Mechanical vs Electromagnetic Waves: The Difference

A mechanical wave needs a medium to travel through and an electromagnetic wave does not. That one difference sets the rest: a mechanical wave takes its speed from the medium, while every electromagnetic wave crosses a vacuum at the same universal speed. Sound stops at a vacuum; light carries on.

AP Physics: Unit 14 (topics 14.1 Properties of Wave Pulses and Waves, 14.4 Electromagnetic Waves). The defining statement is AP Physics 2 essential knowledge 14.1.A.2 in Topic 14.1: mechanical waves or wave pulses require a medium in which to propagate, and electromagnetic waves or wave pulses do not require a medium in which to propagate. Essential knowledge 14.4.A.2 restates the second half. Speed follows from 14.1.A.3, that the speed at which a wave propagates through a medium depends on the type of wave and the properties of the medium, with 14.1.A.3.i giving the vacuum case as a universal physical constant, c = 3.00 x 10^8 m/s, printed in the constants box of the AP Physics 2 equation sheet and read off the rendered appendix page; 14.1.A.3.ii giving the string relation v = sqrt(F_T/(m/l)); and 14.1.A.3.iii giving the qualitative statement that in a given medium the speed of sound waves increases with the temperature of the medium. No speed of sound is printed in the constants box. Topic 14.4 develops the electromagnetic half: 14.4.A.1 says electromagnetic waves consist of oscillating electric and magnetic fields that are mutually perpendicular, 14.4.A.1.i says they are therefore transverse, 14.4.A.1.ii says they are commonly assumed to be plane waves characterized by planar wave fronts, 14.4.A.3 says categories are characterized by their wavelengths, 14.4.A.3.i lists them in order of decreasing wavelength spanning kilometers to picometers as radio waves, microwaves, infrared, visible, ultraviolet, X-rays and gamma rays, 14.4.A.3.ii orders visible light red, orange, yellow, green, blue, violet by decreasing wavelength, and 14.4.A.3.iii notes that electromagnetic waves of all wavelengths are sometimes collectively called light or electromagnetic radiation. The Topic 14.4 boundary statement says AP Physics 2 expects students to know the ordering of the electromagnetic spectrum, including visible light, but students will not be expected to define exact wavelength ranges within the spectrum; Topic 14.1 has no boundary statement. Sound is modelled as a mechanical longitudinal wave by 14.1.A.5.i, and polarization separates the two geometries at 14.3.A.2, 14.3.A.2.i and 14.3.A.2.ii, the last of which says longitudinal waves cannot be polarized. Light slowing inside a material is Topic 13.3, essential knowledge 13.3.A.3, n = c/v. Unit 14 is weighted at 12 to 15 percent of the multiple-choice section over a suggested 14 to 23 class periods. The current AP Physics 1 framework has no waves unit: its eight units are kinematics, force and translational dynamics, work energy and power, linear momentum, torque and rotational dynamics, energy and momentum of rotating systems, oscillations, and fluids, checked against that CED's table of contents, and its equation sheet carries no wavelength symbol, checked by rendering the appendix page.

The distinction, stated once

AP Physics 2 puts both halves in one essential knowledge statement, 14.1.A.2: mechanical waves or wave pulses require a medium in which to propagate, and electromagnetic waves or wave pulses do not require a medium in which to propagate. Essential knowledge 14.4.A.2 repeats the second half on its own.

That is the definition, and it is a definition by requirement rather than by appearance. A mechanical wave is a disturbance of something: a string, a body of air, a body of water. Take the something away and there is nothing left to disturb, so the wave does not slow down or fade, it simply cannot exist. An electromagnetic wave is a disturbance of the electric and magnetic fields themselves, which are present in empty space, so nothing has to be there for it to travel.

