mastering physics hw 7 ch 21 - superposition

HW 7 Ch 21 SuperpositionDue: 11:59pm on Tuesday, October 27, 2015

To understand how points are awarded, read the Grading Policy for this assignment.

Beat Frequency Ranking Task

An all female guitar septet is getting ready to go on stage. The lead guitarist, Kira,who is always in tune, plucks her low E string and the other sixmembers, sequentially, do the same. Each member records the initial beat frequency between her low E string and Kira's low E string.

Part A

Rank each member on the basis of the frequency of her low E string.

Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1. Beat frequency

Beats result when two sound sources are producing waves with slightly different frequencies, and . The combined sound of the twosources will alternately increase and decrease in loudness with a frequency (the beat frequency, ) equal to the difference in frequencybetween the two sources:

.

Hint 2. Find the frequency of Aiko's E string

Assume that Kira's low E string vibrates with a frequency of 83 . What is the frequency of Aiko's low E string?

ANSWER:

ANSWER:

Correct

Because the beat frequency between Kira's guitar and Diane's guitar is 0 , these guitars play the exact same note and are in tune.

fbeat,i

f1 f2fbeat

= | − |fbeat f1 f2

Hz

86

80

either 86 or 80

HzHz

Hz Hz

Hz

f

Zombie Rae

Stamp

To tune an instrument using beats, more information than just the beat frequency is needed. In addition to recording the initial beat frequency , eachmember, except Diane, also records the change in the frequency (increase or decrease) when they increase the tension in their low E string.

Part B

Rank each member on the basis of the initial frequency of their low E string.


Hint 1. Determine the relationship between tension and beat frequency

When the tension in a guitar string is increased, the frequency of that string also increases. This increase in frequency will change the beatfrequency.If the frequency of a guitar's low E string increased but the beat frequency decreased, this means that increasing the frequency in the stringbrought the sting closer in tune with Kira's low E string. Did the frequency of the guitar string start out below or above the frequency of Kira'sstring?

ANSWER:

Hint 2. Determine the initial frequency of Aiko's E string

Assume that Kira's low E string vibrates with a frequency of 83 . If Aiko increases the tension in her E string, and the beat frequencyincreases, what was the initial frequency of Aiko's string?

Express your answer in hertz.

Hint 1. Determine the possible frequencies of Aiko's string

Given that Kira's low E string vibrates with a frequency of 83 and , what are the two possible frequencies of Aiko's lowE string?

Express your answers in hertz separated by a comma.

ANSWER:

ANSWER:

ANSWER:

fbeat,ifbeat

ff

below

above

Hz

Hz = 3 Hzfbeat

, Hz Hz

= fAiko Hz

Correct

Introduction to TwoSource Interference

Learning Goal:

To gain an understanding of constructive and destructive interference.

Consider two sinusoidal waves (1 and 2) of identical wavelength , period , and maximum amplitude . A snapshot of one of these waves taken at acertain time is displayed in the figure below. Let and represent thedisplacement of each wave at position at time . If these waves were to be in the samelocation ( ) at the same time, they would interfere with one another. This would result in a singlewave with a displacement given by

.This equation states that at time the displacement of the resulting wave at position isthe algebraic sum of the displacements of the waves 1 and 2 at position at time . When themaximum displacement of the resulting wave is less than the amplitude of the original waves,that is, when , the waves are said to interfere destructively because the result issmaller than either of the individual waves. Similarly, when , the waves are said tointerfere constructively because the resulting wave is larger than either of the individual waves.Notice that .

Part A

To further explore what this equation means, consider four sets of identical waves that move in the +x direction. A photo is taken of each wave at time and is displayed in the figures below.

Rank these sets of waves on the basis of the maximum amplitude of the wave that results from the interference of the two waves in each set.

Rank from largest amplitude on the left to smallest amplitude on the right. To rank items as equivalent, overlap them.

ANSWER:

λ T A(x, t)y1 (x, t)y2

x tx

y(x, t)

y(x, t) = (x, t) + (x, t)y1 y2t y(x, t) x

x t

< Aymax> Aymax

0 ≤ ≤ 2Aymax

t

Zombie Rae

Stamp

Correct

When identical waves interfere, the amplitude of the resulting wave depends on the relative phase of the two waves. As illustrated by the set ofwaves labeled A, when the peak of one wave aligns with the peak of the second wave, the waves are in phase and produce a wave with thelargest possible amplitude. When the peak of one wave aligns with the trough of the other wave, as illustrated in Set C, the waves are out ofphase by and produce a wave with the smallest possible amplitude, zero!

