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 six members, sequentially, do the same. Each member records the initial beat frequency f beat,i 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, f 1 and f 2 . The combined sound of the two sources will alternately increase and decrease in loudness with a frequency (the beat frequency, f beat ) equal to the difference in frequency between the two sources: f beat = |f 1 −f | 2 . Hint 2. Find the frequency of Aiko's E string Assume that Kira's low E string vibrates with a frequency of 83 Hz. What is the frequency of Aiko's low E string? ANSWER: 86 Hz 80 Hz either 86 Hz or 80 Hz ANSWER: Correct Because the beat frequency between Kira's guitar and Diane's guitar is 0 Hz, these guitars play the exact same note and are in tune. f To tune an instrument using beats, more information than just the beat frequency is needed. In addition to recording the initial beat frequency f beat,i , each member, except Diane, also records the change in the frequency f beat (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. Rank from largest to smallest. To rank items as equivalent, overlap them. 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 beat frequency. If the frequency f of a guitar's low E string increased but the beat frequency decreased, this means that increasing the frequency in the string brought the sting closer in tune with Kira's low E string. Did the frequency f of the guitar string start out below or above the frequency of Kira's string? ANSWER: below above Hint 2. Determine the initial frequency of Aiko's E string Assume that Kira's low E string vibrates with a frequency of 83 Hz. If Aiko increases the tension in her E string, and the beat frequency increases, 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 Hz and f beat E string? Express your answers in hertz separated by a comma. ANSWER: Hz, Hz ANSWER: f Aiko ANSWER: = Hz , what are the two possible frequencies of Aiko's low = 3 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 λ , period T , and maximum amplitude A. A snapshot of one of these waves taken at a certain time is displayed in the figure below. Let y1 (x, t) and y2 (x, t) represent the displacement of each wave at position x at time t . If these waves were to be in the same location (x ) at the same time, they would interfere with one another. This would result in a single wave with a displacement y(x, t) given by y(x, t) = y (x, t) + y (x, t) . 1 2 This equation states that at time t the displacement y(x, t) of the resulting wave at position x is the algebraic sum of the displacements of the waves 1 and 2 at position x at time t . When the maximum displacement of the resulting wave is less than the amplitude of the original waves, that is, when ymax < A, the waves are said to interfere destructively because the result is smaller than either of the individual waves. Similarly, when ymax > A, the waves are said to interfere constructively because the resulting wave is larger than either of the individual waves. Notice that 0 ≤ ymax ≤ 2A. 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 t 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: 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 of waves 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 the largest 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 of phase by λ/2 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 is shifted relative to the other in the pair by the distance shown: A. −(1/2)λ B. 2λ C. −5λ D. (3/2)λ E. 0 F. (17/2)λ G. (6/2)λ 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 to pairs that interfere destructively in alphabetical order separated by a comma. For example if pairs A, B and D interfere constructively and pairs C and F interfere destructively enter ABD,CF. ANSWER: BCEG,ADF Correct Do you notice a pattern? When the phase difference between two identical waves can be written as mC λ, where mC = 0, ±1, ±2, ±3, …, the waves will interfere constructively. When the phase difference can be expressed as mD (λ/2), where mD = ±1, ±3, ±5, …, the waves will 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 all directions. When viewed from above, these waves can be drawn as shown, where the solid lines represent 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 identical waves 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. For example if constructive interference occurs at points A, C and D, and destructive interference 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 wave aligns with the peak (or trough) of the other wave. Destructive interference occurs when waves are λ/2 out of phase so that the peak of one wave aligns with the trough of the other wave. Study the picture to find where each type of interference occurs. ANSWER: ccdd Correct Each wave travels a distance d 1 or d 2 from its source to reach Point B. Since the distance between consecutive peaks is equal to λ , from the picture you can see that Point B is 2λ away from Source 1 and 3λ away from Source 2. The pathlength difference, Δd B , is the difference in the distance each wave travels to reach Point B: . Δd B = d 1 − d 2 = 2λ − 3λ = −1λ Part D What are the pathlength differences at Points A, C, and D (respectively, Δd A , Δd C , and Δd D )? Enter your answers numerically in terms of λ separated by commas. For example, if the pathlength differences at Points A, C, and D are 4λ , λ/2 , and λ , respectively, enter 4,.5,1. ANSWER: Δd A , Δd C , Δd D = 0,1.5,0.5 λ , λ , λ Correct Knowing the pathlength difference helps to confirm what you found in Part C. When the pathlength difference is mC λ, where m C = 0, ±1, ±2, ±3, …, the waves interfere constructively. When the pathlength difference is m D (λ/2), where m D = ±1, ±3, ±5, …, the waves interfere destructively. Part E What are the pathlength differences at Points L to P? Enter your answers numerically in terms of λ separated by commas. For example, if the pathlength differences at Points L, M, N, O, and P are 5λ , 2λ , λ, λ , and 6λ , 3 2 respectively, enter 5,2,1.5,1,6. ANSWER: Δd L , Δd M , Δd N , Δd O , Δd P = 1,1,1,1,1 λ , λ , λ , λ , λ Correct Every point along the line connecting Points L to P corresponds to a pathlength difference Δd line, waves from the two sources interfere constructively. . This means that at every point along this = λ The figure below shows two other lines of constructive interference: One corresponds to a pathlength difference Δd = −λ, and the other corresponds to Δd = 0. 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 Δd = 0 line of the Δd = λ and the Δd = −λ 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 halfway between the two sources; however, the lines along which destructive interference occurs fall midway between adjacent lines of constructive interference. Fundamental Wavelength and Frequency Ranking Task A combination work of art/musical instrument is illustrated. Six pieces of identical piano wire (cut to different lengths) are hung from the same support, and masses are hung from the free end of each wire. Each wire is 1, 2, or 3 units long, and each supports 1, 2, or 4 units of mass. The mass of each wire is negligible compared to the total mass hanging from it. When a strong breeze blows, the wires vibrate and create an eerie sound. Part A Rank each wiremass system on the basis of its fundamental wavelength. Rank from largest to smallest. To rank items as equivalent, overlap them. Hint 1. Identify the fundamental wavelength For any wave on a wire of length L, 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 the mass 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 L? ANSWER: 4L 2L L L/2 L/4 ANSWER: Correct Part B Rank each wiremass system on the basis of its wave speed. Rank from largest to smallest. To rank items as equivalent, overlap them. Hint 1. Factors that determine wave speed The speed v of a wave on a wire depends on only two factors, the tension T in the wire and the linear mass density μ of the wire: − − v = √ T μ . Basically, light wires under high tension carry very fast waves whereas heavy wires under low tension carry slower moving waves. 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. Rank from largest to smallest. To rank items as equivalent, overlap them. Hint 1. Find an equation for the fundamental frequency Let L be the length of the wire, T the tension in the wire, and μ the linear mass density of the wire. Combine the result for the fundamental wavelength, , λ = 2L and the result for determining the wave speed, − − v = √ T μ , with the relationship among these quantities and frequency f , f = v λ , to yield an equation for the fundamental frequency f 0 . Enter an expression for the fundamental frequency in terms of some or all of the variables L, T , and μ . ANSWER: f0 = ANSWER: Correct ± Harmonics of a Piano Wire A piano tuner stretches a steel piano wire with a tension of 765 N. The steel wire has a length of 0.700 m and a mass of 5.25 g . Part A What is the frequency f 1 of the string's fundamental mode of vibration? Express your answer numerically in hertz using three significant figures. Hint 1. How to approach the problem 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 μ for the wire? Express your answer numerically in kilograms per meter using three significant figures. ANSWER: μ = kg/m Hint 3. Equation for the fundamental frequency of a string under tension The fundamental frequency of a string under tension is given by f1 = 1 2L − − √ F μ , where L is the length of the string, F is the tension in the string, and μ is its mass per unit length. ANSWER: f 1 = 228 Hz Correct Part B What is the number n of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to f = 16 kHz? Express your answer exactly. Hint 1. Harmonics of a string The harmonics of a string are given by f n = nf 1 , where f n is the n th harmonic of a string with fundamental frequency f 1 . Be careful if you get a noninteger answer for n , as harmonics can only be integer multiples of the fundamental frequency. ANSWER: n = 70 Correct When solving this problem, you may have found a noninteger value for n , but harmonics can only be integer multiples of the fundamental frequency. 