6Sound test w solutions

PHYSICS TOPICAL: Sound Test Time: 23 Minutes* Number of Questions: 18 * The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit MCAT DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions PERIODIC TABLE OF THE ELEMENTS H 1.0 He 4.0 Li 6.9 Be 9.0 B 10.8 C 12.0 N 14.0 O 16.0 F 19.0 10 Ne 20.2 11 Na 23.0 12 Mg 24.3 13 Al 27.0 14 Si 28.1 15 P 31.0 16 S 32.1 17 Cl 35.5 18 Ar 39.9 19 K 39.1 20 Ca 40.1 21 Sc 45.0 22 Ti 47.9 23 V 50.9 24 Cr 52.0 25 Mn 54.9 26 Fe 55.8 27 Co 58.9 28 Ni 58.7 29 Cu 63.5 30 Zn 65.4 31 Ga 69.7 32 Ge 72.6 33 As 74.9 34 Se 79.0 35 Br 79.9 36 Kr 83.8 37 Rb 85.5 38 Sr 87.6 39 Y 88.9 40 Zr 91.2 41 Nb 92.9 42 Mo 95.9 43 Tc (98) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 Cd 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 55 Cs 132.9 56 Ba 137.3 57 La * 138.9 72 Hf 178.5 73 Ta 180.9 74 W 183.9 75 Re 186.2 76 Os 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 Hg 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 209.0 84 Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra 226.0 89 Ac † 227.0 104 Rf (261) 105 Ha (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 Ce 140.1 59 Pr 140.9 60 Nd 144.2 61 Pm (145) 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 67 Ho 164.9 68 Er 167.3 69 Tm 168.9 70 Yb 173.0 71 Lu 175.0 † 90 Th 232.0 91 Pa (231) 92 U 238.0 93 Np (237) 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) GO ON TO THE NEXT PAGE as developed by Sound Test Passage I (Questions 1–7) A church organ is a musical instrument that creates sound by forcing air through pipes, causing them to resonate Different notes are created by using pipes of different length The intensity of a sound wave is proportional to both the square of the amplitude and the square of the frequency The perceived volume of a sound is indicated by the sound level, measured in decibels: β = 10 log(I/I0), where I is the intensity of the sound, and I0 is a reference intensity, taken to be 10–l2 W/m2 The characteristics of the sound of a note are determined by the relative intensity of the harmonic frequencies For pipes open at one end and closed at the other end, the harmonics are determined by the formula: n λ = 4L, where n = 1, 3, 5, … For pipes open at both ends, the harmonics are determined by the formula: n λ = 2L, where n = 1, 2, 3, … In both cases, λ is the wavelength of the harmonic, and L is the length of the pipe The fundamental frequency determines the note and corresponds to the wavelength where n = Organs can be tuned by adjusting the length of the pipes Tuning is generally done with the aid of a pitch pipe, a small pipe that can be blown to produce notes in a standard scale These notes are then compared to those produced by the organ pipes Consider an organ pipe closed at one end What is the ratio of the frequency of the n = harmonic to the fundamental frequency? A B C D 1:4 1:3 2:1 3:1 The length of an organ pipe open at both ends is doubled, thereby doubling the wavelength of each harmonic it produces What happens to the intensity? (Note: Assume that the amplitude of each harmonic is kept constant.) A B C D It is reduced to 1/4 the original value It remains the same It is doubled It increases for some harmonics and decreases for others An organ produces a G note with an intensity of 10–6 W/m2 What is the sound level of this G note? A dB B 18 dB C 60 dB D 120 dB If an E note is played simultaneously on an organ pipe and on a pitch pipe, beats are heard Which of the following best explains this effect? A The organ pipe is not producing the same frequency as the pitch pipe B The organ pipe is not producing the same intensity as the pitch pipe C The pitch pipe produces more harmonics than the organ pipe D The pitch pipe produces fewer harmonics than the organ pipe Pipe is closed at one end, and Pipe is open at both ends If they each produce the same fundamental wavelength, then Pipe must be: How many wavelengths are there in the organ pipe shown below, and what type of pipe is it? A B C D A B C D one half as long as Pipe equal in length to Pipe two times as long as Pipe four times as long as Pipe 2, closed at one end 2, open at both ends 4, closed at one end 4, open at both ends GO ON TO THE NEXT PAGE KAPLAN MCAT The length of a pitch pipe is much smaller than the length of a typical organ pipe Which of the following best explains how the pitch pipe can be used to tune the organ pipe? A The fundamental frequency of the organ pipe can be tuned to the fundamental frequency of the pitch pipe B The fundamental frequency of the organ pipe can be tuned to one of the higher harmonics of the pitch pipe C One of the higher harmonics of the organ pipe can be tuned to the fundamental frequency of the pitch pipe D The fundamental frequency of the organ pipe can be tuned to the frequency of the beats