Part B

Now consider a wave which is paired with seven other waves into seven pairs. The two waves in each pairing are identical, except that one of them isshifted relative to the other in the pair by the distance shown:

A. B. C. D. E. F. G.

Identify which of the seven pairs will interfere constructively and which will interfere destructively. Each letter represents a pair of waves.

Enter the letters of the pairs that correspond to constructive interference in alphabetical order and the letters of the pairs that correspond topairs that interfere destructively in alphabetical order separated by a comma. For example if pairs A, B and D interfere constructively andpairs C and F interfere destructively enter ABD,CF.

ANSWER:

λ/2

−(1/2)λ2λ−5λ(3/2)λ0(17/2)λ(6/2)λ

Zombie Rae

Stamp

Correct

Do you notice a pattern? When the phase difference between two identical waves can be written as , where ,the waves will interfere constructively. When the phase difference can be expressed as , where , the waveswill interfere destructively.

Consider what water waves look like when you throw a rock into a lake. These waves start at the point where the rock entered the water and travel out in alldirections. When viewed from above, these waves can be drawn as shown, where the solid linesrepresent wave peaks and troughs are located halfway between adjacent peaks.

Part C

Now look at the waves emitted from two identical sources (e.g., two identical rocks that fall into a lake at the same time). The sources emit identicalwaves at the exact same time.

Identify whether the waves interfere constructively or destructively at each point A to D.

For points A to D enter either c for constructive or d for destructive interference. Forexample if constructive interference occurs at points A, C and D, and destructiveinterference occurs at B, enter cdcc.

Hint 1. How to approach the problem

Recall that constructive interference occurs when the two waves are in phase when they interfere, so that the peak (or trough) of one wavealigns with the peak (or trough) of the other wave. Destructive interference occurs when waves are out of phase so that the peak of onewave aligns with the trough of the other wave.

Study the picture to find where each type of interference occurs.

ANSWER:

Correct

Each wave travels a distance or from its source to reach Point B. Since the distance between consecutive peaks is equal to , from the picture youcan see that Point B is away from Source 1 and away from Source 2. The pathlength difference, , is the difference in the distance each wavetravels to reach Point B:

.

BCEG,ADF

λmC = 0, ±1, ±2, ±3, …mC(λ/2)mD = ±1, ±3, ±5, …mD

λ/2

ccdd

d1 d2 λ2λ 3λ ΔdB

Δ = − = 2λ − 3λ = −1λdB d1 d2

Part D

What are the pathlength differences at Points A, C, and D (respectively, , , and )?

Enter your answers numerically in terms of separated by commas. For example, ifthe pathlength differences at Points A, C, and D are , , and , respectively,enter 4,.5,1.

ANSWER:

CorrectKnowing the pathlength difference helps to confirm what you found in Part C. When the pathlength difference is , where

, the waves interfere constructively. When the pathlength difference is , where , the waves interfere destructively.

Part E

What are the pathlength differences at Points L to P?

Enter your answers numerically in terms of separated by commas. For example, ifthe pathlength differences at Points L, M, N, O, and P are , , , , and ,respectively, enter 5,2,1.5,1,6.

ANSWER:

ΔdA ΔdC ΔdD

λ4λ λ/2 λ

, , = 0,1.5,0.5 , , ΔdA ΔdC ΔdD λ λ λ

λmC= 0, ±1, ±2, ±3, …mC (λ/2)mD= ±1, ±3, ±5, …mD

λ5λ 2λ λ3

2 λ 6λ

Correct

Every point along the line connecting Points L to P corresponds to a pathlength difference . This means that at every point along thisline, waves from the two sources interfere constructively.

The figure below shows two other lines of constructive interference: One corresponds to a pathlength difference , and the othercorresponds to . It should make sense that the line halfway between the two sources corresponds to a pathlength difference of zero,since any point on this line is equally far from each source. Notice the symmetry about the line of the and the lines.

A similar figure can be drawn for the lines of destructive interference. Notice that the pattern of lines is still symmetric about the line halfwaybetween the two sources; however, the lines along which destructive interference occurs fall midway between adjacent lines of constructiveinterference.

Fundamental Wavelength and Frequency Ranking Task

A combination work of art/musical instrument is illustrated. Six pieces of identical piano wire (cutto different lengths) are hung from the same support, and masses are hung from the free end ofeach wire. Each wire is 1, 2, or 3 units long, and each supports 1, 2, or 4 units of mass. Themass of each wire is negligible compared to the total mass hanging from it. When a strongbreeze blows, the wires vibrate and create an eerie sound.

