so distant that nothing comes back. Changing the frequency or the amplitude of the hand changes the wave from that moment on, and the change travels along with it.

In a periodic wave, the amplitude \(A\) is the largest displacement of the medium, the period \(T\) is the time of one complete oscillation and the frequency \(f = 1/T\) counts the oscillations per second, in hertz. The wavelength \(\lambda\) is the distance between two neighbouring crests.

In one period the wave moves forward by one wavelength, and from this comes the fundamental equation, \(v = \lambda\,f\). The speed depends on the medium (on the tension and the mass of the rope, for example), and the frequency is set by the source, which here is the hand.

Let's discuss

Follow the red particle. Our first intuition may say that it travels off with the wave, so what does the simulation show?
Double the frequency of the hand and measure the wavelength on the ruler, pausing if you need to. What happened to \(\lambda\), and to the speed?
Increase only the amplitude. Does the wavelength seem to change?
Switch to the spring and look for the compressions. In which direction does the red ring move?

Hand frequency

Amplitude

Frequency f

Period T

Wavelength λ

Speed v

Amplitude A

Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A wave on a rope has a wavelength of 0.5 m and a frequency of 4 Hz. What is its speed?

Show solution
In one period the wave moves forward by one wavelength.
\(v = \lambda\,f = 0.5 \cdot 4 = 2\ \text{m/s}\)
basic
Classify as transverse or longitudinal: the wave on a guitar string, sound in air, light and a 'Mexican wave' of fans in a stadium.

Show solution
On the string and in the Mexican wave, the medium (the string, the people) moves at right angles to the direction in which the wave travels. In sound, the air moves back and forth along the direction of travel. Light is a transverse electromagnetic wave.
The string, light and the Mexican wave are transverse; sound in air is longitudinal.
basic
An FM radio station broadcasts at 100 MHz. What is the wavelength of this electromagnetic wave, which travels at \(3 \cdot 10^8\ \text{m/s}\)?

Show solution
Isolating \(\lambda\) in \(v = \lambda\,f\), with \(100\ \text{MHz} = 10^8\ \text{Hz}\), we get
\(\lambda = \dfrac{3 \cdot 10^8}{10^8} = 3\ \text{m}\)
basic
In films, explosions in space usually make a noise. Why would astronauts, in reality, see the light from a distant explosion and hear nothing?

Show solution
Sound is a mechanical wave and needs a material medium to travel, and the space between spacecraft is almost a vacuum.
Light, being electromagnetic, crosses a vacuum; sound does not.
intermediate
A buoy at sea rises and falls 10 times in 20 s, and the distance between two neighbouring crests is 3 m. What are the period, the frequency and the speed of the waves?

Show solution
The period is \(T = \dfrac{20}{10} = 2\ \text{s}\), and the frequency, \(f = 0.5\ \text{Hz}\).
The distance between crests is the wavelength, \(\lambda = 3\ \text{m}\).
\(v = \lambda\,f = 3 \cdot 0.5 = 1.5\ \text{m/s}\)
intermediate
On a rope where waves travel at 6 m/s, Renata shakes the end at 3 Hz and then at 6 Hz. What is the wavelength in each case?

Show solution
The speed depends on the rope, and not on the hand, so it stays at 6 m/s in both cases.
At 3 Hz, \(\lambda = \dfrac{6}{3} = 2\ \text{m}\).
At 3 Hz, \(\lambda = 2\ \text{m}\); at 6 Hz, \(\lambda = \dfrac{6}{6} = 1\ \text{m}\).
intermediate
The human ear hears sounds from 20 Hz to 20,000 Hz. With sound travelling at 340 m/s in air, what are the longest and the shortest audible wavelengths?

Show solution
The longest wavelength comes from the lowest frequency, \(\lambda = \dfrac{340}{20} = 17\ \text{m}\).
The shortest comes from the highest frequency, \(\lambda = \dfrac{340}{20\,000} = 0.017\ \text{m}\).
From 17 m down to 1.7 cm.
intermediate
A point on a rope takes 0.2 s to go from a crest to the next trough, and the horizontal distance between that crest and that trough is 30 cm. What is the speed of the wave?

Show solution
From a crest to the next trough takes half a period, and so \(T = 0.4\ \text{s}\).
The distance between a crest and the neighbouring trough is half a wavelength, and so \(\lambda = 0.6\ \text{m}\).
\(v = \dfrac{\lambda}{T} = \dfrac{0.6}{0.4}\) \(= 1.5\ \text{m/s}\)
challenge
During a storm, Fábio sees the lightning and hears the thunder 3 s later. How far away did the lightning strike? Why can we neglect the time the light takes to reach him? Use 340 m/s for sound.

Show solution
Light, at \(3 \cdot 10^8\ \text{m/s}\), covers a kilometre in about 3 millionths of a second, a negligible time next to 3 s.
Practically all of the delay comes from the sound.
\(d = 340 \cdot 3 = 1020\ \text{m}\), about 1 km.
challenge
In an earthquake, P waves (longitudinal) travel through rock at 6 km/s and S waves (transverse) at 3.5 km/s. A seismograph records the S waves arriving 20 s after the P waves. How far away was the earthquake?

Show solution
Both waves cover the same distance \(d\), and their times differ by 20 s, \(\dfrac{d}{3.5} - \dfrac{d}{6} = 20\).
Taking \(d\) out as a common factor, \(d\left(\dfrac{1}{3.5} - \dfrac{1}{6}\right) = 20\), that is, \(d \cdot \dfrac{5}{42} = 20\).
\(d = 168\ \text{km}\)

STEP 4

What comes back and what goes through

When a wave reaches the end of a medium, part of it comes back, and we call this reflection. With the end of the rope tied to a wall, the pulse comes back upside down, because the end cannot move up and the wall pulls the rope down as it holds it.