GO ON TO THE NEXT PAGE as developed by Sound Test Passage II (Questions 8–12) When a source of sound is moving toward a stationary detector, the frequency of sound perceived by the detector is not the same as that emitted by the source In order to see this quantitatively, consider the time T elapsed between the emission of a successive pair of spherical wave fronts (see Figure 1) A source moving with speed v s emits the first wave front and then travels a distance vsT, where it emits the second wave front At this point, the first wave front has already traveled a distance vT, where v = 340 m/s is the speed of sound Therefore, the wavelength detected along the direction of motion is given by the difference: λ ′ = vT – vsT Using the fact that λ′f ′ = λf = v and T = l/f, we find that the shifted frequency f ′ perceived by the detector is: f ′ = fv/(v - vs ) vt vst vs Figure A roller skater carrying a portable stereo skates at constant speed past an observer at rest Which of the following accurately represents how the frequency perceived by the observer changes with time? A vT frequency frequency vsT C vs λ' time time The previous equation does not describe situations where vs ≥ v When the source travels faster than the speed of sound, a shock wave is produced by the spherical wave fronts Figure shows the spherical wave fronts produced by such a source at equally spaced positions over an arbitrary time t During this time, the source travels a distance v s t, and the first wave front travels a distance vt However, in this case, the source emits each new wave after traveling beyond the front of the previously emitted wave The wave fronts bunch along the surface of a cone called the Mach cone The resulting rise and fall in air pressure as the surface of the cone passes through a point in space produces a shock wave KAPLAN frequency Figure D frequency B time time A police car moving toward a stationary pedestrian at a speed of 10 m/s operates its siren If the pedestrian perceives the frequency of the siren to be 1030 Hz, what is the frequency emitted by the siren? A 10 Hz B 100 Hz C 1000 Hz D 10,000 Hz GO ON TO THE NEXT PAGE MCAT A bat flies toward a stationary wall with speed v b If the bat emits a signal at frequency f, what is the correct expression for the frequency of the reflected signal that the bat hears? (Note: Any signal of frequency f ′ reflected off of the wall is heard by the bat as: f ′′ = f ′(v + vb)/v.) A f v v − vb B f v - vb v + vb C f v + vb v D f v + vb v − vb The sine of the angle θ in Figure is called the Mach number What physical quantity does it represent? A The ratio of the speed of sound to the speed of light B The ratio of the speed of sound to the speed of the source C The ratio of the speed of the source to the speed of the detector D The ratio of the speed of sound in air to the speed of sound in a vacuum 1 Which of the following graphs represents a plot of the frequency f ′ of a sound wave perceived by a stationary observer versus the speed vs of the source? A C f' f' v vs v B D f' f' v vs v vs vs GO ON TO THE NEXT PAGE as developed by Sound Test Questions 13 through 18 are NOT based on a descriptive passage A vibrating guitar string produces sound that travels through the air to the human ear Why is the wavelength of the sound traveling through the air NOT the same as the wavelength of the wave traveling on the string? A The speed of the sound wave is not the same as the speed of the wave traveling on the guitar string B The frequency of the sound wave is not the same as the frequency of the wave traveling on the guitar string C The amplitude of the sound wave is not the same as the amplitude of the wave traveling on the guitar string D Sound waves are longitudinal waves, and the wave traveling on the guitar string is transverse A sound wave of frequency 300 Hz travels into a pipe of length L that is closed at one end, and the air in the pipe resonates in its fundamental mode If the frequency of the sound wave is increased until the air in the pipe resonates again, what is its new frequency? A B C D 450 Hz 600 Hz 750 Hz 900 Hz A car and a train are traveling parallel to one another, both at a speed of 30 m/s If the train’s whistle blows at 1,000 Hz, what frequency of sound does the driver of the car hear? (Note: Assume the speed of sound in air is 340 m/s.) A 833 Hz B 1,000 Hz C 1,030 Hz D 1,200 Hz When a sound wave passes from medium having density ρl to medium having density ρ2, its velocity increases by 30% What is the ratio of the wavelength in medium to that in medium ? A B C D 0.3 0.8 1.0 1.3 How much more intense is a 30 dB sound than a 10 dB sound? A times more intense B 20 times more intense C 100 times more intense D 3000 times more intense A sound wave traveling in air has a wavelength of 85 cm and a speed of 340 