, , , , = 1,1,1,1,1 , , , , ΔdL ΔdM ΔdN ΔdO ΔdP λ λ λ λ λ

Δd = λ

Δd = −λΔd = 0

Δd = 0 Δd = λ Δd = −λ

Part A

Rank each wiremass system on the basis of its fundamental wavelength.


Hint 1. Identify the fundamental wavelength

For any wave on a wire of length , the fundamental wavelength is the longest wave that "fits" on the wire, with a node at both fixed ends.Recall that a node is a place where the wave has zero amplitude. (Although the bottom of each wire is not absolutely fixed, the inertia of themass hanging from the end causes it to remain relatively fixed compared to the vibration of the wire.)

What is the longest wave that can fit on a wire of length ?

ANSWER:

ANSWER:

Correct

Part B

Rank each wiremass system on the basis of its wave speed.


Hint 1. Factors that determine wave speed

The speed of a wave on a wire depends on only two factors, the tension in the wire and the linear mass density of the wire:

.

Basically, light wires under high tension carry very fast waves whereas heavy wires under low tension carry slower moving waves.

L

L

4L

2L

L

L/2

L/4

v T μ

v = Tμ

−−√

Zombie Rae

Stamp

The piano wires in this question all have identical linear mass density.

Hint 2. Tension in the wires

Since the masses hanging from each wire are in equilibrium, the tension in each wire is equal to the total weight of the masses hanging from it.

ANSWER:

Correct

Part C

Rank each wiremass system on the basis of its fundamental frequency.


Hint 1. Find an equation for the fundamental frequency

Let be the length of the wire, the tension in the wire, and the linear mass density of the wire. Combine the result for the fundamentalwavelength,

,and the result for determining the wave speed,

,

with the relationship among these quantities and frequency ,

,to yield an equation for the fundamental frequency .

Enter an expression for the fundamental frequency in terms of some or all of the variables , , and .

ANSWER:

ANSWER:

L T μ

λ = 2L

v = Tμ

−−√f

f = v

λ

f0

L T μ

= f0

Zombie Rae

Stamp

Correct

± Harmonics of a Piano Wire

A piano tuner stretches a steel piano wire with a tension of 765 . The steel wire has a length of 0.700 and a mass of 5.25 .

Part A

What is the frequency of the string's fundamental mode of vibration?

Express your answer numerically in hertz using three significant figures.


Find the mass per unit length for the wire. Then apply the equation for the fundamental frequency of the wire.

Hint 2. Find the mass per unit length

What is the mass per unit length for the wire?

Express your answer numerically in kilograms per meter using three significant figures.

ANSWER:

Hint 3. Equation for the fundamental frequency of a string under tension

The fundamental frequency of a string under tension is given by

,

where is the length of the string, is the tension in the string, and is its mass per unit length.

ANSWER:

N m g

f1

μ

= μ kg/m

=f11

2LFμ

−−√L F μ

= 228 f1 Hz

Zombie Rae

Stamp

Correct

Part B

What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to = 16 kHz?

Express your answer exactly.

Hint 1. Harmonics of a string

The harmonics of a string are given by , where is the th harmonic of a string with fundamental frequency . Be careful if youget a noninteger answer for , as harmonics can only be integer multiples of the fundamental frequency.

ANSWER:

Correct

When solving this problem, you may have found a noninteger value for , but harmonics can only be integer multiples of the fundamentalfrequency.

Introduction to Wind Instruments

The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside theinstrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of openand closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an openopenpipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an openclosed pipe because themouthpiece of the clarinet is almost completely closed by the reed.

Throughout the problem, take the speed of sound in air to be 343 .

Part A

Consider a pipe of length 80.0 open at both ends. What is the lowest frequency of the sound wave produced when you blow into the pipe?



The lowest frequency that can be produced is the fundamental frequency of the standing wave in the pipe.

Hint 2. Frequencies of standing waves in an openopen pipe

The frequencies possible in an openopen pipe of length are given by the formula

,

where is the speed of sound in the air.

ANSWER:

Correct

If your pipe were a flute, this frequency would be the lowest note that can be produced on that flute. This frequency is also known as thefundamental frequency or first harmonic.

Part B

n f

= nfn f1 fn n f1n

= 70n

n

m/s

cm f

L

= m , m = 1,2,3,4,…fmv

2L

v

= 214 f Hz

A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of thesound wave generated in the pipe is now


Since the hole opens the pipe to the pressure of the surrounding air, the standing wave created in the pipe has an antinode near the hole. Inother words, the presence of a hole in the pipe reduces the length of the column of air that can oscillate in the pipe.

