FFR - Wave-Particles

Original textbook reading.

Enrichment beyond the AP course.

This original chapter is extra exploration beyond the supplied AP Physics 1 course. Keep it if you're curious about where these ideas go next.

In this section, we are going to introduce a new topic and an old one that are very important applications of waves: the Doppler effect which deals with sound waves, and photons.

Doppler Effect

This may be something that you hear on a daily basis, but maybe won’t notice how it works. Ever been in a situation where you were in your car and you heard either an ambulance or a fire truck coming? It’s because they’re blasting their sirens trying to get you to move out of the way!

As a siren approaches, received intensity often rises because its distance decreases. Separately, relative motion changes the detected pitch: approach gives a higher frequency and recession a lower one under the usual subsonic conditions.

These changes in frequencies with respect to motion of a relative object is what is known as the doppler effect. When dealing with this, we will denote two other definitions: the source, $S$ which is the source of where the sound is coming from, and the detector, $D$ which is the object that detects the sound.

A detector moving toward a stationary sound source encounters wavefronts more frequently and hears a higher pitch. Intensity may also rise as it gets closer, but that is a separate distance effect.

The general equation for this is:

$f^{'} = f \frac{v \pm v_{D}}{v \pm v_{S}}$ (Equation 9.4.1)
Where $f^{'}$ is the detected frequency, $f$ is the emitted frequency, $v$ is the speed of sound through the medium (in this case its air), and $v_{D}$ and $v_{S}$ are the speeds of the detector and source relative to the air, respectively. The reason for the “ $\pm$ ” is because you can have multiple scenarios that affect the detected frequency:

Use +u_D in the numerator when the detector’s velocity component relative to the medium points toward the source.

Use −u_S in the denominator when the source’s velocity component relative to the medium points toward the detector.

Use −u_D in the numerator when the detector’s velocity component relative to the medium points away from the source.

Use +u_S in the denominator when the source’s velocity component relative to the medium points away from the detector.

An object stationary relative to the medium has zero velocity in the Doppler formula. The usual expressions here assume collinear subsonic motion.

Example Problem 9.4.1

Ultrasound imaging detects reflected acoustic waves. A Doppler frequency shift can measure a moving reflector’s velocity component along the beam. The following example uses this reflected-wave effect.

The way it works is that it emits waves and produces an image by detecting waves that bounce off the objects. In that case, it forms the image on a screen from the reflected waves it catches back from within the woman’s womb.

In a hypothetical reflected-ultrasound model, a stationary transducer emits 2.30 × 10⁶ Hz and receives 3.0 × 10⁶ Hz from a reflector moving toward it. Sound speed is 1.54 × 10³ m/s. Find the reflector’s speed along the beam. These frequencies are not realistic fetal-heart data.

Step 1: Realize that this is a double doppler question

The tricky part about this question is realizing that you have $2$ different scenarios going on. First thing is that the initial source is the ultrasound machine that sends the waves to the detector which is the heart.

Then the heart emits back a wave for the ultrasound machine to receive and make into an image. This means that at that point the heart acts as the source and the ultrasound machine as the detector.

Step 2: Solve for the detected frequency of the Heart

We will be using our relative frequency equation:

$f^{'} = f \frac{v \pm v_{D}}{v \pm v_{S}}$

But for this case since we’re dealing with multiple doppler effects, let’s change the subscripts to prevent confusion.

The first part to solving this requires us to set up the first part of the doppler system, where the heart acts as the detector (we’ll refer to it as $(2)$ ) and then the ultrasound, which we’ll refer to as $(1)$ , so the equation will look like:

$f_{2} = f_{1} \frac{v \pm v_{2}}{v \pm v_{1}}$

Initially, the ultrasound is the source while the fetus is the detector.

The reflector is moving toward the stationary transducer, so it encounters the incoming wavefronts more frequently. For this first shift, f₂ = f₁(v + u)/v.