Everything else on this page is a consequence of that sentence, and the most useful consequence is about speed. Essential knowledge 14.1.A.3 says the speed at which a wave or wave pulse propagates through a medium depends on the type of wave and the properties of the medium. For a mechanical wave that is the whole story: the medium decides. For an electromagnetic wave in a vacuum there is no medium to decide, and 14.1.A.3.i fills the gap with a constant instead:

c=3.00×108 m/sc = 3.00 \times 10^8 \ \mathrm{m/s}

The CED calls it a universal physical constant, and it is printed in the constants and conversion factors box on the AP Physics 2 equation sheet as the speed of light.

One clarification that prevents most of the errors in this topic. Not requiring a medium is not the same as not entering one. Light passes through glass and water perfectly well, and when it does it slows, by n=c/vn = c/v from essential knowledge 13.3.A.3. The claim in 14.1.A.2 is about what is necessary, not about what is possible.

Side by side

Mechanical waveElectromagnetic wave
Needs a mediumYes, 14.1.A.2No, 14.1.A.2 and 14.4.A.2
What is oscillatingMatter, displaced from equilibriumElectric and magnetic fields, mutually perpendicular, 14.4.A.1
Speed in a vacuumIt has none. The wave does not exist therec=3.00×108 m/sc = 3.00 \times 10^8 \ \mathrm{m/s}, a universal constant, 14.1.A.3.i
Speed set byThe medium and the type of wave, 14.1.A.3Nothing, in vacuum. In a material, by v=c/nv = c/n, 13.3.A.3
Speed formula printed on the sheetvstring=FT/(m/)v_{\text{string}} = \sqrt{F_T/(m/\ell)}, for a stringn=c/vn = c/v, which gives the slowed speed in a material
Transverse or longitudinalEither. A string wave is transverse, sound is longitudinal, 14.1.A.5.iAlways transverse, 14.4.A.1.i
Can be polarizedOnly if transverse. Sound cannot, 14.3.A.2.iiYes, since it is transverse, 14.3.A.2.i
AP examplesWaves on a string, sound wavesRadio, microwave, infrared, visible, ultraviolet, X-ray, gamma, 14.4.A.3.i
Sorted byNothing in particularWavelength, 14.4.A.3
Temperature dependenceYes for sound in a given medium, 14.1.A.3.iiiNone stated
Frequency at a boundaryUnchanged, 14.3.A.1.ivUnchanged, 14.3.A.1.iv
Transfers matterNo, 14.1.A.1No

Two rows deserve a second look.

Speed in a vacuum is where the asymmetry is sharpest. It is tempting to read the mechanical entry as "zero", and that is wrong in a way that costs explanation marks. A wave with zero speed would be a wave that stays put; what actually happens is that the wave has nothing to propagate in, so there is no wave. Sound reaching the edge of the air does not slow to a stop, it stops being sound.

Transverse or longitudinal is the row people forget exists. Mechanical is not a synonym for longitudinal. A wave on a string is mechanical and transverse, and 14.1.A.4 defines transverse by the direction of the disturbance relative to the direction of propagation without mentioning what the wave is made of. Electromagnetic waves, by contrast, have only one option: 14.4.A.1.i says they are transverse because the oscillations of the electric and magnetic fields are perpendicular to the direction of propagation.

The case that separates them: the bell in the vacuum jar

Put a ringing bell inside a glass jar and pump the air out. The demonstration is old and it is still the cleanest test of 14.1.A.2, because two waves leave the bell and only one of them survives.

The sound fades to nothing. Sound is modelled as a mechanical longitudinal wave by 14.1.A.5.i, and the air is what carries it. Remove the air and the disturbance has nowhere to propagate. The bell is still being struck; nothing arrives.

The light does not fade at all. You can still see the hammer moving. Essential knowledge 14.4.A.2 says electromagnetic waves do not need a medium through which to propagate, so the emptier the jar the better the light does.

The experiment isolates the variable properly, which is why it is worth stating carefully. The bell, the frequency, the amplitude and the distance are all unchanged. The only thing removed is the medium, and exactly one of the two waves depended on it.