If the end is free, attached to a ring that slides without friction along a rod, the pulse comes back on the same side it arrived on. In both cases it returns with the same speed and the same width, since it is still in the same medium.

When a thin rope is joined to a thick rope, under the same tension, part of the pulse passes into the thick one and part comes back. Passing into another medium is called refraction, and on the thick rope, which is heavier per metre, the wave travels more slowly.

\(f_1 = f_2\)\(\dfrac{\lambda_1}{\lambda_2} = \dfrac{v_1}{v_2}\)\(v = \sqrt{\dfrac{F}{\mu}}\)\(F\) is the tension and \(\mu\) the mass per metre of the rope. In the simulation the thick rope has four times as much mass per metre as the thin one, under the same tension, and the wave travels on it at half the speed, 1 m/s against 2 m/s. We assume ideal ropes, with no energy loss, and the hand does not send back whatever reaches it.

The frequency does not change in refraction, because each piece of the thick rope is shaken by the piece of thin rope next to it, at the same rhythm. Since \(v = \lambda\,f\), with \(f\) unchanged the wavelength changes in the same proportion as the speed.

From thin to thick, the reflected part comes back inverted, as at the wall, which behaves like an extremely heavy rope. From thick to thin, the reflected part comes back upright, and the transmitted part carries on upright in both cases.

Let's discuss

Send one pulse with the end fixed and another with the end free. In which of the two cases does the pulse come back inverted?
Choose the join from the thin rope to the thick one and compare the width of the transmitted pulse with that of the incident pulse. Does the difference seem to match the speed on each rope?
Switch on the continuous wave at the join and measure the wavelengths on the two ropes. Does their ratio remind you of the ratio between the speeds?
Reverse the join, from thick to thin. What changes in the reflected pulse, and in the transmitted one?

Frequency (continuous wave)

v on 1st rope

v on 2nd rope

Frequency

λ on 1st rope

λ on 2nd rope

Reflected amplitude

Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
An upward pulse travels along a rope to an end tied to a wall. How does it come back? And what if the end were attached to a ring that slides without friction along a rod?

Show solution
At the fixed end, the wall holds the rope and pulls it down, and the pulse comes back inverted.
At the free end, the ring rises with the pulse and sends it back on the same side.
With a fixed end the pulse comes back inverted; with a free end it comes back upright.
basic
A 5 Hz wave passes from a rope on which it travels at 10 m/s to another on which it travels at 5 m/s. What are the frequency and the wavelength on the second rope?

Show solution
In refraction the frequency does not change, and on the second rope it is still 5 Hz.
\(f = 5\ \text{Hz}\) and \(\lambda_2 = \dfrac{5}{5} = 1\ \text{m}\) (on the first rope it was 2 m).
basic
When a wave passes from one medium into another, which of these quantities change: the frequency, the speed and the wavelength?

Show solution
The frequency is imposed by the source, and each piece of the new medium is shaken at the rhythm of the neighbouring piece of the old one. The speed depends on the medium, and \(\lambda = v/f\) follows the speed.
The speed and the wavelength change; the frequency stays the same.
basic
An 8 m rope has its end tied to a wall, and waves travel along it at 4 m/s. How long does a pulse take to go from the hand to the wall and back?

Show solution
The pulse runs along the rope twice, 16 m in all.
\(t = \dfrac{16}{4} = 4\ \text{s}\)
intermediate
A string with a mass of 0.01 kg per metre is under a tension of 100 N. What is the speed of waves on it? And on a string of 0.04 kg per metre, under the same tension?

Show solution
The speed on a string is \(v = \sqrt{F/\mu}\), and on the first one \(v = \sqrt{\dfrac{100}{0.01}}\) \(= \sqrt{10\,000} = 100\ \text{m/s}\).
On the heavier string, \(v = \sqrt{\dfrac{100}{0.04}}\) \(= \sqrt{2500} = 50\ \text{m/s}\), half the previous value.
intermediate
In a ripple tank, waves 2 cm long travel at 0.4 m/s in the deep part. In the shallow part, the speed drops to 0.2 m/s. What is the frequency, and what is the wavelength in the shallow part?

Show solution
In the deep part, \(f = \dfrac{v}{\lambda} = \dfrac{0.4}{0.02}\) \(= 20\ \text{Hz}\), and this frequency does not change on crossing.
\(f = 20\ \text{Hz}\) and \(\lambda = \dfrac{0.2}{20} = 0.01\ \text{m} = 1\ \text{cm}\)
intermediate
Red light has a wavelength of 700 nm in air, where it travels at \(3.0 \cdot 10^8\ \text{m/s}\). In water, it travels at \(2.25 \cdot 10^8\ \text{m/s}\). What is its frequency, and what is its wavelength in water?

Show solution
In air, \(f = \dfrac{3.0 \cdot 10^8}{700 \cdot 10^{-9}}\) \(\approx 4.3 \cdot 10^{14}\ \text{Hz}\), and the same frequency holds in water.
The wavelength drops in the same proportion as the speed, \(\lambda_{\text{water}} = 700 \cdot \dfrac{2.25}{3.0}\).
\(f \approx 4.3 \cdot 10^{14}\ \text{Hz}\) and \(\lambda_{\text{water}} = 525\ \text{nm}\)
intermediate
An upward pulse passes from a thin rope to a thick rope, under the same tension. What do the reflected and the transmitted pulses look like? And if the pulse went from thick to thin?

