Where is the **amplitude** of the waveform shown below?

**Amplitude** is the distance between the average value of the waveform and the extreme value of the waveform.

Don't fall into the trap of thinking the amplitude is the full range from minimum to maximum; that is actually twice the amplitude.

What is the amplitude of the waveform below, if each vertical division represents 1 cm?

The amplitude of the wave is 3 cm.

Don't make the mistake of thinking the amplitude covers the entire wave, from minimum to maximum; that value (6 cm in this case) is twice the actual amplitude.

What is the **period** of the waveform shown below?

The **period** is the amount of time it takes the wave to complete one full oscillation.

Don't make the mistake of measuring from one zero of the waveform to the next zero; that value only captures one-half of the oscillation, and represents one-half of the period.

Period is typically represented by T on the AP Physics exam.

What is the period of the waveform below, if each horizontal division represents 1 s?

The period of the wave is 6.5 s.

Don't make the mistake of thinking the period is measured from one midpoint to the next; that value (3.25 s in this case) is half the period.

Which waveform is oscillating more rapidly, if the period of waveform 1 is twice that of waveform 2?

Waveform 2 is oscillating more rapidly.

The period is the time for the waveform to complete one full oscillation. The larger the period, the more time it takes for an oscillation to complete, the slower the oscillation.

What is the **frequency** of a waveform?

The **frequency** of a waveform is a measure of how rapidly the waveform oscillates.

The frequency of a waveform is calculated as **f = 1/T**, where f is the waveform's frequency, and T is the waveform's period.

The common units of frequency on the AP Physics exam are Hz, where 1 Hz is 1 s^{-1}.

How is the **angular frequency** ω of an oscillating system calculated?

The **angular frequency **ω is calculated as:

ω = 2πf

The units of ω are radian/s.

What is the frequency of the waveform shown below, if each horizontal division represents 1 second?

The waveform's frequency is 0.2 Hz.

The period of the waveform is 4.5 s, and the frequency f = 1/T, or 1/4.5 s^{-1}.

Which waveform is oscillating more rapidly, if the frequency of waveform 1 is 100 Hz, while that of waveform 2 is 200 Hz?

Waveform 2 is oscillating more rapidly.

Since frequency is inversely proportional to period, it has the inverse relationship with wave oscillation speed; the higher a wave's frequency, the more rapidly it oscillates.

the **phase** of a waveform

The **phase** of a waveform is the offset of the waveform relative to its origin (zero value).

Although phase can hold any value, on the AP Physics exam the only commonly-tested values are integer amounts of quarter-wavelengths.

What is the **phase difference** between two waveforms?

The **phase difference** between two waveforms is the value of the phase of the second waveform at the origin of the first waveform.

Ex: The picture below represents a phase difference of one-half of a wave. Notice that point A is the origin of the red waveform, while the black waveform is halfway through a full oscillation.

The two waveforms below differ by one-half of a wave. What is the calculated phase difference in:

- degrees?
- radians?

One-half of a wave corresponds to a phase difference of:

- 180º
- π radians

The two waveforms below differ by one-quarter of a wave. What is the calculated phase difference in:

- degrees?
- radians?

One-quarter of a wave corresponds to a phase difference of:

- 90º
- π/2 radians

**Hooke's Law** for spring force

**Hooke's Law** relates the extension of a spring to the force exerted by the spring.

F_{x} = -kx

Where:

F_{x} = the force exerted by the spring in N

k = the force constant of the spring in N/m

x = the extension of the spring from equilibrium in m

By how much does a vertically-hung spring stretch if its force constant is 1,000 N/m and a 10 kg object is attached to it?

The spring stretches by 10 cm.

According to Hooke's Law, F_{x} = -kx. The spring will stretch until the force equals the weight of the object, 100 N in this case. So:

100 = -1,000 * x

x = 10^{-1} m

**simple harmonic motion (SHM)**

**Simple harmonic motion** is the motion that results when an object at equlibrium is displaced, and feels a restorative force that is proportional to the displacement.

The most common examples of simple harmonic motion seen on the AP Physics exam are the mass on a spring and the motion of a pendulum.

What is the shape of the graph for the motion of any simple harmonic motion system? (assume no frictional forces)

All simple harmonic motion systems have a sinusoidal graph.

Ex: In the below graph, displacement is graphed vs. time for a mass on a spring system.