The same logic scales up. Sunlight crosses roughly 1.5×10111.5 \times 10^{11} metres of near vacuum to reach the ground, which no sound could do. The Sun is an extremely loud object and nothing of it is ever heard.

The reverse case is worth stating too, because it is where the misconception hides. Light entering glass does not stop, or struggle, or need the glass. It slows, by exactly the factor the index of refraction gives, and it emerges on the other side at cc again. A medium is an obstacle to light and a requirement for sound, which is a stronger contrast than "needs one, does not need one" makes it sound.

Speed, and which numbers you are allowed to use

This is the part where an exam question is usually decided, so it is worth knowing which figures the course hands you and which it does not.

For electromagnetic waves in a vacuum you get one number and it never changes. Essential knowledge 14.1.A.3.i: the speed of all electromagnetic waves in a vacuum is a universal physical constant, c=3.00×108c = 3.00 \times 10^8 m/s. The word all is load bearing. A gamma ray and a radio wave travel at the same speed in vacuum despite differing in wavelength by something like fifteen orders of magnitude, which 14.4.A.3.i describes as a range from kilometres down to picometres. Since λ=v/f\lambda = v/f and vv is the same for every one of them, the entire spectrum is one relation between λ\lambda and ff.

For mechanical waves you get one formula and one direction of dependence. The formula is for a string, essential knowledge 14.1.A.3.ii, printed on the sheet:

vstring=FTm/v_{\text{string}} = \sqrt{\frac{F_T}{m/\ell}}

Tighten the string and the wave speeds up; use a heavier string at the same tension and it slows down. The direction of dependence is for sound, essential knowledge 14.1.A.3.iii: in a given medium, the speed of sound waves increases with the temperature of the medium. No number, no formula, just the direction.

There is no printed speed of sound. The AP Physics 2 table of constants and conversion factors carries, among others, the speed of light, Planck's constant, the gas constant, Boltzmann's constant, the three particle masses, the elementary charge, the vacuum permittivity and permeability, Wien's constant, the Stefan-Boltzmann constant, the gravitational constant and gg. Not one of them concerns sound. That was read off the rendered appendix page rather than recalled. So a sound problem must give you a speed, and if it does not, the answer is symbolic.

In a material, an electromagnetic wave gets its speed from the material after all. Essential knowledge 13.3.A.3 says the index of refraction of a given medium is inversely proportional to the speed of light in that medium, with n=c/vn = c/v printed on the sheet. So v=c/nv = c/n, and light in glass of index 1.601.60 moves at 1.88×1081.88 \times 10^8 m/s. This is the one place the two categories touch: a medium is optional for light and, once present, it does set the speed.

What an electromagnetic wave is made of

The mechanical side of this pair is easy to picture and the electromagnetic side is not, which is worth addressing directly rather than leaving as "it is a wave in nothing".

Essential knowledge 14.4.A.1: electromagnetic waves consist of oscillating electric and magnetic fields that are mutually perpendicular. Three directions are in play. The electric field oscillates along one axis, the magnetic field oscillates along a second at right angles to it, and the wave travels along a third at right angles to both. Essential knowledge 14.4.A.1.i draws the conclusion: electromagnetic waves are transverse waves, because the oscillations of the electric and magnetic fields are perpendicular to the direction of propagation.

Essential knowledge 14.4.A.1.ii adds the modelling assumption the course uses: electromagnetic waves are commonly assumed to be plane waves, characterized by planar wave fronts.

Notice what is not being claimed. Nothing is being displaced from an equilibrium position in the sense 14.1.A.6 uses for a mechanical wave, because there is no matter to displace. What has an amplitude is a field. This is also why the CED never gives an electromagnetic amplitude a unit, and why you will never be asked to compute one in AP Physics 2.