Show solution
From thin to thick, the heavier thick rope plays the part of an imperfect wall, and the reflected part comes back inverted. The transmitted part carries on upwards, slower and narrower.
From thick to thin, the thin rope behaves like an almost free end, and the reflected part comes back upright.
Thin to thick, the reflected pulse is inverted; thick to thin, it is upright. The transmitted pulse never inverts.
challenge
A thin rope is joined to a rope with 4 times as much mass per metre, under the same tension. A 10 Hz wave has a wavelength of 0.8 m on the thin rope. What are the speeds on the two ropes, and what is the wavelength on the thick rope?

Show solution
On the thin rope, \(v_1 = \lambda_1 f = 0.8 \cdot 10 = 8\ \text{m/s}\).
Since \(v = \sqrt{F/\mu}\), with \(\mu\) four times larger the speed is divided by \(\sqrt{4} = 2\), and \(v_2 = 4\ \text{m/s}\).
\(v_1 = 8\ \text{m/s}\), \(v_2 = 4\ \text{m/s}\) and \(\lambda_2 = \dfrac{4}{10} = 0.4\ \text{m}\)
challenge
On a 6 m rope with a fixed end, waves travel at 3 m/s. Leonor sends an upward pulse at \(t = 0\) and another identical upward pulse at \(t = 1\ \text{s}\). When and where does the second pulse meet the first, already reflected? What do we see when they meet?

Show solution
The first pulse reaches the wall at \(t = 2\ \text{s}\) and comes back inverted, at position \(x = 6 - 3(t - 2)\). The second is at \(x = 3(t - 1)\).
When they meet, \(6 - 3(t - 2) = 3(t - 1)\), so \(6t = 15\), that is, \(t = 2.5\ \text{s}\), at \(x = 3 \cdot 1.5\).
An upward pulse meets an identical downward one, and for an instant the two shapes cancel out.
At \(t = 2.5\ \text{s}\), 4.5 m from the hand, the rope is straight for an instant.

STEP 5

When two waves meet

Two waves can occupy the same place at the same time. Where they meet, the displacement of the medium is the sum of the displacements each one would produce on its own, and after the meeting each wave carries on as if the other did not exist. This is the principle of superposition.

If a crest meets another crest, the sum is a bigger crest, and we say the interference is constructive. If a crest meets a trough of the same amplitude, the two cancel out, and the interference is destructive.

With two identical sources tapping the water in phase, what decides the outcome at a point is the difference between its distances to the two sources. When that difference is a whole number of wavelengths, the waves arrive in phase and reinforce each other, and when it is an odd number of half wavelengths, they arrive in antiphase and cancel.

\(\Delta r = n\,\lambda\) (constructive)\(\Delta r = \left(n + \tfrac{1}{2}\right)\lambda\) (destructive)\(n = 0, 1, 2, \dots\)\(\Delta r = |r_1 - r_2|\) is the difference between the distances from the point to the two sources. These rules hold for sources in phase; with the sources in antiphase, they swap places. The simulation adds the two waves at each point and, to run smoothly on a phone as well, calculates the tank on a coarser grid and enlarges the image.

The points of cancellation form lines of almost still water, which we can see in the tank, opening out from the sources like fringes. The further apart the sources, or the shorter the wavelength, the more lines appear, and they tend to crowd closer together.

Noise-cancelling headphones rely on destructive interference. A microphone picks up the noise from outside, and the circuit makes the speaker emit a wave of the same shape and opposite phase, which tends to cancel the noise, above all low, steady sounds such as the drone of an aircraft engine.

Let's discuss

Drag the probe onto a line of still water and read \(\Delta r/\lambda\). Is the number close to 0.5, 1.5 or 2.5?
Move the sources apart. Do the fringes get closer together or further apart, and how many cancellation lines can you count?
Reduce the wavelength and count the cancellation lines again.
Put the sources in antiphase and look at the middle line, at the same distance from both. What happens to it, and how does that relate to the headphones?

Distance between the sources

Wavelength λ

Distance r₁

Distance r₂

Δr

Δr / λ

Interference

Amplitude at the probe

Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Two pulses on a rope, one 3 cm high and the other 2 cm high, meet. What is the maximum height at the meeting if both are on the same side? And if they are on opposite sides?

Show solution
By the principle of superposition, the displacements add, sign included.
On the same side, \(3 + 2 = 5\ \text{cm}\); on opposite sides, \(3 - 2 = 1\ \text{cm}\), on the side of the bigger pulse.
basic
After two pulses cross on a rope, do they leave the meeting any different from how they entered it?

Show solution
Superposition only applies while the pulses occupy the same place. After the meeting, each one carries on as if the other had never been there.
No, each pulse leaves with the same shape and the same speed.
basic
Two sources in phase produce waves 2 cm long in a tank. At a point P, the difference between the distances to the two sources is 6 cm. Is the interference at P constructive or destructive?

Show solution
We measure the path difference in wavelengths, \(\dfrac{\Delta r}{\lambda} = \dfrac{6}{2} = 3\), a whole number.
The waves arrive in phase, and the interference is constructive.
basic
With the same sources as in the previous exercise (\(\lambda = 2\ \text{cm}\), in phase), the path difference at a point Q is 5 cm. What happens at Q?

Show solution
Now \(\dfrac{\Delta r}{\lambda} = \dfrac{5}{2} = 2.5\), a whole number of wavelengths plus a half.
The waves arrive in antiphase, and the interference is destructive.
intermediate
Two sources in phase make waves 4 cm long on water. A point P is 20 cm from one source and 26 cm from the other. Is the water at P choppy or almost still?

Show solution
The path difference is \(\Delta r = 26 - 20 = 6\ \text{cm}\), and \(\dfrac{\Delta r}{\lambda} = \dfrac{6}{4} = 1.5\).
With a difference of 1.5 wavelengths, the interference is destructive and the water at P is almost still.
intermediate
A pair of noise-cancelling headphones has to cancel a 170 Hz hum. With what delay, relative to the hum, must the headphones emit an identical wave for the two to cancel? Use 340 m/s for sound.