How would the graph for a mass on a spring look different if the mass begins to oscillate more quickly?

The peaks of the graph will move closer together.

Ex: In the graph below, the red line represents a system which is oscillating more rapidly.

When is the point of maximum kinetic energy for a system oscillating with simple harmonic motion?

The system has maximum kinetic energy when the object is at the equilibrium position.

Kinetic energy is proportional to the square of the velocity, and velocity is the slope of the line tangent to the position curve, shown below. The slope maximizes as the line crosses through equilibrium, and is at a minimum (0) at the extreme positions.

When is the point of maximum potential energy for a system oscillating with simple harmonic motion?

The system has maximum potential energy when the object is at the maximum distance from equilibrium, or at the amplitude.

Since the system's total energy remains constant, the maximum potential energy will occur when the kinetic energy is at a minimum.

What is the formula for the angular frequency of an object oscillating on a spring?

For an object oscillating on a spring, the angular frequency is:

ω = (k/m)^{½}

Where:

k = force constant of the spring (N/m)

m = the mass of the object

Two objects are oscillating on identical springs. Object 1 has a mass of 5 kg, while the mass of object 2 is 10 kg. Which one oscillates more rapidly?

Object 1 oscillates at the higher frequency.

Remember: the frequency of an object oscillating on a spring is (k/m)^{½}. The frequency is inversely proportional to the square root of the mass, so the larger the mass, the lower the frequency.

Two identical objects are oscillating on springs. Spring 1 has a force constant of 5,000 N/m, while the force constant of spring 2 is 10,000 N/m. Which one oscillates more rapidly?

The mass on spring 2 oscillates at the higher frequency.

Remember: the frequency of an object oscillating on a spring is (k/m)^{½}. The frequency is directly proportional to the square root of the force constant, so the larger the force constant, the higher the frequency.

Two identical objects are suspended from identical springs. One is displaced 10 cm from equilibrium, the other is displaced 20 cm from equilibrium, and both are allowed to oscillate. Which one oscillates more rapidly?

The two springs oscillate at the same frequency.

Remember: the frequency of an object oscillating on a spring is (k/m)^{½}. The frequency depends only on the force constant of the spring and the mass of the object, not the amplitude of the oscillation.

What is the potential energy of a spring displaced from equilibrium?

The potential energy of any spring stretched a distance x from equlibrium is:

U = ½kx^{2}

Where k is the force constant of the spring.

What is the maximum value of the potential energy of a mass oscillating on a spring?

The maximum value of the potential energy for a mass oscillating on a spring is U_{max} = ½kA^{2}.

Remember: the potential energy for the system for any value x of the spring's extension is ½kx^{2}. The maximum value of x is A, the amplitude. Plugging in A for x gives the maximum value of U.

What is the kinetic energy of a mass oscillating on a spring?

The kinetic energy of a mass oscillating on a spring is:

KE = ½mv^{2}

Where:

m = the mass of the object oscillating

v = the velocity of the object at that moment

What is the maximum value of the kinetic energy of a mass oscillating on a spring?

KE_{max} = ½kA^{2}.

Since the energy of the system is constant, the maximum value of the kinetic energy is equal to the maximum value of the potential energy at full extension. At the equilibrium position, all the energy in the system is in the form of kinetic energy.

What is the maximum value of the velocity of a mass oscillating on a spring?

v_{max} = A(k/m)^{½} = Aω.

The maximum value of the kinetic energy is ½kA^{2}. Setting this equal to ½mv^{2} and solving for v yields the final answer.

What is the angular frequency of an oscillating pendulum?

f = (g/L)^{½}

Where:

f = the pendulum oscillation frequency in Hz

g = gravitational acceleration in m/s^{2}

L = length of the string of the pendulum

What is the difference in the oscillation frequency of two different pendulums, if pendulum 1 has a string four times the length of pendulum 2?

Pendulum 2 has a frequency twice as high as the frequency of pendulum 1.

From the equation f = (g/L)^{½}, the frequency of a pendulum's oscillation is inversely proportional to the square root of the string's length. Since pendulum 1 has a string 4 times as long, its frequency is 1/√4 that of pendulum 2, or half as much.

**Transverse wave**

In a** transverse wave**, the particles oscillating move with displacement perpendicular to the direction of propagation.