The consequence that gets examined is polarization. Essential knowledge 14.3.A.2 says transverse waves that are reflected from a surface, refracted through a medium, or pass through specific openings may be polarized, 14.3.A.2.i says transverse waves can be polarized and oscillate in a single plane, and 14.3.A.2.ii says longitudinal waves cannot be polarized. Because every electromagnetic wave is transverse, every electromagnetic wave can be polarized. Because sound is longitudinal, sound cannot be, ever. Polarizing sunglasses exist and polarizing earplugs do not.

That gives a clean one-way test. If something can be polarized, it is transverse. It does not tell you whether it is mechanical, because a wave on a string is transverse and mechanical, and it can be polarized in the same sense.

When it costs a mark

Saying an electromagnetic wave has no speed in a vacuum, or slows down there. It has the largest speed it will ever have there, c=3.00×108c = 3.00 \times 10^8 m/s, and 14.1.A.3.i calls that a universal physical constant.

Saying a mechanical wave travels slowly through a vacuum. It does not travel through a vacuum at all. Write "cannot propagate", not "travels slowly", because the second answer describes a wave that still exists.

Treating mechanical as a synonym for longitudinal. A wave on a string is mechanical and transverse. Essential knowledge 14.1.A.4 and 14.1.A.5 classify by the direction of the disturbance, which is a separate question from whether a medium is needed.

Saying light needs a medium because it slows in glass. Slowing in a medium and requiring one are different claims. Essential knowledge 14.4.A.2 settles the requirement, and 13.3.A.3 handles the slowing.

Saying different colours of light travel at different speeds in a vacuum. Essential knowledge 14.1.A.3.i says the speed of all electromagnetic waves in a vacuum is the same constant. They differ in wavelength and frequency, not in vacuum speed.

Quoting a speed of sound from memory as though the sheet printed it. It does not. Take it from the question, or leave the answer symbolic.

Ordering the electromagnetic spectrum by frequency when the CED orders it by wavelength. Essential knowledge 14.4.A.3 says the categories are characterized by their wavelengths, and 14.4.A.3.i lists them in order of decreasing wavelength: radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays. Both orderings describe the same sequence, since higher frequency means shorter wavelength at fixed cc, but state which one you are using.

Learning wavelength ranges for the spectrum. The Topic 14.4 boundary statement says AP Physics 2 expects students to know the ordering of the electromagnetic spectrum, including visible light, and adds that students will not be expected to define exact wavelength ranges within the spectrum. Learn the order, not the numbers.

Saying either kind transfers matter. Neither does. Essential knowledge 14.1.A.1 says waves transfer energy between two locations without transferring matter between those locations, and it is stated for waves in general.

Where they behave alike, and why that lulls you

Once a medium is present, the two kinds obey the same rules, and that is what makes the distinction easy to forget.

The same relation connects speed, frequency and wavelength. λ=v/f\lambda = v/f is printed once on the sheet and used for both. Nothing in it knows which kind of wave it is describing.

The same boundary rules apply. Essential knowledge 14.3.A.1 covers a wave travelling from one medium to another, and its statements are written for waves generally: both reflected and transmitted waves result (14.3.A.1.i), the reflected wave is inverted if the transmitted wave slows (14.3.A.1.ii) and not inverted if it speeds up (14.3.A.1.iii), and the frequency does not change (14.3.A.1.iv). Light entering glass and a pulse crossing onto a heavier rope obey the same four statements.

Neither transfers matter, and both transfer energy. Essential knowledge 14.1.A.1 again, stated once for both.

Both can interfere and diffract. The relations on the sheet using mλm\lambda are written for waves in general, and the course applies them to sound and to light.

So a student can work correctly through most of Unit 14 without ever using the distinction, which is precisely why the questions that do use it feel like they came from somewhere else. They are the questions about a vacuum, about polarization, and about what happens when you take the medium away.

The test that separates the pair in one move: could this wave cross the space between the Earth and the Moon? If yes it is electromagnetic. If no it is mechanical, and 14.1.A.2 is the reason.