Show solution
To cancel the hum, the wave from the headphones must be in antiphase, delayed by half a period, \(\dfrac{T}{2} = \dfrac{1}{2f}\).
This is the same as a path difference of half a wavelength, \(\dfrac{\lambda}{2} = \dfrac{340}{2 \cdot 170} = 1\ \text{m}\).
\(\dfrac{T}{2} = \dfrac{1}{2 \cdot 170} \approx 0.0029\ \text{s}\), about 2.9 ms.
intermediate
Two loudspeakers in phase play the same 340 Hz note. Gabriel is 3.0 m from one and 3.5 m from the other. Does he hear a loud sound or a faint one? Use 340 m/s for sound.

Show solution
The wavelength is \(\lambda = \dfrac{340}{340} = 1\ \text{m}\), and the path difference is \(\Delta r = 0.5\ \text{m}\), half a wavelength.
The interference is destructive, and Gabriel hears a very faint sound at that spot.
intermediate
In destructive interference, two waves cancel at a point. Does that mean the energy of the two waves has disappeared?

Show solution
Energy is conserved. It is redistributed in space, and what is missing on the cancellation lines turns up as extra in the regions of constructive interference, where the amplitude doubles and the intensity becomes up to four times greater.
No, the energy is redistributed between the regions of reinforcement and those of cancellation.
challenge
Two sources in phase, 4 cm apart, produce waves 1.5 cm long in a tank. How many cancellation lines (nodal lines) cross the segment joining the two sources?

Show solution
Along the segment, the path difference \(\Delta r\) runs from \(-4\ \text{cm}\) to \(+4\ \text{cm}\).
There is cancellation where \(|\Delta r| = \left(n + \tfrac{1}{2}\right)\lambda\), that is, \(|\Delta r| = 0.75\), \(2.25\) and \(3.75\ \text{cm}\), all less than 4 cm. The next one, 5.25 cm, no longer fits.
There are 3 values on each side of the perpendicular bisector, and so 6 nodal lines.
challenge
Two sources in a tank vibrate in antiphase, with \(\lambda = 2\ \text{cm}\). At a point P, 10 cm from one source and 13 cm from the other, is the interference constructive or destructive? And at the points on the perpendicular bisector?

Show solution
With the sources in antiphase the rules swap, because the waves already set off half a period out of step.
At P, \(\dfrac{\Delta r}{\lambda} = \dfrac{3}{2} = 1.5\), and the extra half wavelength makes up for the phase difference between the sources.
At P the interference is constructive, and on the perpendicular bisector, with \(\Delta r = 0\), it is destructive.

STEP 6

The guitar string and the waves that stay put

On a string fixed at both ends, waves travel back and forth, reflecting at the ends, and overlap. At certain frequencies the sum forms a pattern that does not travel, with points that stay still, the nodes, and points that oscillate with the largest amplitude, the antinodes, and we call this pattern a standing wave.

The fixed ends have to be nodes, and so only whole numbers of half wavelengths fit on the string, \(L = n\,\lambda/2\). Each value of \(n\) gives a harmonic, with frequency \(f_n = n\,v/(2L)\), and the first, the lowest, is called the fundamental.

On a guitar, the wave speed depends on the tension and on the mass per metre of the string. To tune it, we tighten the tuning peg, and the tension and the speed rise together with the note. Pressing the string against a fret shortens the part that vibrates and the frequency goes up as well, and the lowest strings are the thickest ones.

\(L = n\,\dfrac{\lambda_n}{2}\)\(f_n = n\,\dfrac{v}{2L}\)\(f_n = n\,f_1\)\(v = \sqrt{\dfrac{F}{\mu}}\)\(n = 1, 2, 3, \dots\) Harmonic \(n\) has \(n\) antinodes and \(n + 1\) nodes, counting the ends. In the simulation the string has \(\mu = 1.0\ \text{g/m}\), and the motion is shown in slow motion. With the vibration generator on, we see the vibration that settles in after a while on a string with a little damping.

When we shake a body at a frequency equal to one of its natural frequencies, each push arrives at the right moment and the amplitude grows with every cycle. This is called resonance, and it is why a singer can, in principle, shatter a crystal wine glass by holding the right note loudly enough.

The Tacoma Narrows Bridge, in the United States, which collapsed in a strong wind in 1940, is often given as an example of resonance. Engineering studies indicate a more complicated mechanism, in which the motion of the bridge itself changed the wind around it and fed the oscillation (what engineers call flutter), and it may be better to treat it as a relative of simple resonance.

Let's discuss

Go through harmonics 1 to 6 and count the nodes and the antinodes in each. What rule links these numbers to \(n\)?
Tighten the tuning peg until the tension is four times larger. What happens to the fundamental frequency?
Halve the length of the string, as when pressing the middle fret on a guitar, and compare the new note with the old one.
Switch on the vibration generator and sweep the frequency slowly. At which frequencies does the string seem to come alive, and how do they compare with \(f_1\)?

Length L

Tension F

Generator frequency

Speed v

Fundamental f₁

Frequency

Wavelength λ

Nodes

Antinodes

Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A 0.6 m string, fixed at both ends, vibrates in its fundamental mode. What is the wavelength?

Show solution
In the fundamental, half a wavelength fits between the ends, \(L = \dfrac{\lambda}{2}\).
\(\lambda = 2L = 1.2\ \text{m}\)
basic
The fundamental frequency of a string is 100 Hz. What are the frequencies of the second and third harmonics?

Show solution
On a string fixed at both ends, \(f_n = n\,f_1\).
\(f_2 = 200\ \text{Hz}\) and \(f_3 = 300\ \text{Hz}\)
basic
How many nodes (counting the ends) and how many antinodes does a string vibrating in the 4th harmonic have?