What are some classic examples of transverse waves?

Transverse waves include:

- light waves (electromagnetic waves)
- string waves
- pond waves*
- stadium waves

*Technically water waves also fall under the classification of "surface waves", but the AP Physics exam does not require that definition.

**Longitudinal wave**

In a **longitudinal wave**, the particles oscillating move with displacement parallel to the direction of propagation.

What are some classic examples of longitudinal waves?

Logitudinal waves include:

- sound waves
- stretched slinky waves (spring waves)

Calculate the period of the wave shown below:

The period is 5 seconds.

A full wave cycle is any 360 degree segment and is usually easiest to see as peak-to-peak, or valley-to-valley. Notice that the wave valley hits exactly at 0, 5, 10, etc.

Calculate the frequency of the wave below:

The frequency is .2 Hz.

Since frequency = 1/T, and the period is 5 seconds, this gives f = 1/5 = .2

Calculate the amplitude of the wave below:

The amplitude is 17 meters.

Amplitude is the greatest positive displacement from zero of a wave, measured at its peak.

Calculate the wavelength of the wave below:

The wavelength is 12.5 meters.

A full wave cycle is any peak-to-peak, or valley-to-valley distance. Notice that the wave valley hits exactly at 0 and 25, but this requires two oscillations to cover this distance. Hence, wavelength is 25/2 = 12.5 for just one.

What is the speed of the wave below, if its wavelength is exactly 5 meters?

1 m/s

wave speed = λ * f = 5 (1/5) = 1

**Intensity of sound**

**Sound intensity** is the power per area being delivered by the medium. This is the same as the wave pressure (force per area), times the velocity of the wave.

The standard reference value for sound intensity is I_{o} = 10^{-12} W/m^{2}.

How is does the intensity of a wave relate to its amplitude?

A wave's intensity is proportional to the square of its amplitude:

I α A^{2}

**Constructive interferenc****e**

**Constructive interference** occurs when two waves overlap and the new amplitude is greater than the amplitude of either initial wave.

**Totally constructive interference** occurs when the waves overlap with exactly the same speed, wavelength and frequency. This causes the final amplitude to be exactly the sum of the two starting amplitudes.

Two waves with amplitude 4m and 3m interfere and a wave with amplitude 7m is the result. What type of interference occurred?

Totally constructive interference.

The resultant wave has amplitude exactly the sum of the amplitudes of the two initial waves.

**Destructive interference**

**Destructive interference** occurs when two waves overlap and the new amplitude is less than the amplitude of either initial wave.

**Totally destructive interference** occurs when the waves are 180 degrees out of phase with exactly the same speed, wavelength and frequency. This causes the final amplitude to be exactly the difference of the two starting amplitudes.

Two waves with amplitude 4m and 7m interfere and a wave with amplitude 3m is the result. What type of interference occurred?

Totally destructive interference.

The resultant wave has amplitude exactly the difference between the amplitudes of the two initial waves.

**resonance **of an oscillating system

An oscillating system is in **resonance** when it is driven to oscillate at its natural frequency.

Ex: a mass on a spring system resonates when an external force drives it at a frequency of (k/m)^{½}.

When a system is oscillating in resonance, its amplitude tends to increase dramatically, a concept familiar to any small child on a playground swing.

**standing wave**

A **standing wave** is one that exists at a fixed length in a given medium.

An example of this is a guitar string that is stationary at both ends, but can have waves propogate between the two ends. Below is a guitar string vibrating at the 4th harmonic.

**Nodes**

A **node **is the point on a standing wave that experiences zero displacement. **Nodes **remain fixed in position.

In the image below, the black dots are the nodes.

What condition must be met, in order for a node to exist at the end of a standing wave?

For a node to exist at the end of the standing wave, that end must be fixed in position.

A guitar string tied down on both ends has nodes at both ends. An organ pipe with one end closed has a node at the closed end only.

**antinode**

An **antinode **is the set of positions on a standing wave that experience the greatest displacement. A wave will fluctuate between **antinode **positions.

In the image below, the black dots are the antinodes.

What condition must be met, in order for an antinode to exist at the end of a standing wave?

For an antinode to exist at the end of the standing wave, that end must be open (not fixed).

The pipe on a wind chime with both ends open has antinodes at both ends. An organ pipe with one end open has an antinode at the open end only.