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 the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an openopen pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an openclosed pipe because the mouthpiece of the clarinet is almost completely closed by the reed. Throughout the problem, take the speed of sound in air to be 343 m/s . Part A Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe? Express your answer in hertz. Hint 1. How to approach the problem 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 L are given by the formula f m =m v 2L , m = 1, 2, 3, 4, …, where v is the speed of sound in the air. ANSWER: f = 214 Hz 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 the fundamental frequency or first harmonic. Part B 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 the sound wave generated in the pipe is now Hint 1. How to approach the problem 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. In other 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: the same as before. lower than before. higher than before. 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 and higher 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 f of the sound that can be produced in the pipe? Express your answer in hertz. Hint 1. How to approach the problem Repeat the same calculation you did in Part A, this time using half the length of the pipe. ANSWER: f ′ = 429 Hz 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 hole halfway down its length? Hint 1. How to approach the problem 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 harmonics that have an antinode halfway down the pipe will still be present. ANSWER: Only the odd multiples of the fundamental frequency Only the even multiples of the fundamental frequency All integer multiples of the fundamental frequency Correct Part E f What length of openclosed pipe would you need to achieve the same fundamental frequency f as the openopen pipe discussed in Part A? Hint 1. Frequencies on an openclosed pipe The frequencies possible in an openclosed pipe of length L are given by fm = m v 4L , m = 1, 3, 5, …, where v is the speed of sound in the air. ANSWER: 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 Correct Part F ′′ What is the frequency f of the first possible harmonic after the fundamental frequency in the openclosed pipe described in Part E? Express your answer in hertz. Hint 1. How to approach the problem Recall that possible frequencies of standing waves that can be generated in an openclosed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third harmonic. ANSWER: f ′′ = 643 Hz Correct ± Standing Waves on a Guitar String Learning Goal: To understand standing waves, including calculation of λ and f , 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 ms . The guitar string is 60 cm 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 indicated with 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 sinusoidal traveling wave passes through an identically shaped wave moving in the opposite direction on the same guitar string. The string is momentarily flat when the underlying traveling waves are exactly out of phase. The shape is sinusoidal with twice the original amplitude when the underlying waves are momentarily in phase. This pattern is called a standing wave because no wave features travel down the length of the string. 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 the fingerboard. 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 λ 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 usually measured from one peak to the next peak. What is the wavelength λ of this sinusoidal pattern? Express your answer in centimeters. ANSWER: λ = cm ANSWER: λ = 40 cm 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 between nodes are antinodes, where the string moves up and down the most. This standing wave pattern has three antinodes, at x = 10 cm, 30 cm, and 50 cm. The pattern also has four nodes, at x = 0 cm, 20 cm, 40 cm, and 60 cm. 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 L tied down at both ends. A pattern's number n is the number of antinodes it contains. The wavelength of the n th pattern is denoted λ n . The n th pattern has n halfwavelengths along the length of the string, so n Thus the wavelength of the n th pattern is = 2L λn 2 = L. λn = 2L n . Part B What is the wavelength of the longest wavelength standing wave pattern that can fit on this guitar string? Express your answer in centimeters. Hint 1. How to approach the problem Look at the figure to determine the pattern number n for the longest wavelength pattern, then calculate its wavelength λ n . Recall that the guitar string is 60 cm long. Hint 2. Determine n for the longest wavelength pattern What is the pattern number n for the longest wavelength standing wave pattern? ANSWER: 1 n = 2 3 some other