Show solution
In harmonic \(n\), \(n\) half wavelengths fit on the string, each with an antinode in the middle and nodes at its ends.
With \(n = 4\), there are 4 antinodes and 5 nodes.
basic
Why can a singer shatter a crystal glass with her voice, but only if she hits one particular note?

Show solution
The glass has natural frequencies of vibration. When the voice holds one of them, each push arrives at the right moment and the amplitude grows with every cycle, which we call resonance.
Only a note equal to a natural frequency of the glass makes it resonate.
intermediate
The vibrating part of a guitar string is 0.65 m long, and waves travel along it at 286 m/s. What is the fundamental frequency?

Show solution
In the fundamental, \(\lambda = 2L = 1.3\ \text{m}\).
\(f_1 = \dfrac{v}{2L} = \dfrac{286}{1.3}\) \(= 220\ \text{Hz}\)
intermediate
Tuning a string that sounded at 110 Hz, Tiago tightens the tuning peg until the tension is 4 times larger. What is the new fundamental frequency?

Show solution
Since \(v = \sqrt{F/\mu}\), with 4 times the tension the speed doubles. The length has not changed, and \(f_1 = v/(2L)\) doubles too.
\(f_1 = 220\ \text{Hz}\), an octave higher.
intermediate
An open 0.64 m string plays 330 Hz. Pressed at the 12th fret, its vibrating part drops to half. What is the new frequency?

Show solution
The speed, which depends on the tension and on the string, does not change. With \(L\) halved, \(f_1 = v/(2L)\) doubles.
\(f = 660\ \text{Hz}\), also an octave higher.
intermediate
A 1.5 m string, fixed at both ends, vibrates with 3 antinodes at a frequency of 60 Hz. What is the wavelength, and what is the speed of the waves on the string?

Show solution
With 3 antinodes, 3 half wavelengths fit, \(1.5 = 3 \cdot \dfrac{\lambda}{2}\), and so \(\lambda = 1\ \text{m}\).
\(\lambda = 1\ \text{m}\) and \(v = \lambda\,f = 1 \cdot 60 = 60\ \text{m/s}\)
challenge
A 0.5 m string, with a mass of 0.005 kg per metre, is under a tension of 200 N. What are the frequencies of the fundamental and of the second harmonic?

Show solution
The speed is \(v = \sqrt{\dfrac{200}{0.005}}\) \(= \sqrt{40\,000} = 200\ \text{m/s}\).
The fundamental is \(f_1 = \dfrac{v}{2L} = \dfrac{200}{1.0}\) \(= 200\ \text{Hz}\).
\(f_1 = 200\ \text{Hz}\) and \(f_2 = 400\ \text{Hz}\)
challenge
In a microwave oven with the turntable taken out, Helena heats a dish covered in cheese and sees that it melts in spots 6.1 cm apart. The oven works at 2.45 GHz. What wave speed can she estimate, assuming the melted spots are neighbouring antinodes of a standing wave?

Show solution
Neighbouring antinodes are half a wavelength apart, and so \(\lambda = 2 \cdot 6.1 = 12.2\ \text{cm}\).
With \(2.45\ \text{GHz} = 2.45 \cdot 10^9\ \text{Hz}\), \(v = \lambda\,f = 0.122 \cdot 2.45 \cdot 10^9\).
\(v \approx 3.0 \cdot 10^8\ \text{m/s}\), the speed of light, as we expect for an electromagnetic wave.

STEP 7

Sound is a pressure wave

Sound is a longitudinal mechanical wave. A loudspeaker pushes and pulls the air in front of it, forming slightly compressed and slightly rarefied layers that travel outwards, with the pressure oscillating around atmospheric pressure. In air at about 20 °C, sound travels at about 340 m/s, and in water and in solids it usually travels much faster. In a vacuum there is no sound.

We hear frequencies between about 20 Hz and 20,000 Hz, with infrasound below this range and, above it, ultrasound, which bats use to find their way and which ultrasound scanners turn into images. The pitch of a sound is its frequency, and in physics a 'high' sound is a high-pitched one, although in everyday speech we also turn the volume 'up' when we just mean louder.

Intensity measures the energy the wave carries, per second, through each square metre. The ear copes with an enormous range, from the threshold of hearing, \(10^{-12}\ \text{W/m}^2\), up to about \(1\ \text{W/m}^2\), where sound starts to hurt, and that is why we use a logarithmic scale, the sound level in decibels, in which every extra 10 dB multiplies the intensity by 10.

\(\beta = 10\log\dfrac{I}{I_0}\)\(I_0 = 10^{-12}\ \text{W/m}^2\)\(f^{\prime} = f\,\dfrac{v \pm v_o}{v \mp v_s}\)\(v_{\text{sound}} \approx 340\ \text{m/s}\)\(\Delta t_{\text{echo}} = \dfrac{2d}{v}\)In the Doppler formula we use the upper signs when the observer and the source move towards each other (\(+v_o\) and \(-v_s\)) and the lower ones when they move away from each other. In the simulation the air is still, time runs in slow motion and we draw only one wavefront every 0.1 s.

Timbre is what tells apart the same note played on a guitar and on a flute, which produce the same fundamental frequency mixed with harmonics in different proportions. An echo is sound reflected off an obstacle, and to hear it separately from the original sound the obstacle needs to be about 17 m away or more.

When an ambulance approaches, its siren sounds higher, and when it moves away, lower. This is the Doppler effect. The moving source keeps catching up with the wavefronts it has itself emitted, which bunch up in front of it and spread out behind, and an observer running towards the source meets more wavefronts per second.