**beat frequency**

**Beat frequency** is the frequency of the new wave that occurs when two waves with different frequencies interfere. Specifically, this term refers to the frequency of the "envelope" - the amplitude oscillation of the new wave.

It is possible to have beat frequencies with any number of waves at once (3, 10, etc) but the AP Physics exam only tests beats with two waves.

When two waves with frequencies f_{1} and f_{2} interfere, what is frequency of the beats?

The frequency of the beats (amplitude oscillation envelope) will be the absolute value of their difference.

f_{beat} = |f_{1} - f_{2}|

Tone A: 220Hz and Tone B: 224Hz are played simultaneously, creating a beat frequency. If tone A is increased by 12 Hz and tone B is increased by 4 Hz, how will the new beat frequency compare to the old one?

The new beat frequency will be exactly the same.

Old beat frequency = |220-224| = 4

New beat frequency = |232-228| = 4

**refraction** of a wave

A wave **refracts **whenever it transitions from one medium to another. Typically, the wave will have different properties in the incident medium than in the propogating medium.

Refracting waves can change speed, amplitude, and direction, but their frequency is always conserved.

A wave refracts from medium 1 to medium 2 at an angle to the boundary. The wave moves much slower in medium 2. How does the angle of the wave vary in medium 1 vs. in medium 2?

The angle will be smaller (closer to the normal) in medium 2.

Waves always travel closer to the normal in the medium where they travel more slowly. This is due to the fact that the momentum of the component of the wave parallel to the normal must be conserved. If the wave is moving more slowly, it must move closer to the normal to conserve this quantity.

**diffraction** of a wave

**Diffraction** occurs when a wave is incident on a narrow gap in a barrier. The gap acts as a point source, and the wave on the far side of the gap propogates in circular waveforms.

In Young's Double Slit Experiment, why does light incident on two narrow slits in a barrier form a fringe pattern on a far screen?

Young's results are explained by diffraction. The two slits each act as point sources, and the waves interfere with one another depending on their path length difference, creating the classic fringe pattern.

What type of wave is sound?

Longitudinal.

Waves that oscillate in the diection of propagation are longitudinal.

How does the sound wave physically propogate, as described by looking at specific particles of the medium?

Particles pulse outward from some initiation point, towards their neighbors, creating a high density region.

That high density wave-front continues outward, while the original particles settle back into the now much lower density region that they initially occupied.

This high-and-then-low density pulse pair is a sound wave, and can exist in any medium.

What two properties of the medium most affect speed of sound?

Elasticity and density.

Recall that sound is both a pulse forward and the resultant pull backward on a particle. Mediums that can bounce back faster and have less mass/volume to move will allow sound to travel at a greater rate.

Which should have a faster speed of sound: nickel or bronze, assuming that both have relatively equal density but nickel is stiffer?

Sound travels faster in nickel.

Sound will travel 1.6 times faster in nickel than in bronze, due to the greater stiffness.

Though it may seem counterintuitive, the *elastic modulus* (elasticity) measures how well a medium's particles bounce back to their original position. A stiff material will bounce back much faster than a spongy material.

Which should have a faster speed of sound: protium (^{1}_{1}H) or deuterium (^{2}_{1}H), assuming the same number of particles per volume?

Sound travels faster in protium.

Sound travels about 1.4 times faster in protium than in deuterium. Since deuterium has twice the mass, it will have twice the density for a given volume. All other physical properties being equal, density is the determining factor.

Please rank the phases of matter in terms of highest to lowest speed of sound, according to elasticity.

Since elasticity is a measure of a substance's ability to bounce back quickly into an original position following a pulse:

Gases will have the lowest speed (air ~330 m/s).

Liquids will be faster (water ~1,480 m/s).

Solids will be the fastest speed (iron ~5,120 m/s).

What is the speed of sound in air at STP?

v_{c} =~330 m/s at STP.

v_{c} or c_{air} are often used to represent the constant speed of sound in air.

Though most universal constants are given on the AP Chemistry exam, this is so often used in problems that it bears memorizing.

**intensity **of sound

**Intensity** of sound is the power per unit area of that wave.

**Units** of intensity are** W/m ^{2}**, and the standard reference value (or baseline value) is 10

^{-12}W/m

^{2}.

Define the relationship between intensity of sound and the decibel scale.