$f_{2} = f_{1} \frac{v \pm v_{2}}{v \pm v_{1}}$ which simplifies to $f_{1} \frac{v + v_{2}}{v}$

Step 3: Set up our second doppler equation

This time we are looking at the heart as the source since it will be reflecting waves back for the ultrasound to detect. We will designate the ultrasound as ( $3$ ) since it is a new frequency that the ultrasound will detect. The heart is still partially moving, so it still has it’s velocity but this time the waves are moving away from the heart, so the source is essentially moving

On return, the reflector acts as a source moving toward the stationary transducer, so the source denominator is v − u.

$f_{3} = f_{2} \frac{v}{v - v_{2}}$

Step 4: Find the equation for the velocity of the heart

We have $2$ equations with $2$ unknowns so we’ll have to use substitution to eliminate $f_{2}$ . Rearranging $f_{3}$ to solve for $f_{2}$ gives us:

$f_{2} = f_{3} (\frac{v - v_{2}}{v})$ and we already know that $f_{2}$ also equals $f_{1} \frac{v + v_{2}}{v}$ therefore we can set these $2$ equal to each other to give us the following equation:

$f_{3} (\frac{v - v_{2}}{v}) = f_{1} (\frac{v + v_{2}}{v})$

Canceling the $v$ and solving for $v_{2}$ doing a little more algebra gives us the our final equation for the velocity of the heart:

$v_{2} = \frac{v (f_{3} - f_{1})}{f_{1} + f_{3}}$

Step 5: Plug in all the values

We know all the frequencies and the velocity of the speed of sound through a tissue is given to us. They are:

$\begin{matrix} f_{1} = 2.3 MHz \\ f_{3} = 3.0 MHz \\ v = 1540 m/s \end{matrix}$

So plugging these values in gives us:

Photon Momentum

$v_{2} = \frac{1540 m/s (3.0 MHz - 2.3 MHz)}{2.3 MHz + 3.0 MHz}$

$v_{2} = 203.4 m/s$

This sounds weird, we know but it’s actually very applicable in many situations. Remember from the previous section, you learned about light and how they emit waves.

Well in this part we tie that together by saying that light waves are actually made up of many “light particles” which we designate as photons.

A laser produces a beam through stimulated emission and optical amplification, often with high coherence and a narrow range of frequencies. Focusing can concentrate its optical power into a small area. Photons have zero rest mass and carry momentum p = E/c = h/λ in vacuum.

For photons in vacuum, use p = E/c. The nonrelativistic expression p = mv does not apply to a massless photon.

A photon is a quantum of the electromagnetic field. This drawing is a representation, not a picture of a photon emitting its own light wave.

A cutting laser deposits concentrated energy in a material through absorption, producing heating, melting, vaporization, or ablation depending on the material and pulse conditions. Light also exerts radiation pressure, but photon momentum is not the usual explanation of laser cutting.

Wave Particle Duality

Proved through the double slit experiment. The basic principle of it was that it proved energy and matter could display characteristics of both waves and particles.

How the experiment was run was a single stream of light (a laser) was passed through a double slitted plate where the light would pass through and on the other side there would be a screen to view the light coming through.

A double slit produces an interference pattern when the alternatives remain coherent. Individual photon or electron detections are localized events; many such events can build an interference pattern. A model using only independent classical trajectories lacks the same interference pattern.

The interference pattern depends on whether which-path information is physically available, not on whether a person looks at the screen. A detector that distinguishes the slit paths can destroy their coherence. Detection at the screen is still localized while the accumulated distribution can show interference.

Take a look at the gallery on the next page. Hopefully that will clear things up.

If electrons were modeled only by independent classical paths, no coherent two-path interference term would appear. Quantum electrons are detected as localized events, but their accumulated distribution can contain fringes.

Read every step and figure

Step 1

If electrons were modeled only by independent classical paths, no coherent two-path interference term would appear. Quantum electrons are detected as localized events, but their accumulated distribution can contain fringes.