Consider the formula used in Part A and use the fact that the length of the vibrating column of air is now shorter.

ANSWER:

Correct

By opening successive holes closer and closer to the opening used to blow air into the pipe, the pipe can be made to produce sound at higher andhigher frequencies. This is what flutists do when they open the tone holes on the flute.

Part C

If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound thatcan be produced in the pipe?



Repeat the same calculation you did in Part A, this time using half the length of the pipe.

ANSWER:

Correct

Part D

What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a holehalfway down its length?


Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonicsthat have an antinode halfway down the pipe will still be present.

ANSWER:

Correct

Part E

the same as before.

lower than before.

higher than before.

f ′

= 429 f ′ Hz

Only the odd multiples of the fundamental frequency

Only the even multiples of the fundamental frequency

All integer multiples of the fundamental frequency

f

What length of openclosed pipe would you need to achieve the same fundamental frequency as the openopen pipe discussed in Part A?

Hint 1. Frequencies on an openclosed pipe

The frequencies possible in an openclosed pipe of length are given by

,

where is the speed of sound in the air.

ANSWER:

Correct

Part F

What is the frequency of the first possible harmonic after the fundamental frequency in the openclosed pipe described in Part E?



Recall that possible frequencies of standing waves that can be generated in an openclosed pipe include only odd harmonics. Then the firstpossible harmonic after the fundamental frequency is the third harmonic.

ANSWER:

Correct

± Standing Waves on a Guitar String

Learning Goal:

To understand standing waves, including calculation of and , and to learn the physical meaning behind some musical terms.

The columns in the figure show the instantaneous shape of a vibrating guitar string drawn every 1. The guitar string is 60 long.

The left column shows the guitar string shape as a sinusoidal traveling wave passes through it.Notice that the shape is sinusoidal at all times and specific features, such as the crest indicatedwith the arrow, travel along the string to the right at a constant speed.

The right column shows snapshots of the sinusoidal standing wave formed when this sinusoidaltraveling wave passes through an identically shaped wave moving in the opposite direction on thesame guitar string. The string is momentarily flat when the underlying traveling waves are exactlyout of phase. The shape is sinusoidal with twice the original amplitude when the underlying wavesare momentarily in phase. This pattern is called a standing wave because no wave features traveldown the length of the string.

f

L

= m , m = 1,3,5,…fmv

4L

v

Half the length of the openopen pipe

Twice the length of the openopen pipe

Onefourth the length of the openopen pipe

Four times the length of the openopen pipe

The same as the length of the openopen pipe

f ′′

= 643 f ′′ Hz

λ f

ms cm

Standing waves on a guitar string form when waves traveling down the string reflect off a point where the string is tied down or pressed against thefingerboard. The entire series of distortions may be superimposed on a single figure, like this ,indicating different moments in time using traces of different colors or line styles.

Part A

What is the wavelength of the standing wave shown on the guitar string?

Express your answer in centimeters.

Hint 1. Identify the wavelength of a sinusoidal shape

The wavelength of a sinusoidal shape is the distance from a given feature to the next instance of that same feature. Wavelengths are usuallymeasured from one peak to the next peak. What is the wavelength of this sinusoidal pattern?


ANSWER:

ANSWER:

Correct

Nodes are locations in the standing wave pattern where the string doesn't move at all, and hence the traces on the figure intersect. In betweennodes are antinodes, where the string moves up and down the most.

This standing wave pattern has three antinodes, at , , and . The pattern also has four nodes, at , , , and . Notice that the spacing between adjacent antinodes is only half of one wavelength, not one full wavelength. The same is true

for the spacing between adjacent nodes.

This figure shows the first three standing wave patterns that fit on any string with length tied down at both ends. A pattern's number is the number ofantinodes it contains. The wavelength of the th pattern is denoted . The th pattern has halfwavelengths along the length of the string, so

.Thus the wavelength of the th pattern is

λ

λ

= λ cm

= 40 λ cm

x = 10 cm 30 cm 50 cm x = 0 cm 20 cm40 cm 60 cm

L nn λn n n

n = Lλn

2n

= 2L

.

Part B

What is the wavelength of the longest wavelength standing wave pattern that can fit on this guitar string?



Look at the figure to determine the pattern number for the longest wavelength pattern, then calculate its wavelength . Recall that the guitarstring is 60 long.

Hint 2. Determine for the longest wavelength pattern

What is the pattern number for the longest wavelength standing wave pattern?

ANSWER:

ANSWER:

Correct

This longest wavelength pattern is so important it is given a special name—the fundamental. The wavelength of the fundamental is always givenby for a string that is held fixed at both ends.