There is a logarithmic relationship between decibels (L), intensity (I'), and baseline intensity (I_{0}), such that:

L = 10*log(I' / I_{0})

The number L in decibels is technically a "dimensionless" quantity, since it's describing a ratio relationship between the baseline value and the observed value, instead of a physical quantity.

What is the shortcut to calculate the log of any number with a positive exponent?

If the number Z = n x 10^{e}

then log(Z) = {e}.{n}

Ex: if Z = 6.2 x 10^{4}, then n = 6.2, and e = 4, so the log(Z) can be approximated as equal to

[4].[6.2] = 4.62

Note: the real value of log(Z) = 4.79, which is within the acceptable error for the AP Physics exam answer set.

What decibel value corresponds to an intensity reading of 10^{-10} W/m^{2}?

20 dB

L = 10*log(I' / I_{0}) = 10*log(10^{-10} / 10^{-12})

= 10*log(10^{2}) = 10*2 = 20dB

This is nearly the lowest intensity of sound that humans can hear, and corresponds to a low whisper.

What decibel value corresponds to an intensity reading of 10^{2} W/m^{2}?

140 dB

L = 10*log(I' / I_{0}) = 10*log(10^{2} / 10^{-12})

= 10*log(10^{14}) = 10*14 = 140dB

This is nearly the highest intensity of sound that humans can hear without going instantly deaf, and corresponds to being at the source of a jet engine on take-off.

What is the increase in intensity between a shout at 50dB and a loud stereo at 80dB?

1000X more intensity.

Recall that the dB scale is logarithmic, so that every increase of 10dB corresponds to 10X intensity increase. 30dB increase therefore must be 10X10X10=1000.

**attenuation** of sound

**Attenuation** is the gradual drop in intensity of sound as it passes through any real medium. This is generally due to absorption of wave energy by the medium.

Attenuation = α * *l* * f

Where:

α = attenuation coefficient in dB/MHz*cm

*l* = length of medium passed through in cm

f = frequency of wave in MHz

How will the attenuation amount differ between water α=.002 and blood α=.2, assuming equal frequency waves pass through equal lengths of both?

Attenuation will be 100X greater in blood, due to the higher α value.

From: attenuation = α * l * f

Since attenuation is proportional to α, and all other terms are constant, an α 100X larger will product an attenuation 100X larger.

How much dentin α=80 will a wave have to pass through in order to show the same attenuation as if it were passing through 2cm of enamel α=120?

3 cm

From: Attenuation = α * *l* * f

Enamel = 120*2*f

Dentin = 80**l**f

setting equal: 240f = 80*l*f

240 = 80*l
l = *3 cm

**Pitch**

**Pitch** is the tonal property comparing different frequency sounds to each other (one will necessarily be "higher pitch" and one "lower pitch").

Usually there is at least one pitch or tone set as standard to gauge the rest against. The pitch "A above middle C" is usually paired with the frequency 440Hz and is such a reference.

Which is a higher pitch, of two off-pitch A notes at 448 Hz and 465 Hz?

465 Hz is the higher pitch.

Pitch is always compared between tones; hence 465 Hz is high compared to 448 Hz.

What is the relationship between wave speed, wavelength and frequency?

v_{c} = f * λ

This relationship is often referred to as the **wave formula.**

What is the change in wavelength, if the frequency of a sound wave traveling through air doubles?

Wavelength will decrease to 1/2.

From: v_{c} = f * λ

Frequency and wavelength are inversely proportional, since the speed of sound in air is constant. As frequency doubles, wavelength must halve.

What is the change in wavelength, if sound at one frequency enters a new medium that causes the speed to halve?

Wavelength will decrease to 1/2.

From: v_{c} = f * λ

Speed and wavelength are directly proportional, since the frequency of sound is held constant. As speed halves, wavelength must halve.

What does the **Doppler effect **describe, and what formula can be used to calculate it?

The** Doppler effect** describes the change in observed frequency of a wave, compared to the original emitted frequency.

f' = f_{o} (v_{c} ± v_{d}) / (v_{c} ± v_{s})

Where:

f_{o }= emitted frequency in Hz

f' = new observed frequency in Hz

v_{c} = constant speed of sound in m/s

v_{d} = speed of detector in m/s

v_{s} = speed of source in m/s

If both the source and the detector are moving in the same direction at the same speed, what will be the observed change from emitted frequency?