Step 2

$This image shows what happens in wave theory, a diffraction pattern that occurs when waves interfere with each other thus causing the alternating fringes or shadows that are shown on the screen. If the electrons behaved as a wave, as denoted by the squiggly lines, then interference would occur after going through the slits thus causing the band pattern on the screen where there are dark and light spots.$

This image shows what happens in wave theory, a diffraction pattern that occurs when waves interfere with each other thus causing the alternating fringes or shadows that are shown on the screen. If the electrons behaved as a wave, as denoted by the squiggly lines, then interference would occur after going through the slits thus causing the band pattern on the screen where there are dark and light spots.

Step 3

When the electrons are not measured before they hit the screen, then even though they are fired off one at a time, they will act like waves, showing an interference pattern. The detector is placed on the screen so that the screen lights up as each electron hits it.

Step 4

Obtaining reliable which-slit information can suppress interference by disturbing coherence between the alternatives. This is not a choice made by a person’s act of looking, and the remaining pattern need not be two perfectly sharp bars.

Chapter 9 Equations List

Pause & recall

Pop Quiz 9.4.1 Doppler Effects and Light

Question 1 of 5 · 0 correct

Extra exploration · Read the worked solution

A photon behaves like a particle and like a wave.

Read every full question without answers

Question 1

A photon behaves like a particle and like a wave.

True
False

Separate worked answer

Question 2

What chiefly enables a focused high-power laser to cut a material?

Photons acquiring a nonzero rest mass.
Mechanical photon impulse alone, without energy absorption.
An increase in light speed above c in vacuum.
Absorption of optical energy concentrated into a small area.

Separate worked answer

Question 3

A police siren is behind a robber’s car, and both move forward through still air. Identify the source and detector. Use f′ = f(v − u_D)/(v − u_S), with both u_D and u_S positive in the forward direction. Match −u_D to the detector and −u_S to the source.

1234

Match these labels to the numbered figure targets. Write your matches before opening the separate worked answer.

−u_S
−u_D
Detector
Source

Separate worked answer

Question 4

A detector moves directly away from a stationary alarm at 25 m/s while the stipulated sound speed is only 5 m/s. What problem arises when applying the usual subsonic Doppler model to ask for a lower observed frequency?

The usual subsonic formula yields a negative signed frequency ratio and does not provide the requested lower-frequency answer.
$3$ times
$1.5$ times
$10$ times

Separate worked answer

Question 5

The police siren source has speed 12.5 m/s in a medium with stipulated sound speed 8 m/s. Can the usual subsonic source–detector Doppler formula be used to infer a unique robber speed from a doubled detected frequency?

Question diagram; use the accompanying question and labels.

$21 m/s$
No valid unique speed follows from the usual subsonic model with these data.
$13 m/s$
$46 m/s$

Separate worked answer

Section 1:

$y (x, t) = A \cdot \sin (k \cdot x - ω \cdot t + ϕ)$ (Equation 9.1.1)

$ω = 2 \cdot π \cdot f$ (Equation 9.1.2)

$k = \frac{2 π}{λ}$ (Equation 9.1.3)

$T = \frac{1}{f}$ (Equation 9.1.4)

$y = Σ y_{m}$ (Equation 9.1.5)

$y (x, t) = A_{1} \cdot \sin (k_{1} \cdot x - ω_{1} \cdot t + ϕ_{1}) + A_{2} \cdot \sin (k_{2} \cdot x - ω_{2} \cdot t + ϕ_{2})$ (Equation 9.1.6)

$y (x, t) = 2 A \cdot \cos (\frac{1}{2} ϕ) \sin (k \cdot x - ω \cdot t - \frac{1}{2} ϕ)$ (Equation 9.1.7)

Section 2:

$v = \sqrt{\frac{B}{ρ}}$ (Equation 9.2.1)

$B = - \frac{Δ p}{Δ V / V}$ (Equation 9.2.2)

$s (x, t) = s_{m} \cos (k x - ω t)$ (Equation 9.2.3)

$Δ p (x, t) = Δ p_{m} \sin (k x - ω t)$ (Equation 9.2.4)

$Δ p_{m} = (v ρ ω) s_{m}$ (Equation 9.2.5)

$I = \frac{P}{A}$ (Equation 9.2.6)

$β = (10 dB) \log \frac{I}{I_{0}}$ (Equation 9.2.7)

