Periodic Motion, Waves & Sound

Question 1

Where is the amplitude of the waveform shown below?

Answer 1

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.

Question 2

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

Answer 2

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.

Question 3

What is the period of the waveform shown below?

Answer 3

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.

Question 4

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

Answer 4

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.

Question 5

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

Answer 5

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.

Question 6

What is the frequency of a waveform?

Answer 6

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.

Question 7

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

Answer 7

The angular frequency ω is calculated as:

ω = 2πf

The units of ω are radian/s.

Question 8

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

Answer 8

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.

Question 9

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

Answer 9

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.

Question 10

Define:

the phase of a waveform

Answer 10

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.

Question 11

What is the phase difference between two waveforms?

Answer 11

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.

Question 12

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

• degrees?
• radians?

Answer 12

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

• 180º
• π radians

Question 13

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

• degrees?
• radians?

Answer 13

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

• 90º
• π/2 radians

Question 14

Define:

Hooke’s Law for spring force

Answer 14

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

Question 15

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?

Answer 15

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

Question 16

Define:

simple harmonic motion (SHM)

Answer 16

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.

Question 17

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

Answer 17

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.

Question 18

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

Answer 18

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.

Question 19

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

Answer 19

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.

Question 20

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

Answer 20

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.

Question 21

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

Answer 21

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

Question 22

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?

Answer 22

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.

Question 23

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?

Answer 23

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.

Question 24

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?

Answer 24

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.

Question 25

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

Answer 25

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

U = ½kx²

Where k is the force constant of the spring.

Question 26

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

Answer 26

The maximum value of the potential energy for a mass oscillating on a spring is U_max = ½kA².

Remember: the potential energy for the system for any value x of the spring’s extension is ½kx². The maximum value of x is A, the amplitude. Plugging in A for x gives the maximum value of U.

Question 27

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

Answer 27

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

KE = ½mv²

Where:
m = the mass of the object oscillating
v = the velocity of the object at that moment

Question 28

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

Answer 28

KE_max = ½kA².

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.

Question 29

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

Answer 29

v_max = A(k/m)^½ = Aω.

The maximum value of the kinetic energy is ½kA². Setting this equal to ½mv² and solving for v yields the final answer.

Question 30

What is the angular frequency of an oscillating pendulum?

Answer 30

f = (g/L)^½

Where:
f = the pendulum oscillation frequency in Hz
g = gravitational acceleration in m/s²
L = length of the string of the pendulum

Question 31

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?

Answer 31

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.

Question 32

Define:

Transverse wave

Answer 32

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

Question 33

What are some classic examples of transverse waves?

Answer 33

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.

Question 34

Define:

Longitudinal wave

Answer 34

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

Question 35

What are some classic examples of longitudinal waves?

Answer 35

Logitudinal waves include:

• sound waves
• stretched slinky waves (spring waves)

Question 36

Calculate the period of the wave shown below:

Answer 36

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.

Question 37

Calculate the frequency of the wave below:

Answer 37

The frequency is .2 Hz.

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

Question 38

Calculate the amplitude of the wave below:

Answer 38

The amplitude is 17 meters.

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

Question 39

Calculate the wavelength of the wave below:

Answer 39

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.

Question 40

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

Answer 40

1 m/s

wave speed = λ * f = 5 (1/5) = 1

Question 41

Define:

Intensity of sound

Answer 41

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/m².

Question 42

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

Answer 42

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

I α A²

Question 43

Define:

Constructive interference

Answer 43

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.

Question 44

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

Answer 44

Totally constructive interference.

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

Question 45

Define:

Destructive interference

Answer 45

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.

Question 46

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

Answer 46

Totally destructive interference.

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

Question 47

Define:

resonance of an oscillating system

Answer 47

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.

Question 48

Define:

standing wave

Answer 48

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.

Question 49

Define:

Nodes

Answer 49

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.

Question 50

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

Answer 50

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.

Question 51

Define:

antinode

Answer 51

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.

Question 52

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

Answer 52

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.

Question 53

Define:

beat frequency

Answer 53

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.

Question 54

When two waves with frequencies f₁ and f₂ interfere, what is frequency of the beats?

Answer 54

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

f_beat = |f₁ - f₂|

Question 55

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?

Answer 55

The new beat frequency will be exactly the same.

Old beat frequency = |220-224| = 4
New beat frequency = |232-228| = 4

Question 56

Define:

refraction of a wave

Answer 56

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.

Question 57

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?

Answer 57

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.

Question 58

Define:

diffraction of a wave

Answer 58

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.

Question 59

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?

Answer 59

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.

Question 60

What type of wave is sound?

Answer 60

Longitudinal.

Waves that oscillate in the diection of propagation are longitudinal.

Question 61

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

Answer 61

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.

Question 62

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

Answer 62

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.

Question 63

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

Answer 63

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.

Question 64

Which should have a faster speed of sound: protium (¹₁H) or deuterium (²₁H), assuming the same number of particles per volume?

Answer 64

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.

Question 65

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

Answer 65

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).

Question 66

What is the speed of sound in air at STP?

Answer 66

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.

Question 67

Define and give the units for:

intensity of sound

Answer 67

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

Units of intensity are W/m², and the standard reference value (or baseline value) is 10^-12 W/m².

Question 68

Define:

the relationship between intensity of sound and the decibel scale.

Answer 68

There is a logarithmic relationship between decibels (L), intensity (I’), and baseline intensity (I₀), such that:

L = 10*log(I’ / I₀)

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.

Question 69

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

Answer 69

If the number Z = n x 10^e
then log(Z) = {e}.{n}

Ex: if Z = 6.2 x 10⁴, 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.

Question 70

What decibel value corresponds to an intensity reading of 10^-10 W/m²?

Answer 70

20 dB

L = 10*log(I’ / I₀) = 10*log(10^-10 / 10^-12)
= 10*log(10²) = 10*2 = 20dB

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

Question 71

What decibel value corresponds to an intensity reading of 10² W/m²?

Answer 71

140 dB

L = 10*log(I’ / I₀) = 10*log(10² / 10^-12)
= 10*log(10¹⁴) = 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.

Question 72

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

Answer 72

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.

Question 73

Define and give the formula for:

attenuation of sound

Answer 73

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

Question 74

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

Answer 74

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.

Question 75

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?

Answer 75

3 cm

From: Attenuation = α * l * f
Enamel = 120*2*f
Dentin = 80*l*f

setting equal: 240f = 80lf
240 = 80l
l = 3 cm

Question 76

Define:

Pitch

Answer 76

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.

Question 77

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

Answer 77

465 Hz is the higher pitch.

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

Question 78

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

Answer 78

v_c = f * λ

This relationship is often referred to as the wave formula.

Question 79

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

Answer 79

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.

Question 80

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

Answer 80

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.

Question 81

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

Answer 81

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

Question 82

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?

Answer 82

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

Question 83

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

Answer 83

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

Question 84

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

Answer 84

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

Question 85

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

Answer 85

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

Question 86

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

Answer 86

A node.

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

Question 87

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

Answer 87

Antinodes.

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

Question 88

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

Answer 88

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.

Question 89

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

Answer 89

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).

Question 90

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

Answer 90

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.

Question 91

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

Answer 91

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.

Question 92

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

Answer 92

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.

Question 93

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

Answer 93

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

Question 94

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

Answer 94

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).

Question 95

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

Answer 95

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

Question 96

Define:

Ultrasound

Answer 96

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.

Question 97

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

Answer 97

No.

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