Let's discuss

Set the ambulance to 30 m/s and compare the spacing of the wavefronts in front of it and behind it. Check the frequency heard against the formula.
Move the observer to the other side. Does the siren sound higher or lower, and by how much does it change?
Take the source up to about 340 m/s and then a little beyond. What do you imagine will happen to the wavefronts in front of it? Then check.
Stop the ambulance and set the observer running towards it. Does the frequency heard change, even with the source standing still?

Siren frequency

Source speed

Observer speed (towards the source)

Emitted f

Heard f

λ in front

λ behind

Source in km/h

v_source / v_sound

Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Standing in front of a cliff, Caio shouts and hears the echo 2 s later. How far away is the cliff? Use 340 m/s for sound.

Show solution
The sound goes to the cliff and comes back, covering twice the distance, \(2d = 340 \cdot 2 = 680\ \text{m}\).
\(d = 340\ \text{m}\)
basic
A flute and a violin play the same note at the same intensity. Which quality of sound lets us tell the two instruments apart?

Show solution
The same note means the same fundamental frequency, the same pitch. Each instrument mixes in harmonics in different proportions, and the shape of the wave changes.
Timbre.
basic
A vacuum cleaner produces 80 dB and a conversation 60 dB. How many times greater is the sound intensity of the vacuum cleaner than that of the conversation?

Show solution
On the decibel scale, every extra 10 dB multiplies the intensity by 10. The difference here is 20 dB.
\(10 \cdot 10 = 100\) times.
basic
An ambulance passes Sofia with its siren on. How does Sofia hear the siren while the ambulance approaches, and after it has moved away? Does the frequency emitted by the siren change?

Show solution
Because of the Doppler effect, the wavefronts arrive closer together while the source approaches and further apart when it moves away. The siren keeps emitting the same frequency.
Higher as it approaches and lower as it moves away; the emitted frequency does not change.
intermediate
A 680 Hz siren approaches a stationary observer at 20 m/s. What frequency does he hear? And after the source has passed and moves away at the same speed? Use 340 m/s for sound.

Show solution
With the source approaching, we use the upper sign, \(v - v_s\) in the denominator, \(f^{\prime} = 680 \cdot \dfrac{340}{340 - 20}\) \(= 680 \cdot \dfrac{340}{320}\).
Moving away, \(f^{\prime} = 680 \cdot \dfrac{340}{340 + 20}\) \(= 680 \cdot \dfrac{340}{360}\).
722.5 Hz as it approaches and about 642 Hz as it moves away.
intermediate
A stationary alarm emits 500 Hz. Marcelo rides towards it at 34 m/s, on a bicycle, down a hill. What frequency does he hear? Use 340 m/s for sound.

Show solution
The observer moves towards the source and meets more wavefronts per second, so we use \(+v_o\) in the numerator.
\(f^{\prime} = 500 \cdot \dfrac{340 + 34}{340}\) \(= 500 \cdot 1.1 = 550\ \text{Hz}\)
intermediate
At a concert, the sound intensity near the speakers reaches \(10^{-4}\ \text{W/m}^2\). What is the sound level, in decibels? The threshold of hearing is \(I_0 = 10^{-12}\ \text{W/m}^2\).

Show solution
The sound level is \(\beta = 10\log\dfrac{I}{I_0}\) \(= 10\log\dfrac{10^{-4}}{10^{-12}}\) \(= 10\log 10^{8}\).
\(\beta = 10 \cdot 8 = 80\ \text{dB}\)
intermediate
A boat's sonar sends an ultrasound pulse down to the seabed and receives the echo 0.8 s later. How deep is the water, if sound travels at 1500 m/s in water?

Show solution
The pulse goes to the bottom and back, and so \(2d = 1500 \cdot 0.8 = 1200\ \text{m}\).
\(d = 600\ \text{m}\)
challenge
A car at 20 m/s and an ambulance at 20 m/s, with a 700 Hz siren, are driving along a straight road towards each other. What frequency does the driver of the car hear? And after they pass each other and move apart? Use 340 m/s for sound.

Show solution
As they approach, both signs raise the frequency, \(f^{\prime} = 700 \cdot \dfrac{340 + 20}{340 - 20}\) \(= 700 \cdot \dfrac{360}{320} = 787.5\ \text{Hz}\).
As they move apart, both lower it, \(f^{\prime} = 700 \cdot \dfrac{340 - 20}{340 + 20}\) \(= 700 \cdot \dfrac{320}{360}\).
About 788 Hz before they pass and about 622 Hz afterwards.
challenge
2 m from a loudspeaker, a sound meter reads 80 dB. Assuming the sound spreads equally in all directions, without being absorbed or reflected, what would the meter read at 20 m?

Show solution
The power spreads over a sphere of area \(4\pi r^2\), and the intensity falls off as \(1/r^2\). At 10 times the distance, the intensity becomes \(10^2 = 100\) times smaller.
Dividing the intensity by 100 takes \(10\log 100 = 20\ \text{dB}\) off the sound level.
\(80 - 20 = 60\ \text{dB}\)

WRAP-UP

Challenges

Mass–spring

A 2 kg block oscillates on a spring of 200 N/m, with an amplitude of 5 cm. What is the period, and what is the block's maximum speed? (\(\pi \approx 3.14\))

Show solution

The angular frequency is \(\omega = \sqrt{\dfrac{200}{2}} = 10\ \text{rad/s}\), and the period, \(T = \dfrac{2\pi}{10} \approx 0.63\ \text{s}\).
The maximum speed is \(v_{\text{max}} = \omega A = 10 \cdot 0.05 = 0.5\ \text{m/s}\), at the centre.