Section 3:

$v = f \cdot λ$ (Equation 9.3.1)

$n = \frac{c}{v}$ (Equation 9.3.2)

$\frac{\sin θ_{1}}{\sin θ_{2}} = \frac{n_{2}}{n_{1}}$ (Equation 9.3.3)

$c = f \cdot λ$ (Equation 9.3.4)

$E = h \cdot f$ (Equation 9.3.5)

Section 4:

$f^{'} = f \frac{v \pm v_{D}}{v \pm v_{S}}$ (Equation 9.4.1)

Chapter 9 Problems

Work it out

Original problem 1

The source calls this “amplitude,” but a value at one position and time is displacement. It omits the coefficients’ displacement unit, so metres cannot be asserted for y.

Problem text

Two waves on a string are y₁ = 4U sin(9x − 4t) and y₂ = 6U sin(4x − t), where x is in metres, t in seconds, and U denotes the common displacement unit omitted in the source. Find the resultant displacement at x = 4 m and t = 4 s. Use radians.

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Work it out

Original problem 2

The requested quantity is displacement, not a time-independent resultant amplitude. The source omits the displacement unit.

Problem text

Two waves on a string are y₁ = 6U sin(x − t) and y₂ = 8U sin(x + t), where x is in metres, t in seconds, and U is the common displacement unit omitted in the source. Find the resultant displacement at x = 1 m and t = 6 s. Use radians.

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IMG 0092 cropped — details in the accompanying text.

103017569c — details in the accompanying text.

Section breakdown

Doppler Effect
Example 9.4.1- Ultrasounds
Photon Momentum
Wave-Particle Duality
Chapter Equations
Chapter Problems

Reviewed physics notes (31)

These corrections keep the original lesson’s purpose while fixing errors in the source edition.

The source introduces intensity changes as if they were frequency changes. — Reference 1
Clarifies that velocities are measured relative to the medium; relative car separation alone is insufficient. — Reference 1
Makes the velocity convention consistent with the source/detector quiz. — Reference 1
Clarifies the velocity convention. — Reference 1
Stationary relative to the ground is not enough if the medium moves. — Reference 1
Basic echo imaging is not itself a Doppler effect. — Reference 1
The computed 203.4 m/s is mathematically consistent but physiologically implausible; keep the source numbers as a hypothetical reflector. — Reference 1
The sign is determined by reflector motion relative to the medium, not merely by the fact that waves move toward it. — Reference 1
The source prose calls it receding but uses the approaching-source sign. — Reference 1
Removes “densified light” and the false claim that concentrating photons gives them inertial/rest mass. — Reference 1, Reference 2
Replaces the incorrect mass-times-velocity derivation of photon momentum. — Reference 1
The source wrongly attributes cutting hard materials to a large mechanical impulse from photons. — Reference 1
Corrects the classical-particle comparison and removes the claim that all waves vanish on contact. — Reference 1
Corrects observation-based switching between particle and wave and the confusion between slit and screen detection. — Reference 1
The existing Doppler caption mentions only louder sound. — Reference 1
A photon is not a source that emits a separate wave. — Reference 1
Corrects the caption’s mass claim and the confusion between beam intensity and individual photon energy. — Reference 1
Corrects the overly literal classical-particle stream description. — Reference 1
The source claims a particle changes behavior based simply on when it is observed and ignores single-slit diffraction. — Reference 1
Replaces a quiz built around the incorrect densified-light, inertial-mass, and momentum-only explanation. — Reference 1
Incorrect mass claim becomes an explicit distractor. — Reference 1
Momentum-only cutting claim becomes an explicit distractor. — Reference 1
A third unambiguous distractor preserves the four-choice structure. — Reference 1
Corrects the marked All of the Above answer. — Reference 1
The source/detector signs cannot be assigned from “getting closer” alone when both move in the same direction. Explicit direction removes the ambiguity. — Reference 1
The forward-moving detector uses v − u_D in this convention. Distinct labelled velocity terms keep the association choices unambiguous. Preserve the detector-sign association target. — Reference 1
The source behind the forward-moving detector uses v − u_S in this explicit positive-forward convention. Preserve the source-sign association target. — Reference 1
The source data are outside the ordinary subsonic model. They are retained as a model-validity check. — Reference 1
The formal ratio (5 − 25)/5 = −4 is not a lower positive frequency by a factor of four. No silent change of the stipulated speed is made. — Reference 1
The source is faster than the stated sound speed; the ordinary subsonic formula and insufficient geometry do not yield a valid unique answer. — Reference 1
Replaces the invalid marked 33 m/s without fabricating a new speed of sound or car speed. — Reference 1