integer ANSWER: λ1 = 120 cm Correct This longest wavelength pattern is so important it is given a special name—the fundamental. The wavelength of the fundamental is always given by λ 1 = 2L for a string that is held fixed at both ends. Waves of all wavelengths travel at the same speed v on a given string. Traveling wave velocity and wavelength are related by , v = λf where v is the wave speed (in meters per second), λ is the wavelength (in meters), and f is the frequency [in inverse seconds, also known as hertz (Hz)]. Since only certain wavelengths fit properly to form standing waves on a specific string, only certain frequencies will be represented in that string's standing wave series. The frequency of the n th pattern is fn = Note that the frequency of the fundamental is f 1 v λn = v (2L/n) =n v 2L . , so f n can also be thought of as an integer multiple of f 1 : f n = v/(2L) Part C The frequency of the fundamental of the guitar string is 320 Hz. At what speed v do waves move along that string? Express your answer in meters per second. = nf 1 . Hint 1. How to approach the problem The velocity of waves on the string equals the product of the wavelength and the corresponding frequency for any standing wave pattern: v = λ 1 f = λ 2 f = …. 1 2 You know enough about the fundamental to calculate v . ANSWER: v = 384 m/s Correct Notice that these transverse waves travel slightly faster than the speed of sound waves in air, which is about 340 m/s . We are now in a position to understand certain musical terms from a physics perspective. The standing wave frequencies for this string are f 1 = 320 Hz, f 2 = 2f 1 = 640 Hz, f 3 = 3f 1 = 960 Hz, etc. This set of frequencies is called a harmonic 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 f 2 is one octave above f 1 , f 3 is a perfect fifth above f 2 , and so on. Standing wave patterns with frequencies higher than the fundamental frequency are called overtones. The n = 2 pattern is called the first overtone, the n = 3 pattern is called 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 n ? ANSWER: overtone number = pattern number overtone number = pattern number + 1 overtone number = pattern number − 1 There is no strict relationship between overtone number and pattern number. 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 a number, 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 that is 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 if they 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 plus some of its overtones. Interestingly enough, it is possible to fool someone into identifying a frequency that is not present by playing just its overtones. For example, consider a sound containing pure tones at 450 Hz, 600 Hz, and 750 Hz. Here 600 Hz and 750 Hz are not integer multiples of 450 Hz, so 450 Hz would not be considered the fundamental with the other two as overtones. However, because all three frequencies are consecutive overtones of 150 Hz a listener might claim to hear 150 Hz , over an octave below any of the frequencies present. This 150 Hz is called a virtual pitch or a missing fundamental. Part E A certain sound contains the following frequencies: 400 Hz, 1600 Hz, and 2400 Hz. 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, but only if all the other frequencies are integer multiples of it. ANSWER: This is a pure tone. This is a complex tone with a fundamental of 400 Hz, plus some of its overtones. This is a complex tone with a virtual pitch of 800 Hz. These frequencies are unrelated, so they are probably pure tones from three different sound sources. Correct These concepts of fundamentals and overtones can be applied to other types of musical instruments besides string instruments. Hollowtube instruments, such as brass instruments and reed instruments, have standing wave patterns in the air within them. Percussion instruments, such as 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 being sounded. 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 number of such modes, each one with a distinctive frequency f i and associated pattern of oscillation. Consider an example of a system with normal modes: a string of length L held fixed at both ends, located at x = 0 and x = L. Assume that waves on this string propagate with speed v . 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 y (x, t) = Ai sin(2π i x λi ) sin(2πf i t). 