Pendulum

How long must a pendulum be for it to make one complete swing per second? (\(g = 10\ \text{m/s}^2\), \(\pi^2 \approx 10\))

Show solution

With \(T = 1\ \text{s}\), \(L = \dfrac{g\,T^2}{4\pi^2} = \dfrac{10 \cdot 1}{40}\) \(= 0.25\ \text{m}\), or 25 cm.

v = λ·f

An AM radio station broadcasts at 1000 kHz. How long are its waves, which travel at \(3 \cdot 10^8\ \text{m/s}\)?

Show solution

With \(f = 10^6\ \text{Hz}\), \(\lambda = \dfrac{3 \cdot 10^8}{10^6} = 300\ \text{m}\), nearly three football pitches.

Refraction

A 6 Hz wave passes from a rope on which it travels at 12 m/s to another on which it travels at 4 m/s. What are the wavelengths on the two ropes?

Show solution

The frequency does not change on crossing. On the first rope, \(\lambda_1 = \dfrac{12}{6} = 2\ \text{m}\).
On the second, \(\lambda_2 = \dfrac{4}{6} \approx 0.67\ \text{m}\), three times smaller, like the speed.

Interference

Two loudspeakers in phase play 680 Hz. Lívia is 4.00 m from one and 4.25 m from the other. Does she hear the sound reinforced or weakened? (sound at 340 m/s)

Show solution

The wavelength is \(\lambda = \dfrac{340}{680} = 0.5\ \text{m}\), and the path difference, \(\Delta r = 0.25\ \text{m}\), is half a wavelength.
The interference is destructive, and the sound is much fainter at that spot.

Standing waves

A 0.8 m string, fixed at both ends, vibrates with 4 antinodes at 500 Hz. What is the speed of the waves, and what is the fundamental frequency?

Show solution

With 4 antinodes, \(0.8 = 4 \cdot \dfrac{\lambda}{2}\), and \(\lambda = 0.4\ \text{m}\). The speed is \(v = 0.4 \cdot 500 = 200\ \text{m/s}\).
This is the 4th harmonic, and \(f_1 = \dfrac{500}{4} = 125\ \text{Hz}\).

Doppler effect

A train approaches a station at 34 m/s, sounding its horn at 900 Hz. What frequency does a person standing on the platform hear? (sound at 340 m/s)

Show solution

The source is approaching, and we use \(v - v_s\) in the denominator, \(f^{\prime} = 900 \cdot \dfrac{340}{340 - 34}\) \(= 900 \cdot \dfrac{340}{306} = 1000\ \text{Hz}\).

Think it through, no maths

On the Moon, an astronaut strikes a rock with a hammer, and her colleague, 10 m away, hears nothing. If he presses his helmet against the rock, will he start to hear it?

Show solution

Without air there is no medium to carry sound between the two of them, and the colleague does not hear the blow.
With his helmet against the rock, the vibration can reach him through the ground and the helmet, which are solid media, and he starts to hear something.

ENEM-style questions

The ENEM is Brazil's national secondary-school exam, which most students sit to get into university, and we wrote the five questions below in its format, with an everyday context, a question and five options, each one returning to a different step of the lesson. Further down we list real exam questions on the same topics, which may well be the best practice once these are done.

Simple pendulum · Step 2
Rafael restores antique clocks and wants to build a wall clock whose pendulum takes 4 s to complete each swing there and back. Before cutting the rod, he measured the period of pendulums of several lengths, keeping the swings small in every measurement, and recorded the results in the table.

Rafael's measurements
Length L (m) Period T (s)

0.25 1.0

0.50 1.4

1.00 2.0

2.25 3.0

If the trend in the table holds, how long should the pendulum of the new clock be?
1. 1.4 m
2. 2.0 m
3. 3.5 m
4. 4.0 m
5. 0.25 m
Show solution

Answer: D.
The table shows that when the length becomes 4 times larger (from 0.25 m to 1.00 m), the period only doubles, as we expect from \(T = 2\pi\sqrt{L/g}\), where \(L\) sits inside the square root.
To double the period again, from 2 s to 4 s, the length has to become 4 times larger once more, \(L = 4 \cdot 1.00 = 4.0\ \text{m}\).
Checking with \(g = 10\ \text{m/s}^2\) and \(\pi^2 \approx 10\), \(L = \dfrac{g\,T^2}{4\pi^2} = \dfrac{10 \cdot 16}{40} = 4\ \text{m}\).
Assuming the period is proportional to the length leads to 2.0 m, and extending a straight line through the last two points of the table leads to 3.5 m. The 1.4 m comes from swapping the square root for a square, and the 0.25 m from thinking that a longer pendulum swings faster.
What a wave is · Step 3
In ultrasound scans, a probe held against the skin emits pulses of ultrasound and picks up the echoes that come back from the organs. A typical scanner works at 5 MHz, and in the body's soft tissue sound travels at about 1540 m/s. As a rule of thumb, details smaller than one wavelength tend not to show up in the image.

Roughly what is the wavelength of this ultrasound in soft tissue?
1. 0.031 mm
2. 0.068 mm
3. 0.31 mm
4. 3.1 mm
5. \(7.7 \cdot 10^9\ \text{m}\)
Show solution

Answer: C.
Isolating \(\lambda\) in \(v = \lambda\,f\), with \(5\ \text{MHz} = 5 \cdot 10^6\ \text{Hz}\), we get
\(\lambda = \dfrac{1540}{5 \cdot 10^6}\) \(\approx 3.1 \cdot 10^{-4}\ \text{m} = 0.31\ \text{mm}\)
That is why an ultrasound scan can separate structures a few tenths of a millimetre across, and higher-frequency scanners show even smaller details.
Options A and D get the power of ten wrong when converting to millimetres. Option B uses the speed of sound in air, 340 m/s, which does not apply inside the body, and E multiplies the speed by the frequency.
Interference · Step 5
Juliana flies a lot and has bought a pair of headphones with active noise cancellation. A microphone in the headphones picks up the drone of the engines, which inside the cabin is dominated by a low 200 Hz sound, and an electronic circuit makes the speaker emit a second sound wave along with the music. Let us assume the drone is a pure tone, with constant frequency and amplitude.