m/s What is its frequency? A 40 Hz B 400 Hz C 440 Hz D 800 Hz END OF TEST KAPLAN MCAT ANSWER KEY: D A B C A B C C A 10 D 11 12 13 14 15 A B A D C 16 B 17 D 18 B as developed by Sound Test EXPLANATIONS Passage I (Questions 1—7) D The first thing to note when approaching this question is that it refers to the fundamental frequency The second paragraph of the passage says that the fundamental frequency corresponds to the wavelength where n = The other mode of interest corresponds to the wavelength where n = The question stem states that the pipe is closed at one end; the relevant formula determining the harmonics is therefore nλ = 4L, or rearranging to solve for the wavelength λ: λ= 4L n The wavelength of the n = harmonic is therefore 1/3 that of the fundamental For this question, however, it is the ratio of the frequencies, not of the wavelength, that we are interested in Frequency is inversely proportional to the wavelength via the familiar relation: f = v/λ v, the velocity of the sound wave, is the same for the two modes The frequency of the third harmonic must therefore be three times the frequency of the fundamental, since is the inverse of 1/3 B According to the diagram shown in the question, there are two complete sinusoids from one end of the pipe to the other There are thus two wavelengths’ equivalent in the pipe To determine whether the pipe is closed at one end or open at both ends, one can use one of two methods First, one may just know that a closed end is always a point of zero displacement, i.e a node, while an open end is an antinode—the amplitude of the wave reaches a maximum In the diagram, the two ends are points of maximum displacement These are thus antinodes and the pipe has to be open at both ends Alternatively, one may examine the formulas given in the passage closely—if the pipe is closed at one end, then, from the equation given in the passage, the length of the pipe L would be related to the wavelength by: n λ, where n = 1, 3, 5, … i.e L = λ, λ, λ, etc 4 L= λ) exactly A pipe closed at one end cannot give rise to the situation depicted in the diagram; the pipe will have to be open at both ends There is thus no way that L can contain 2λ (= A In the question stem, we are given both the change in wavelength of the harmonics and the change in length of the pipe and asked to find the change in intensity In the passage, we are not given any relationship between the intensity and either the wavelength or length of the pipe directly What we are given is the relationship between intensity and the frequency Specifically, we are told that the intensity is proportional to the amplitude squared and to the frequency squared, i.e.: I = kA2f2 where I is the intensity, A the amplitude of the wave, f its frequency, and k some proportionality constant The question stem states that all the relevant amplitudes remain constant, and so we need only concern ourselves with the effects of changing the frequency All the wavelengths are doubled, and so the frequencies, inversely proportional to the wavelengths, will be halved 1 The intensity, dependent on the square of the frequency, will therefore be ( )2 = that of before 4 C The sound level β, measured in decibels (dB), is given by the formula: I β = 10 log( ) I0 KAPLAN MCAT I = 10–6–(–12) = 106 One of the properties of the I0 logarithmic function that you should know is that log(10x) = x Here, therefore, we have: where I0 = 10–12 W/m2 The intensity I in this case is 10–6 W/m2, and so β = 10 log(106) = 10 × = 60 A Each musical note corresponds to a distinct wavelength or a discrete frequency Beats occur when two notes of slightly different frequencies are played simultaneously The frequency of the beats equals the difference between the two frequencies played Choice B is incorrect because even though the intensity of a sound wave is proportional to the square of the frequency, it is also proportional to the square of the amplitude Therefore, a difference in frequency need not translate into a difference in intensity, if there is a compensating difference in amplitude In other words, beats can exist even when the intensities are the same; conversely, beats can be absent even when the intensities are different Therefore choice B cannot be a correct explanation to the phenomenon Choices C and D are incorrect because the number of harmonics has no necessary bearing on beats A Let us define L1 as the length of pipe 1, and L2 the length of pipe Pipe is closed at one end, and its allowed wavelengths are given by nλ = 4L1, where n = 1, 3, 5, … Pipe is open at both ends, and so the allowed wavelengths are given by nλ = 2L2, where n = 1, 2, 3, … The fundamental wavelengths for pipes and are hence 4L1 and 2L2 respectively These two wavelengths are actually the same, as stated in the passage Hence: 4L1 = 2L2 L1 = L 