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 the string is correct? Hint 1. Normal mode constraints The key constraint with normal modes is that there are two spatial boundary conditions, yi (0, t) string being fixed at its two ends. and yi (L, t) = 0 ANSWER: 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 λ i . The wavelength λ i can have only certain specific values if the boundary conditions are to be satisfied. The wave does not satisfy the boundary condition yi (0; t) Correct Part B Which of the following statements are true? ANSWER: . = 0 = 0 , which correspond to the The system can resonate at only certain resonance frequencies f i and the wavelength λ i must be such that yi (0; t) Ai . = y i (L; t) = 0 must be chosen so that the wave fits exactly on the string. Any one of Ai or λ i or f i can be chosen to make the solution a normal mode. Correct The key factor producing the normal modes is that there are two spatial boundary conditions, yi (0, t) only for particular values of λ i . and yi (L, t) = 0 = 0 , that are satisfied Part C Find the three longest wavelengths (call them λ 1 , λ 2 , and λ 3 ) that "fit" on the string, that is, those that satisfy the boundary conditions at x x = L. These longest wavelengths have the lowest frequencies. and = 0 Express the three wavelengths in terms of L. List them in decreasing order of length, separated by commas. Hint 1. How to approach the problem The nodes of the wave occur where sin(2π This equation is trivially satisfied at one end of the string (with x x λi ) = 0. ), since sin(0) = 0 . = 0 The three largest wavelengths that satisfy this equation at the other end of the string (with x the three smallest, nonzero values of z that satisfy the equation sin(z) = 0. ) are given by 2πL/λ i = L , where the z i are = zi Hint 2. Values of z that satisfy sin(z) = 0 The spatial part of the normal mode solution is a sine wave. Find the three smallest (nonzero) values of z (call them z 1 , z 2 , and z 3 ) that satisfy sin(z) = 0. Express the three nonzero values of z as multiples of π. List them in increasing order, separated by commas. ANSWER: z1 , z 2 , z 3 = 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 those at the end of the string.) The letter A is written over the antinodes, which are where the oscillation amplitude is maximum. ANSWER: λ1 , λ 2 , λ 3 = 2L, L, 2 3 L 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 λ i . Find the frequency f i of the ith normal mode. Express f i in terms of its particular wavelength λ i and the speed of propagation of the wave v . Hint 1. Propagation speed for standing waves Your expression will involve v , the speed of propagation of a wave on the string. Of course, the normal modes are standing waves and do not travel along the string the way that traveling waves do. Nevertheless, the speed of wave propagation is a physical property that has a well defined 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 speed of propagation v . ANSWER: f i = v λi Correct The frequencies f i are the only frequencies at which the system can oscillate. If the string is excited at one of these resonance frequencies it will respond by oscillating in the pattern given by yi (x, t) , that is, with wavelength λ i associated with the f i at which it is excited. In quantum mechanics these frequencies are called the eigenfrequencies, which are equal to the energy of that mode divided by Planck's constant h . In SI units, Planck's constant has the value h = 6.63 × 10 −34 J ⋅ s. Part E Find the three lowest normal mode frequencies f 1 , f 2 , and f 3 . Express the frequencies in terms of L, v , and any constants. List them in increasing order, separated by commas. ANSWER: f 1 , f 2 , f 3 = v 2L , v L , 3v 2L Correct Note that, for the string, these frequencies are multiples of the lowest frequency. For this reason the lowest frequency is called the fundamental and 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 normal frequencies 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 higher notes. 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. 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 each sequence. 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 each point to determine the displacement of that section of the string. Note that the pulses pass through the same section of the string only after each 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 of the 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: a b c d ANSWER: a b c d e f Correct Part B Consider the point where the two pulses start to overlap, point O in . What is the displacement of point O as these pulses interfere? Hint 1. How to approach the problem 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 Δt be a short interval of time after the two pulses have begun to overlap. If only the pulse coming from the right were present, the displacement of point O at time Δt would be +y0 . What would its displacement be if only the other pulse were present? Express your answer in terms of y0 . Hint 1. Inverted pulses Keep in mind that the two pulses are identical but inverted; thus their displacements are equal magnitude but opposite. ANSWER: 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. Correct Part C Why does destructive interference occur when the two pulses overlap instead of constructive interference? ANSWER: 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 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 this moment? ANSWER: The energy of the string is zero. The string is not moving either up or down. The string has only kinetic energy. 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 string oscillate. 