Where this sits on the AP exam

This distinction is AP Physics 2, and it is worth saying plainly why: the current AP Physics 1 framework has no waves unit at all. Its eight units run kinematics, force and translational dynamics, work energy and power, linear momentum, torque and rotational dynamics, energy and momentum of rotating systems, oscillations, fluids. That was checked against the AP Physics 1 CED's own table of contents, and its equation sheet carries no wavelength symbol anywhere, checked by rendering the appendix page. So a mechanical against electromagnetic question on an AP Physics 1 exam would be off syllabus.

In AP Physics 2 the pair sits in Unit 14, Waves, Sound, and Physical Optics, weighted at 12 to 15 percent of the multiple-choice section over a suggested 14 to 23 class periods. The defining statement is in Topic 14.1 and the electromagnetic half is developed in Topic 14.4. Polarization is in Topic 14.3, and the slowing of light inside a material is Topic 13.3, Refraction, in the geometric optics unit.

Both wave topics list the same four suggested skills for 14.4: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Topic 14.1 differs in one place, listing 1.C, create qualitative sketches of graphs, in place of 1.A.

Topic 14.4 carries the only boundary statement relevant here, and it limits the work usefully: AP Physics 2 expects students to know the ordering of the electromagnetic spectrum, including visible light, but students will not be expected to define exact wavelength ranges within the spectrum. Topic 14.1 has no boundary statement.

For the two definitions on their own, see mechanical wave and electromagnetic wave, and for the ordering itself see electromagnetic spectrum. For the classification that cuts across this one, see transverse vs longitudinal waves. For the speed that a medium fixes and a source does not, see wave speed vs particle speed, and for what survives a boundary, frequency vs wavelength.

Thunder and lightning: one kilometre, two waves

A lightning strike happens 1.00 km1.00 \ \mathrm{km} away. Take the speed of light as the sheet value c=3.00×108c = 3.00 \times 10^8 m/s and take the speed of sound in the air that day as 340 m/s340 \ \mathrm{m/s}, which the question has to supply. (a) Find the time for the light to arrive. (b) Find the time for the sound to arrive. (c) Find the ratio of the two times. (d) State which of the two speeds is a property of the air and which is not.

  1. (a) tlight=d/c=1000/(3.00×108)=3.33×106 st_{\text{light}} = d/c = 1000/(3.00 \times 10^8) = 3.33 \times 10^{-6} \ \mathrm{s}, about three microseconds.

  2. (b) tsound=d/v=1000/340=2.94 st_{\text{sound}} = d/v = 1000/340 = 2.94 \ \mathrm{s}.

  3. (c) 2.94/(3.33×106)=8.8×1052.94/(3.33 \times 10^{-6}) = 8.8 \times 10^5. The sound takes almost a million times longer, which is why counting seconds after the flash estimates the distance and counting after the thunder does not.

  4. (d) The 340 m/s340 \ \mathrm{m/s} is a property of that air, and essential knowledge 14.1.A.3.iii says it would be larger on a warmer day. The 3.00×108 m/s3.00 \times 10^8 \ \mathrm{m/s} is a universal physical constant by 14.1.A.3.i and would be identical with the air removed.

  5. The removal is the point. Take the air away and part (b) has no answer at all, because 14.1.A.2 says a mechanical wave requires a medium in which to propagate. Part (a) is unaffected.

The light takes 3.33×106 s3.33 \times 10^{-6} \ \mathrm{s} and the sound takes 2.94 s2.94 \ \mathrm{s}, a ratio of about 8.8×1058.8 \times 10^5. The speed of sound belongs to the air; the speed of light does not, and it is the value that survives if the air is removed.

Same frequency, wildly different wavelengths

A wave on a string travels at 50 m/s50 \ \mathrm{m/s} and an electromagnetic wave travels through vacuum. Both have a frequency of 500 Hz500 \ \mathrm{Hz}. (a) Find the wavelength of each. (b) Find the ratio of the two wavelengths. (c) Explain, using the CED statements, why the ratio came out as the number it did.