Waves of all wavelengths travel at the same speed on a given string. Traveling wave velocity and wavelength are related by

,

where is the wave speed (in meters per second), is the wavelength (in meters), and is the frequency [in inverse seconds, also known as hertz ( )].

Since only certain wavelengths fit properly to form standing waves on a specific string, only certain frequencies will be represented in that string's standingwave series. The frequency of the th pattern is

.

Note that the frequency of the fundamental is , so can also be thought of as an integer multiple of : .

Part C

The frequency of the fundamental of the guitar string is 320 . At what speed do waves move along that string?

Express your answer in meters per second.

=λn2Ln

n λncm

n

n

=

1

2

3

some other integer

n

= 120 λ1 cm

= 2Lλ1

v

v = λf

v λ f Hz

n

= = = nfnv

λn

v

(2L/n)v

2L

= v/(2L)f1 fn f1 = nfn f1

Hz v


The velocity of waves on the string equals the product of the wavelength and the corresponding frequency for any standing wave pattern:.

You know enough about the fundamental to calculate .

ANSWER:

Correct

Notice that these transverse waves travel slightly faster than the speed of sound waves in air, which is about 340 .

We are now in a position to understand certain musical terms from a physics perspective.The standing wave frequencies for this string are , , , etc. This set of frequencies is called aharmonic series and it contains common musical intervals such as the octave (in which the ratio of frequencies of the two notes is 2:1) and the perfect fifth(in which the ratio of frequencies of the two notes is 3:2). Here is one octave above , is a perfect fifth above , and so on. Standing wavepatterns with frequencies higher than the fundamental frequency are called overtones. The pattern is called the first overtone, the pattern iscalled the second overtone, and so on.

Part D

How does the overtone number relate to the standing wave pattern number, previously denoted with the variable ?

ANSWER:

Correct

The overtone number and the pattern number are easy to confuse but they differ by one. When referring to a standing wave pattern using anumber, be explicit about which numbering scheme you are using.

When you pluck a guitar string, you actually excite many of its possible standing waves simultaneously. Typically, the fundamental is the loudest, so thatis the pitch you hear. However, the unique mix of the fundamental plus overtones is what makes a guitar sound different from a violin or a flute, even ifthey are playing the same note (i.e., producing the same fundamental). This characteristic of a sound is called its timbre (rhymes with amber).

A sound containing just a single frequency is called a pure tone. A complex tone, in contrast, contains multiple frequencies such as a fundamental plussome of its overtones. Interestingly enough, it is possible to fool someone into identifying a frequency that is not present by playing just its overtones. Forexample, consider a sound containing pure tones at 450 , 600 , and 750 . Here 600 and 750 are not integer multiples of 450 , so 450

would not be considered the fundamental with the other two as overtones. However, because all three frequencies are consecutive overtones of 150 a listener might claim to hear 150 , over an octave below any of the frequencies present. This 150 is called a virtual pitch or a missing

fundamental.

Part E

A certain sound contains the following frequencies: 400 , 1600 , and 2400 . Select the best description of this sound.

Hint 1. How to identify a fundamental within a series of frequencies

All frequencies in a harmonic series are integer multiples of the fundamental frequency. The lowest frequency listed will be the fundamental, butonly if all the other frequencies are integer multiples of it.

ANSWER:

v = = = …λ1 f1 λ2 f2

v

= 384 v m/s

m/s

= 320 Hzf1 = 2 = 640 Hzf2 f1 = 3 = 960 Hzf3 f1

f2 f1 f3 f2n = 2 n = 3

n

There is no strict relationship between overtone number and pattern number.

overtone number = pattern numberovertone number = pattern number + 1overtone number = pattern number − 1

Hz Hz Hz Hz Hz HzHzHz Hz Hz

Hz Hz Hz

Correct

These concepts of fundamentals and overtones can be applied to other types of musical instruments besides string instruments. Hollowtubeinstruments, such as brass instruments and reed instruments, have standing wave patterns in the air within them. Percussion instruments, suchas bells and cymbals, often exhibit standing wave vibrations in the solid material of their bodies. Even the human voice can be analyzed this way,with the fundamental setting the pitch of the voice and the presence or absence of overtones setting the unique vowel or consonant beingsounded.

Normal Modes and Resonance Frequencies

Learning Goal:

To understand the concept of normal modes of oscillation and to derive some properties of normal modes of waves on a string.

A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general there are an infinite numberof such modes, each one with a distinctive frequency and associated pattern of oscillation.

Consider an example of a system with normal modes: a string of length held fixed at both ends, located at and . Assume that waves onthis string propagate with speed . The string extends in the x direction, and the waves are transverse with displacement along the y direction.