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 y represents the transverse displacement of a string that extends along the x direction. Here is a common mathematical form for such a wave: y(x, t) = A sin(kx) sin(ωt), where A is the maximum transverse displacement of the string (the amplitude of the wave), which is assumed to be nonzero, k is the wave number, ω is the angular frequency of the wave, and t is time. Part A Which one of the following statements about such a wave as described in the problem introduction is correct? ANSWER: This wave is traveling in the +x direction. This wave is traveling in the −x direction. This wave is oscillating but not traveling. This wave is traveling but not oscillating. 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 t , what is the displacement of the string y(x, 0)? = 0 Express your answer in terms of A, k , and other previously introduced quantities. ANSWER: y(x, 0) = 0 Correct Part C What is the displacement of the string as a function of x at time T /4 , where T is the period of oscillation of the string? Express the displacement in terms of A, x , k , and other constants; that is, evaluate ω T 4 and substitute it in the expression for y(x, t). Hint 1. Find the period in terms of ω What is the period T in terms of the angular frequency ω? Express your answer in terms of ω and constants like π. ANSWER: T = Hint 2. sin(ω T 4 ) What is the value of the timedependent trigonometric function sin(ωt) at the time T /4 ? Express your answer in terms of the previously introduced quantities. ANSWER: sin(ω T 4 ) = ANSWER: y(x, T 4 ) = Asin(kx) Correct Part D At which three points x 1 , x 2 , and x 3 closest to x but with x = 0 will the displacement of the string y(x, t) be zero for all times? These are the > 0 first three nodal points. Express the first three nonzero nodal points in terms of the wavelength λ . List them in increasing order, separated by commas. You should enter only the factors that multiply λ . Do not enter λ for each one. Hint 1. What is equal to zero? If y(x, t) = 0, then A sin(kx) sin(ωt) = 0. Because A > 0, either sin(kx) = 0 or sin(ωt) = 0 (or both) if the displacement of the string is to be zero for all times. The frequency ω is a constant for a given wave and t may take any positive value. Therefore, sin(ωt) cannot equal zero for all times. Hence you need to find solutions to the equation sin(kx) = 0. Express the first three nonzero values of kx (call them (kx) 1 , (kx) 2 , (kx) 3 ) for which y(x, t) = 0 as multiples of π. List them in increasing order, separated by commas. You should not type π each time. You should enter only the factors that multiply π. ANSWER: (kx) 1 , (kx) 2 , (kx) 3 = π Hint 2. What is λ in terms of k ? What is the wavelength λ of the standing wave in terms of k ? Express your answer in terms of k and constants like π. ANSWER: λ = Hint 3. Putting it all together The nodes of the wave are points for which sin(kx) λ for which sin(kx) = 0. . If you substitute x i = 0 using the previous results, you will find the multiples α of = αλ ANSWER: x1 , x 2 , x 3 = 0.500,1,1.50 λ Correct Problem 21.9 Part A What are the three longest wavelengths for standing waves on a 222cmlong string that is fixed at both ends? Enter your answers in descending order separated by commas. ANSWER: λ1 , λ 2 , λ 3 = 4.44,2.22,1.48 m Correct Part B If the frequency of the secondlongest wavelength is 60 Hz , what is the frequency of the thirdlongest wavelength? Express your answer using two significant figures. ANSWER: f3 = 90 Hz Correct Problem 21.11 A heavy piece of hanging sculpture is suspended by a 90 cmlong, 5.0 g steel wire. When the wind blows hard, the wire hums at its fundamental frequency of 61 Hz . Part A What is the mass of the sculpture? Express your answer to two significant figures and include the appropriate units. ANSWER: m = 6.8 kg Correct Problem 21.31 A 2.0mlong string vibrates at its secondharmonic frequency with a maximum amplitude of 3.0 cm . One end of the string is at x = Part A Find the oscillation amplitude at x = 10 cm. Express your answer to two significant figures and include the appropriate units. ANSWER: = = 0.93 cm a(x = 10 cm) Correct Part B Find the oscillation amplitude at x = 20 cm. Express your answer to two significant figures and include the appropriate units. ANSWER: = = 1.8 cm a(x = 20 cm) Correct Part C Find the oscillation amplitude at x = 30 cm. Express your answer to two significant figures and include the appropriate units. ANSWER: = = 2.4 cm a(x = 30 cm) . 0 cm Correct Part D Find the oscillation amplitude at x = 40 cm. Express your answer to two significant figures and include the appropriate units. ANSWER: = = 2.9 cm a(x = 40 cm) Correct Part E Find the oscillation amplitude at x = 50 cm. Express your answer to two significant figures and include the appropriate units. ANSWER: = = 3.0 cm a(x = 50 cm) Correct Score Summary: Your score on this assignment is 50.0%. You received 6 out of a possible total of 12 points.