  1. (a) Use λ=v/f\lambda = v/f, printed on the AP Physics 2 sheet, for both. String: λ=50/500=0.10 m\lambda = 50/500 = 0.10 \ \mathrm{m}. Electromagnetic wave in vacuum: λ=c/f=(3.00×108)/500=6.00×105 m\lambda = c/f = (3.00 \times 10^8)/500 = 6.00 \times 10^5 \ \mathrm{m}, which is 600 km600 \ \mathrm{km}.

  2. (b) (6.00×105)/0.10=6.0×106(6.00 \times 10^5)/0.10 = 6.0 \times 10^6. The electromagnetic wavelength is six million times longer.

  3. (c) The frequency was the same in both calculations, so the entire ratio came from the speeds: (3.00×108)/50=6.0×106(3.00 \times 10^8)/50 = 6.0 \times 10^6, the same number. The string speed came from the medium by 14.1.A.3, and the vacuum speed came from 14.1.A.3.i, where no medium is involved.

  4. A useful sanity check on the electromagnetic answer: 500 Hz500 \ \mathrm{Hz} is an audio frequency, and an electromagnetic wave at that frequency has a wavelength of hundreds of kilometres. Essential knowledge 14.4.A.3.i places the longest-wavelength category, radio waves, at the kilometre end of the spectrum, so this wave is at the far radio end and is nothing anybody would call light. It is still an electromagnetic wave, and it still travels at cc.

0.10 m0.10 \ \mathrm{m} on the string and 6.00×105 m6.00 \times 10^5 \ \mathrm{m} in vacuum, a ratio of 6.0×1066.0 \times 10^6, which is exactly the ratio of the two wave speeds because the frequency was common to both.

Light does not need glass, and glass still slows it

A beam of light crosses a 3.0 cm3.0 \ \mathrm{cm} thickness of glass with index of refraction n=1.60n = 1.60. Take c=3.00×108c = 3.00 \times 10^8 m/s. (a) Find the speed of the light inside the glass. (b) Find the time to cross the glass. (c) Find the time to cross the same 3.0 cm3.0 \ \mathrm{cm} in vacuum, and the extra time the glass costs. (d) Reconcile this with the statement that electromagnetic waves do not require a medium.

  1. (a) Essential knowledge 13.3.A.3 gives n=c/vn = c/v, printed on the sheet, so v=c/n=(3.00×108)/1.60=1.875×108 m/sv = c/n = (3.00 \times 10^8)/1.60 = 1.875 \times 10^8 \ \mathrm{m/s}, which rounds to 1.9×108 m/s1.9 \times 10^8 \ \mathrm{m/s}.

  2. (b) t=d/v=0.030/(1.875×108)=1.6×1010 st = d/v = 0.030/(1.875 \times 10^8) = 1.6 \times 10^{-10} \ \mathrm{s}.

  3. (c) In vacuum, t=0.030/(3.00×108)=1.0×1010 st = 0.030/(3.00 \times 10^8) = 1.0 \times 10^{-10} \ \mathrm{s}. The glass costs an extra 1.6×10101.0×1010=6.0×1011 s1.6 \times 10^{-10} - 1.0 \times 10^{-10} = 6.0 \times 10^{-11} \ \mathrm{s}, and the ratio of the two times is 1.601.60, which is the index itself.

  4. (d) Essential knowledge 14.4.A.2 says electromagnetic waves do not need a medium through which to propagate, which is a statement about necessity. It does not say a medium has no effect. When a medium is there, 13.3.A.3 says it slows the light in inverse proportion to its index, and the light emerges on the far side travelling at cc again.

  5. Contrast the mechanical case to see how different the two statements are. A sound wave in the same 3.0 cm3.0 \ \mathrm{cm} of vacuum has no transit time to compute, because 14.1.A.2 says it cannot propagate there at all. For light, removing the glass gave the fastest crossing rather than no crossing.