In this problem, you will investigate the shape of the normal modes and then their frequency.

The normal modes of this system are products of trigonometric functions. (For linear systems, the time dependance of a normal mode is always sinusoidal,but the spatial dependence need not be.) Specifically, for this system a normal mode is described by

Part A

The string described in the problem introduction is oscillating in one of its normal modes. Which of the following statements about the wave in thestring is correct?

Hint 1. Normal mode constraints

The key constraint with normal modes is that there are two spatial boundary conditions, and , which correspond to thestring being fixed at its two ends.

ANSWER:

Correct

Part B

Which of the following statements are true?

ANSWER:

This is a pure tone.

This is a complex tone with a fundamental of 400 , plus some of its overtones.

This is a complex tone with a virtual pitch of 800 .

These frequencies are unrelated, so they are probably pure tones from three different sound sources.

HzHz

fi

L x = 0 x = Lv

(x,t) = sin(2π )sin(2π t).yi Aixλi

fi

(0, t) = 0yi (L, t) = 0yi

The wave is traveling in the +x direction.

The wave is traveling in the x direction.

The wave will satisfy the given boundary conditions for any arbitrary wavelength .

The wavelength can have only certain specific values if the boundary conditions are to be satisfied.

The wave does not satisfy the boundary condition .

λi

λi

(0; t) = 0yi

Correct

The key factor producing the normal modes is that there are two spatial boundary conditions, and , that are satisfiedonly for particular values of .

Part C

Find the three longest wavelengths (call them , , and ) that "fit" on the string, that is, those that satisfy the boundary conditions at and . These longest wavelengths have the lowest frequencies.

Express the three wavelengths in terms of . List them in decreasing order of length, separated by commas.


The nodes of the wave occur where

.

This equation is trivially satisfied at one end of the string (with ), since .

The three largest wavelengths that satisfy this equation at the other end of the string (with ) are given by , where the arethe three smallest, nonzero values of that satisfy the equation .

Hint 2. Values of that satisfy

The spatial part of the normal mode solution is a sine wave. Find the three smallest (nonzero) values of (call them , , and ) thatsatisfy .

Express the three nonzero values of as multiples of . List them in increasing order, separated by commas.

ANSWER:

Hint 3. Picture of the normal modes

Consider the lowest four modes of the string as shown in the figure.

The letter N is written over each of the nodes defined as places where the string does not move. (Note that there are nodes in addition to thoseat the end of the string.) The letter A is written over the antinodes, which are where the oscillation amplitude is maximum.

ANSWER:

The system can resonate at only certain resonance frequencies and the wavelength must be such that .

must be chosen so that the wave fits exactly on the string.

Any one of or or can be chosen to make the solution a normal mode.

fi λi (0; t) = (L; t) = 0yi yi

Ai

Ai λi fi

(0, t) = 0yi (L, t) = 0yi

λi

λ1 λ2 λ3 x = 0x = L

L

sin(2π ) = 0xλi

x = 0 sin(0) = 0

x = L 2πL/ =λi zi ziz sin(z) = 0

z sin(z) = 0

z z1 z2 z3sin(z) = 0

z π

, , = z1 z2 z3

, , = λ1 λ2 λ3 2L,L, L23

Correct

The procedure described here contains the same mathematics that leads to quantization in quantum mechanics.

Part D

The frequency of each normal mode depends on the spatial part of the wave function, which is characterized by its wavelength .

Find the frequency of the ith normal mode.

Express in terms of its particular wavelength and the speed of propagation of the wave .

Hint 1. Propagation speed for standing waves

Your expression will involve , the speed of propagation of a wave on the string. Of course, the normal modes are standing waves and do nottravel along the string the way that traveling waves do. Nevertheless, the speed of wave propagation is a physical property that has a welldefined value that happens to appear in the relationship between frequency and wavelength of normal modes.

Hint 2. Use what you know about traveling waves

The relationship between the wavelength and the frequency for standing waves is the same as that for traveling waves and involves the speedof propagation .

ANSWER:

CorrectThe frequencies are the only frequencies at which the system can oscillate. If the string is excited at one of these resonance frequencies it willrespond by oscillating in the pattern given by , that is, with wavelength associated with the at which it is excited. In quantummechanics these frequencies are called the eigenfrequencies, which are equal to the energy of that mode divided by Planck's constant . In SIunits, Planck's constant has the value

Part E

Find the three lowest normal mode frequencies , , and .

Express the frequencies in terms of , , and any constants. List them in increasing order, separated by commas.