About this edition and its source ↗

Glossary and deeper reading · 9 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Wave
A wave is a propagating disturbance that can transfer energy without transporting the medium as a whole. Mechanical waves require a material medium, while electromagnetic waves can propagate through vacuum. For example, a pulse travels along a rope while individual rope segments move locally around their equilibrium positions.
Read more: Mechanical waves →
Photon
A photon is a quantum of electromagnetic radiation with energy proportional to its frequency. It has zero rest mass and no electric charge but carries energy and momentum. For example, increasing light’s frequency increases the energy of each photon, while increasing the number of same-frequency photons increases the total energy.
Read more: Energy of light →
Frequency
Frequency is the number of complete repeating cycles per unit time. It is measured in hertz, meaning cycles per second, and is the reciprocal of the period. For example, a vibrating string with a higher frequency produces more cycles each second.
Read more: Frequency in detail →
Velocity
Velocity describes how position changes with time and includes direction. Instantaneous velocity describes one moment, while average velocity is displacement divided by elapsed time. For example, a car turning a corner can change velocity while maintaining the same speed.
Read more: Variables of motion →
Coherence
Coherence describes a sufficiently stable phase relationship between waves for a sustained interference pattern. It depends on the sources and on the time or distance over which their phase relationship remains predictable. For example, illuminating two slits with one suitable monochromatic source produces a stable relative phase at the slits.
Read more: Coherence in detail →
Linear momentum
Linear momentum is a vector describing an object’s motion and equals mass times velocity in the nonrelativistic model. The vector sum of momenta is conserved when the net external impulse on the system is zero. For example, two identical carts moving equally fast in opposite directions have zero total linear momentum but nonzero total kinetic energy.
Read more: Momentum →
Heat
Heat is energy transferred between systems because of a temperature difference. It describes a transfer rather than a substance or energy store contained in an object. For example, when a warmer mug transfers energy to a cooler hand, the mug loses internal energy and the hand gains it.
Read more: Heat in detail →
Thermal radiation
Thermal radiation is electromagnetic energy emitted by matter because of its temperature. It can transfer energy through a vacuum, and net radiative transfer between ordinary bodies is from hotter to cooler. For example, sunlight warming a surface does not need air to carry the energy across space.
Read more: Radiation →
Wave interference
Wave interference is the combination of overlapping wave disturbances. In a linear medium, the signed disturbances add at each location and time without permanently destroying the passing waves. For example, two overlapping upward rope pulses temporarily make a larger upward disturbance.
Read more: Wave interference in detail →

FFR - Wave-Particles

Doppler Effect

Gallery 9.4.1 Doppler Effect

Step 1

Example Problem 9.4.1 Which is the Detector and which is the Source?

Step 1

Step 2

Photon Momentum

Gallery 9.4.2 Photons

Step 1

Step 2

Wave Particle Duality

Gallery 9.4.3 Wave Particle Duality Theory

Step 1

Step 2

Step 3

Step 4

Chapter 9 Equations List

Pop Quiz 9.4.1 Doppler Effects and Light

Question 1

Question 2

Question 3

Question 4

Question 5

Original problem 1

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Original problem 2

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Original problem 3

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Original problem 4

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Original problem 6

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Original problem 5

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Original problem 8

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Original problem 7

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Original problem 10

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Original problem 9

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Wave

Photon

Frequency

Velocity

Coherence

Linear momentum

Heat

Thermal radiation

Wave interference

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