1.9×108 m/s1.9 \times 10^8 \ \mathrm{m/s} in the glass, crossing in 1.6×1010 s1.6 \times 10^{-10} \ \mathrm{s} against 1.0×1010 s1.0 \times 10^{-10} \ \mathrm{s} in vacuum, an extra 6.0×1011 s6.0 \times 10^{-11} \ \mathrm{s}. A medium slows light and is not required by it, which is the opposite of the mechanical case in both respects.

Frequently asked questions

What is the difference between mechanical and electromagnetic waves?

A mechanical wave requires a medium in which to propagate and an electromagnetic wave does not. AP Physics 2 states both halves in essential knowledge 14.1.A.2, and 14.4.A.2 repeats the second. The consequences follow from there: a mechanical wave takes its speed from the medium by 14.1.A.3, while every electromagnetic wave crosses a vacuum at the universal constant c = 3.00 times ten to the eighth metres per second by 14.1.A.3.i. A mechanical wave can be transverse or longitudinal; an electromagnetic wave is always transverse.

Why can sound not travel through a vacuum?

Because sound is a mechanical wave, modelled in AP Physics 2 as a mechanical longitudinal wave at essential knowledge 14.1.A.5.i, and 14.1.A.2 says mechanical waves require a medium in which to propagate. A sound wave is a pattern of compressions and rarefactions in matter, so with no matter present there is nothing to compress and no wave. The correct wording matters on an exam: the sound does not travel slowly through a vacuum, it cannot propagate there at all.

Do all electromagnetic waves travel at the same speed?

In a vacuum, yes. Essential knowledge 14.1.A.3.i in AP Physics 2 says the speed of all electromagnetic waves in a vacuum is a universal physical constant, c = 3.00 times ten to the eighth metres per second. A radio wave and a gamma ray travel at the same speed there despite wavelengths that 14.4.A.3.i describes as spanning kilometres down to picometres. Inside a material the speed drops to c divided by the index of refraction, by essential knowledge 13.3.A.3.

Are all mechanical waves longitudinal?

No. Mechanical describes what the wave needs, and transverse or longitudinal describes which way the disturbance points, so the two labels are independent. A wave on a string is mechanical and transverse; a sound wave is mechanical and longitudinal, by essential knowledge 14.1.A.5.i. Essential knowledge 14.1.A.4 defines a transverse wave as one where the direction of the disturbance is perpendicular to the direction of propagation, without saying anything about what the wave is made of.

Can sound be polarized?

No. Essential knowledge 14.3.A.2.ii in AP Physics 2 says longitudinal waves cannot be polarized, and 14.1.A.5.i models sound as a mechanical longitudinal wave. Polarization restricts an oscillation to a single plane, by 14.3.A.2.i, and a longitudinal disturbance already lies along one line, so there is no plane to restrict. Electromagnetic waves can always be polarized, because 14.4.A.1.i says they are transverse.

Is light a mechanical wave?

No. Light is electromagnetic, and essential knowledge 14.4.A.1 says electromagnetic waves consist of oscillating electric and magnetic fields that are mutually perpendicular, with 14.4.A.2 stating that they do not need a medium through which to propagate. Nothing material is displaced. Light does slow down when it passes through a material, by n = c over v from essential knowledge 13.3.A.3, but slowing in a medium is a different claim from requiring one.

Which AP course covers mechanical and electromagnetic waves?

AP Physics 2. The distinction sits in Unit 14, Waves, Sound, and Physical Optics, weighted at 12 to 15 percent of the multiple-choice section over a suggested 14 to 23 class periods, with the defining statement in Topic 14.1 and the electromagnetic material in Topic 14.4. The current AP Physics 1 framework has no waves unit at all; its eight units run from kinematics through fluids, and its equation sheet carries no wavelength symbol. So a waves question on an AP Physics 1 exam would be off syllabus.