ANSWER:

CorrectNote that, for the string, these frequencies are multiples of the lowest frequency. For this reason the lowest frequency is called the fundamentaland the higher frequencies are called harmonics of the fundamental. When other physical approximations (for example, the stiffness of the string)are not valid, the normal mode frequencies are not exactly harmonic, and they are called partials. In an acoustic piano, the highest audible normalfrequencies for a given string can be a significant fraction of a semitone sharper than a simple integer multiple of the fundamental. Consequently,the fundamental frequencies of the lower notes are deliberately tuned a bit flat so that their higher partials are closer in frequency to the highernotes.

Two Identical Pulses along a String

Two identical pulses are moving in opposite directions along a stretched string that has one fixed end and the other movable, as shown in .

Above each pulse a green arrow indicates the direction of motion of the pulse.

λi

fi

fi λi v

v

v

= fi

vλi

fi(x, t)yi λi fi

hh = 6.63 × J ⋅ s.10−34

f1 f2 f3

L v

, , = f1 f2 f3 , ,v2L

vL

3v2L

Part A

The two pulses reflect off the boundaries of the string, and at some later time, they pass through the middle of the string and interfere.

Below are six different sequences of snapshots taken as the two pulses meet in the middle of the string. Time increases from top to bottom in eachsequence. Which sequence correctly represents the displacement of the string as the pulses interfere?

Hint 1. Superposition of waves

When two pulses move through the same section of a string and overlap, you need to add the displacements of the individual pulses at eachpoint to determine the displacement of that section of the string. Note that the pulses pass through the same section of the string only aftereach reaches a boundary and is reflected.

Hint 2. Reflection of waves at the boundary

When each pulse reaches one end of the string, a reflected pulse forms. Which of the following sketches represents the reflected pulses?

Hint 1. Reflection of a wave pulse at a fixed end

When a wave pulse reaches the end of the string, it is reflected and travels in the opposite direction from the initial pulse. If the end ofthe string is tied to a support and cannot move, the pulse becomes inverted upon reflection.

Hint 2. Reflection of a wave pulse at a free end

If the end of the string is free to move, the direction of displacement of the reflected pulse is the same as that of the initial pulse.

ANSWER:

ANSWER:

Correct

Part B

Consider the point where the two pulses start to overlap, point O in . What is thedisplacement of point O as these pulses interfere?


At any time, the displacement of point O is given by the algebraic sum of the displacements of the two pulses at that point.

Hint 2. Displacement of point O if only one pulse were present

Let be a short interval of time after the two pulses have begun to overlap. If only the pulse coming from the right were present, thedisplacement of point O at time would be . What would its displacement be if only the other pulse were present?

Express your answer in terms of .

Hint 1. Inverted pulses

Keep in mind that the two pulses are identical but inverted; thus their displacements are equal magnitude but opposite.

ANSWER:

a

b

c

d

a

b

c

d

e

f

ΔtΔt +y0

y0

ANSWER:

Correct

Part C

Why does destructive interference occur when the two pulses overlap instead of constructive interference?

ANSWER:

Correct

Part D

As the pulses interfere destructively there is a point in time when the string is perfectly straight. Which of the following statements is true at thismoment?

ANSWER:

Correct

When you apply a force to a string to produce a pulse, work is done on the string and energy is stored in it. As the pulse travels along the string,this energy is transported. In particular, this energy is converted back and forth between kinetic and potential energy as the particles in the stringoscillate. When destructive interference occurs and the string is momentarily straight, it does not mean that the string has zero energy. Rather,the energy transported by the pulses has been completely converted into kinetic energy. A short time later, the pulses will be reconstituted.

Nodes of a Standing Wave (Sine)

The nodes of a standing wave are points at which the displacement of the wave is zero at all times. Nodes are important for matching boundary conditions,for example, that the point at which a string is tied to a support has zero displacement at all times (i.e., the point of attachment does not move).Consider a standing wave, where represents the transverse displacement of a string that extends along the x direction. Here is a common mathematicalform for such a wave:

where is the maximum transverse displacement of the string (the amplitude of the wave), which is assumed to be nonzero, is the wave number, isthe angular frequency of the wave, and is time.

Part A

Which one of the following statements about such a wave as described in the problem introduction is correct?

ANSWER:

It varies with time.

It remains zero.

It depends on the (identical) amplitude of the pulses.

It is zero only when the pulses begin to overlap.

because the pulses are traveling in opposite directions

because a pulse is inverted upon reflection

because the pulses are identical and cancel each other out

because constructive interference occurs only when the pulses have the same amplitude

The energy of the string is zero.