2 Pipe (closed at one end) node antinode L1 Pipe (open at both ends) antinode l = 4L l = 2L2 antinode L2 C One way to approach this problem is to examine each answer choice individually to determine whether it provides a plausible explanation Choice A states that the fundamental frequency of the organ pipe can be tuned to match the fundamental frequency of the pitch pipe From the formulas in the passage, we know that the wavelength of the fundamental frequency (n = 1) is proportional to the length of the pipe: λ = 4L or 2L depending on which pipe we are talking about Furthermore, since frequency is inversely proportional to the wavelength, the fundamental frequency is inversely proportional to the length of the v v pipe: f1 = or Therefore, the pitch pipe, which is said to be much shorter than the organ pipe, will have a much higher 4L 2L 10 as developed by Sound Test fundamental frequency than the organ pipe So its fundamental frequency cannot be used to tune the fundamental frequency of the organ pipe, and choice A is incorrect Choice B states that the fundamental frequency of the organ pipe can be tuned to one of the higher harmonics of the pitch pipe We already determined when considering choice A that the fundamental frequency of the pitch pipe is much higher than the fundamental frequency of the organ pipe The frequency of a higher harmonic will only be higher: for the nth harmonic, 4L 2L v vn the wavelength is λ = or ; the frequency is therefore f = = or The frequency is therefore proportional to the n n λ 4L 2L number of the harmonic So a higher harmonic of the pitch pipe will have a higher frequency than its fundamental, which already has a higher frequency than the organ pipe fundamental to begin with There is then no way to match the fundamental frequency of the organ pipe to a higher harmonic of the pitch pipe, and choice B is incorrect Choice C reverses the situation given in choice B and uses the pitch pipe fundamental to tune a higher harmonic of the organ pipe The frequency of a higher harmonic of the organ pipe is some multiple of the fundamental We have already determined that the fundamental of the pitch pipe has a higher frequency than the organ pipe fundamental: It is then possible for the pitch pipe fundamental frequency to also be some multiple of the organ pipe fundamental If such is the case, then one of the higher harmonics of the organ pipe would have the same frequency as the pitch pipe fundamental So choice C presents a plausible way to tune the organ For the sake of completeness let us also consider choice D, which states that the fundamental frequency of the organ pipe can be tuned to the frequency of the beats Beats occur when two tones that are close in frequency are played simultaneously The frequency of the beats is then equal to the difference in frequencies of the two tones In order to tune the fundamental to the beat frequency, however, the difference between the organ pipe and pitch frequencies must be that of the organ pipe fundamental This, however, contradicts what we just said about the two tones having to be close in frequency to give rise to beats in the first place So choice D is incorrect, and choice C is the answer Passage II (Questions 8—12) B We have a sound source that is moving past an observer and are asked to predict how the perceived frequency changes Instead of turning to the formula immediately, let us see how far we can go by qualitative reasoning When a sound is emitted from a source that is approaching the observer, the wave crests are more “bunched up” and arrive at the observer more frequently The observer consequently hears a higher frequency than if there were no motion On the other hand, if the source is moving away, each successively emitted wave crest takes a longer time interval to reach the observer, and so the perceived frequency of the wave is lower In this case, the stationary person is going to hear a higher frequency as the skater approaches, and a lower frequency as the skater is moving away From this, we can eliminate choices A and C It is then necessary to distinguish between choices B and D Choice D may look tempting as it shows the perceived frequency gradually increasing then gradually decreasing This is however wrong The perceived frequency depends on the speed of the sound and the speed of the source The speed of the skater is said to be constant in the question stem: it does not change as he approaches Throughout the time that the skater is approaching, then, the person hears one frequency that is higher than that from the emitter’s perspective After the skater has gone past, the person will hear one constant, lower frequency, as the speed of the skater is still the same; the only difference being that the sign in front of vs has been reversed in the formula Choice B