There will be no difference between emitted frequency and observed frequency at the detector.

Source and detector moving in the same direction will be addition in the numerator and denominator of the Doppler formula.

f' = f_{o} (v_{c} ± v_{d}) / (v_{c} ± v_{s})

Since v_{d} = v_{s} = value this gives :

f' = f_{o} (v_{c} + value) / (v_{c} + value)

= f_{o} * (1) = f_{o}

If the source is moving towards the detector, but the detector remains motionless, what will be the observed change from emitted frequency?

The observed frequency at the detector will be higher than the emitted frequency.

Source moving in the direction of the detector will be subtraction in the denominator of the Doppler formula.

f' = f_{o} (v_{c} ± v_{d}) / (v_{c} ± v_{s})

Since v_{d} = 0 and v_{s} = some value, this gives :

f' = f_{o} (v_{c}) / (v_{c} - value)

= f_{o} (v_{c})/(smaller) = larger than f_{o}

If the detector is moving in the direction of the source, what will be the observed change from emitted frequency?

The observed frequency at the detector will be higher than the emitted frequency.

Detector moving in the direction of source will be addition in the numerator of the Doppler formula.

f' = f_{o} (v_{c} ± v_{d}) / (v_{c} ± v_{s})

Since vd = some value and vs = 0 this gives :

f' = f_{o} (v_{c} + value) / (v_{c})

= f_{o} (larger) / (v_{c}) = greater than f_{o}

If both the source and the detector are moving away from one another, what will be the observed change from emitted frequency?

The observed frequency at the detector will be lower than the emitted frequency.

Source and detector moving in opposite directions will be subtraction in the numerator and addition in the denominator of the Doppler formula.

f' = f_{o} (v_{c} ± v_{d}) / (v_{c} ± v_{s})

Since v_{d} = some value and v_{s} = some value, this gives :

f' = f_{o} (vc - value) / (v_{c} + value)

= f_{o} (smaller) / (larger) = smaller than f_{o}

What physical concept, relating to sound, is the arrow pointing to?

A node.

Nodes are the part of a standing wave that remains fixed in position.

What physical concept, relating to sound, are the arrows pointing to?

Antinodes.

Antinodes are the positions of greatest displacement for a standing wave.

Describe the specific standing wave that can exist in a pipe with both ends open.

A pipe with both ends open must have antinodes at both ends, and at least one node between them.

The wave below is the 4th harmonic, since it has 4 nodes.

Describe the specific standing wave that can exist in a pipe with only one end open and the other closed.

A pipe with only one end open must have an antinode at that end, and a node at the closed end.

The wave below is the 7th harmonic, since it must pass through 7 nodes (3 in, 1 on end, 3 back out).

Describe the specific standing wave that can exist on a string with both ends fixed in place.

A string with both ends fixed must have nodes at both ends, and at least one antinode between them.

The wave below is the 3rd harmonic, since it has 3 antinodes.

How is the harmonic number determined, for different types of standing waves?

For any standing wave with at least one end open, count the total number of nodes that one wave contains.

For a standing wave with both ends fixed, count the total number of antinodes that one wave contains.

Without actually seeing a standing wave, what information besides the shape and type of medium is necessary to fully describe it?

The harmonic number.

Given the shape of the medium (closed, open, etc), type of medium (air, water, etc) and the harmonic number - all other values can be calculated.

What will the fourth harmonic look like for a wind chime pipe that has both ends open?

Antinodes at both ends, and four nodes (with corresponding antinodes) between them.

What will the fifth harmonic look like for an organ pipe that has only one end open?

An antinode at the open end and a node at the closed end, and two nodes (with corresponding antinodes) between them.

There are 5 nodes total that the wave passes through (2 in, 1 on the end, 2 back out).

What will the third harmonic look like for a guitar string that has both ends fixed in place?

Nodes at both ends, and three antinodes (with corresponding nodes) between them.

**Ultrasound**

Ultrasound is the region of sound just above human perception.

Specifically, this is the region 20,000 Hz and higher. Medical applications use sound into the megaherz and gigaherz.

If a bat can hear sounds up to 22,000 Hz, does that change the range of ultrasound?

No.

Ultrasound is relative to human hearing, and is the range of sound above 20,000 Hz.