The string is not moving either up or down.

The string has only kinetic energy.

y

y(x, t) = A sin(kx) sin(ωt),A k ω

t

Correct

Each part of the string oscillates with the same phase, so the wave does not appear to move left or right; rather, it oscillates up and down only.

Part B

At time , what is the displacement of the string ?

Express your answer in terms of , , and other previously introduced quantities.

ANSWER:

Correct

Part C

What is the displacement of the string as a function of at time , where is the period of oscillation of the string?

Express the displacement in terms of , , , and other constants; that is, evaluate and substitute it in the expression for .

Hint 1. Find the period in terms of

What is the period in terms of the angular frequency ?

Express your answer in terms of and constants like .

ANSWER:

Hint 2.

What is the value of the timedependent trigonometric function at the time ?

Express your answer in terms of the previously introduced quantities.

ANSWER:

ANSWER:

Correct

Part D

At which three points , , and closest to but with will the displacement of the string be zero for all times? These are the

This wave is traveling in the direction.

This wave is traveling in the direction.

This wave is oscillating but not traveling.

This wave is traveling but not oscillating.

+x

−x

t = 0 y(x, 0)

A k

= 0y(x, 0)

x T/4 T

A x k ω T4

y(x, t)

ω

T ω

ω π

= T

sin(ω )T4

sin(ωt) T/4

= sin(ω )T4

= y(x, )T4

Asin(kx)

x1 x2 x3 x = 0 x > 0 y(x, t)

first three nodal points.

Express the first three nonzero nodal points in terms of the wavelength . List them in increasing order, separated by commas. You shouldenter only the factors that multiply . Do not enter for each one.

Hint 1. What is equal to zero?

If , then . Because , either or (or both) if the displacement of the string isto be zero for all times. The frequency is a constant for a given wave and may take any positive value. Therefore, cannot equalzero for all times. Hence you need to find solutions to the equation .

Express the first three nonzero values of (call them , , ) for which as multiples of . List them inincreasing order, separated by commas. You should not type each time. You should enter only the factors that multiply .

ANSWER:

Hint 2. What is in terms of ?

What is the wavelength of the standing wave in terms of ?

Express your answer in terms of and constants like .

ANSWER:

Hint 3. Putting it all together

The nodes of the wave are points for which . If you substitute using the previous results, you will find the multiples of for which .

ANSWER:

Correct

Problem 21.9

Part A

What are the three longest wavelengths for standing waves on a 222 long string that is fixed at both ends?

Enter your answers in descending order separated by commas.

ANSWER:

Correct

Part B

If the frequency of the secondlongest wavelength is 60 , what is the frequency of the thirdlongest wavelength?

Express your answer using two significant figures.

ANSWER:

λλ λ

y(x, t) = 0 A sin(kx) sin(ωt) = 0 A > 0 sin(kx) = 0 sin(ωt) = 0ω t sin(ωt)

sin(kx) = 0

kx (kx)1 (kx)2 (kx)3 y(x, t) = 0 ππ π

, , = (kx)1 (kx)2 (kx)3 π

λ k

λ k

k π

= λ

sin(kx) = 0 = αλxi αλ sin(kx) = 0

, , = 0.500,1,1.50 x1 x2 x3 λ

cm

, , = 4.44,2.22,1.48 λ1 λ2 λ3 m

Hz

Correct

Problem 21.11

A heavy piece of hanging sculpture is suspended by a 90 long, 5.0 steel wire. When the wind blows hard, the wire hums at its fundamental frequencyof 61 .

Part A

What is the mass of the sculpture?

Express your answer to two significant figures and include the appropriate units.

ANSWER:

Correct

Problem 21.31

A 2.0 long string vibrates at its secondharmonic frequency with a maximum amplitude of 3.0 . One end of the string is at .

Part A

Find the oscillation amplitude at


ANSWER:

Correct

Part B



ANSWER:

Correct

Part C



ANSWER:

= 90 f3 Hz

cm gHz

= 6.8 m kg

m cm x = 0 cm

x = 10 cm.

= = 0.93 a(x = 10 cm) cm

x = 20 cm.

= = 1.8 a(x = 20 cm) cm

x = 30 cm.

= = 2.4 a(x = 30 cm) cm

Correct

Part D



ANSWER:

Correct

Part E



ANSWER:

Correct

Score Summary:Your score on this assignment is 50.0%.You received 6 out of a possible total of 12 points.

x = 40 cm.

= = 2.9 a(x = 40 cm) cm

x = 50 cm.

= = 3.0 a(x = 50 cm) cm

mastering physics hw 7 ch 21 - superposition

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