For the drone to practically disappear in Juliana's ear, what must the wave emitted by the circuit be like?
1. With the same frequency as the drone and in phase with it, so that the two waves add up.
2. With the same frequency and amplitude as the drone, in antiphase, which is equivalent to a delay of 2.5 ms.
3. With the same frequency and amplitude as the drone, delayed by 5 ms.
4. With twice the frequency of the drone, so that each of its crests meets a trough.
5. With an ultrasound frequency, which the ear cannot hear and which muffles the drone.
Show solution

Answer: B.
Cancellation is destructive interference, and for crest to meet trough at every instant the two waves need the same frequency, the same amplitude and opposite phase.
Opposite phase means a delay of half a period, \(\dfrac{T}{2} = \dfrac{1}{2 \cdot 200}\) \(= 0.0025\ \text{s} = 2.5\ \text{ms}\)
Option A describes constructive interference, which would double the drone. Option C delays the wave by a whole period, which puts it back in phase with the drone. With twice the frequency, as in D, crests only coincide with troughs now and then, and an ultrasound, as in E, does not interfere usefully with audible sound.
Real headphones cancel low, steady sounds like this drone better than voices and noises that change quickly.
Standing waves · Step 6
On a guitar, the lowest string, the E, has 65 cm of vibrating length between the nut and the bridge and, played open, sounds a note of 82 Hz. To play a 110 Hz A on this same string, Valéria presses the string against a fret, without changing the tension. Let us assume the string vibrates in its fundamental mode.

How long must the part of the string that keeps vibrating be?
1. 48 cm
2. 87 cm
3. 36 cm
4. 65 cm
5. 17 cm
Show solution

Answer: A.
With the same tension and the same string, the wave speed does not change, and by the formula \(f_1 = \dfrac{v}{2L}\) the frequency is inversely proportional to the length.
\(L^{\prime} = 65 \cdot \dfrac{82}{110}\) \(\approx 48\ \text{cm}\)
Option B uses the ratio upside down and lengthens the string, which would make the note lower. Option C squares the ratio, as if the frequency depended on the square of the length, and D forgets that pressing on a fret shortens the vibrating part. The 17 cm of E is the distance between the nut and the fret, and not the piece that vibrates.
Sound and the Doppler effect · Step 7
Pedro is waiting to cross an avenue when a car travelling at 20 m/s sounds its horn as it approaches and keeps sounding it after it has passed him. The horn emits a 400 Hz sound, and the air is still. Let us take 340 m/s for the speed of sound.

Which frequencies does Pedro hear while the car approaches and after it moves away, respectively?
1. 400 Hz and 400 Hz
2. 378 Hz and 425 Hz
3. 424 Hz and 376 Hz
4. 420 Hz and 380 Hz
5. 425 Hz and 378 Hz
Show solution

Answer: E.
It is the source that moves, and Pedro stands still. As it approaches, the source partly catches up with the wavefronts it has emitted, and the denominator becomes \(v - v_s\).
\(f_{\text{towards}} = 400 \cdot \dfrac{340}{340 - 20}\) \(= 425\ \text{Hz}\)
\(f_{\text{away}} = 400 \cdot \dfrac{340}{340 + 20}\) \(\approx 378\ \text{Hz}\)
Option A ignores the Doppler effect, and B swaps approaching for moving away. Option C uses the formula for a moving observer, which gives similar but different values, and D adds the car's speed to the frequency and subtracts it directly.

Rafael's measurements
Length L (m)	Period T (s)
0.25	1.0
0.50	1.4
1.00	2.0
2.25	3.0

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where we can look each question up by year, day, booklet colour and number.

ENEM 2013, Day 1, blue booklet, question 65. A Mexican wave in a stadium crowd is treated as a travelling wave, and from the speed of the wave and the distance between the people we calculate its frequency.
ENEM 2015, Day 1, blue booklet, question 86. A flute and a piano play the same note, and the question asks which characteristic of sound lets us tell the two instruments apart.
ENEM 2016, Day 1, blue booklet, question 49. An ambulance passes a stationary observer, and we need to choose the graph that shows how the frequency heard changes over time.
ENEM 2016, Day 1, blue booklet, question 86. In a microwave oven without its turntable, butter melts at spots lying in a line, and the question asks us to recognise those spots in a standing wave.
ENEM 2017, Day 2, blue booklet, question 112. In a Quincke tube, sound splits into two paths and fades out at the exit when one of them is lengthened, and from the path difference we work out the frequency of the source.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Simple harmonic motion\(x = A\cos(\omega t + \varphi_0)\)

Mass–spring\(T = 2\pi\sqrt{m/k}\)

Pendulum (small angle)\(T = 2\pi\sqrt{L/g}\)

Energy and speed in SHM\(E = \tfrac{1}{2}k\,A^2\) · \(v_{\text{max}} = \omega A\)

Frequency and angular frequency\(f = 1/T\) · \(\omega = 2\pi f\)

Wave equation\(v = \lambda\,f\)

Speed on a string\(v = \sqrt{F/\mu}\)

Refraction\(f\) unchanged · \(\lambda_1/\lambda_2 = v_1/v_2\)

Interference (sources in phase)\(\Delta r = n\,\lambda\) · \(\left(n + \tfrac{1}{2}\right)\lambda\)

String fixed at both ends\(f_n = n\,v/(2L)\)

Sound level\(\beta = 10\log(I/I_0)\)

Doppler effect\(f^{\prime} = f\,\dfrac{v \pm v_o}{v \mp v_s}\)