is therefore correct Be sure to distinguish between volume and the frequency—the volume will increase and then decrease gradually, but the frequency perceived does not C This is a straightforward application of the Doppler formula The one slightly tricky part is that the frequency given in the stem is the perceived frequency, f’ We can first of all eliminate choice D since the source is approaching the observer: the emitted frequency must be lower than the perceived one You may in fact choose choice C simply by evaluating the magnitude of the numbers: the source speed, which is the velocity of the police car, is vs = 10 m/s This is a relatively small percentage of the speed of sound, v = 340 m/s as given in the passage It is unreasonable, therefore, to expect the perceived frequency to change by factors of 10 and 100, as suggested by choices B and A respectively The minor adjustment given in choice C seems much more reasonable The actual set-up is: f × 340 m/s 34 f’ = 1030 Hz = = f (340 m/s – 10 m/s) 33 33 f= × 1030 Hz 34 KAPLAN 11 MCAT 33/34 is only slightly smaller than one, and so f should be slightly below 1030 Hz It would be a much too inefficient use of time to actually carry out the calculation, not to mention the possibility of making arithmetic errors 10 D The bat flies towards a stationary wall with speed vb and emits a frequency f The wall, then, “detects” a frequency f’ as given by the Doppler formula, with vs = vb: f’ = fv (v – v b ) This frequency is reflected back to the bat Now the wall is acting as a stationary source, “emitting” a frequency of f’ to a moving observer The relevant formula for this scenario is given in the question stem, and so the frequency perceived by the bat, f ”, is given by: f”= f’ (v + v b ) (v + v b ) (v + v b ) fv = × =f v (v – v b ) v (v – v b ) 11 A This is a standard example of a graphical analysis problem We want to come up with the graph that illustrates the formula given in the passage: f’ = fv (v – v s ) Generally, for this type of question there are only a few points to consider to arrive at the correct graph First, it is always a good idea to look at the behavior of the curve as the independent variable (the one plotted on the x-axis, in this case vs) goes to infinity This particular formula, however, is described in the passage as being valid only for vs < v In other words, the velocity of the source is always less than the velocity of the sound As vs gets closer and closer to v, we see that the denominator gets closer and closer to zero (while remaining positive since vs < v) Something divided by a very small number gives a very large number, and so the entire quotient on the right hand side of the formula increases rapidly This by itself should be sufficient for us to choose A as the correct answer, but let us just note in addition that when vs = 0, the formula boils down to f’ = f: both the source and the observer are stationary and so there is no reason for the perceived frequency to be different from the emitted one The graph in choice A is the only one that shows f’ having a nonzero, finite value at vs = 12 B Although seemingly difficult, the answer can be obtained just by examining Figure and using trigonometry, instead of any scientific principle The angle θ is part of a right triangle, with the hypotenuse being vst The side opposite to the angle has length vt Do not be confused by the fact that the triangle is drawn upside down and mistake the adjacent side as the hypotenuse: look at which angle is labeled as a right angle The sine of θ, defined as the opposite over the hypotenuse, is therefore: vt v sin θ = = = v:vs vst vs This is the ratio of the speed of sound to the speed of the source Independent Questions 13 A When the string vibrates, it generates perturbations among the air molecules and causes them to vibrate Specifically, the string vibrations excite compression waves in the air which travel to the human ear and are detected by the eardrum The frequency of all these excitations is the same as the frequency of the string vibrations Choice B is thus incorrect 12 as developed by Sound Test Since we are told that the wavelengths are not the same, while knowing that the frequencies are, then we conclude that the speeds must be different Choice C is incorrect because the amplitude of a wave is independent of its wavelength Choice D is incorrect because even though the statement is true, it is irrelevant here 14 D Again, it is important to keep in mind that as a wave goes from one medium to another, its speed and wavelength may change, but its frequency will remain constant As the wave in this question moves from medium to medium 2, it is said that its velocity increases by 30% Converting this percentage into numbers, we can write that the velocity of the wave in medium is 1.3 times that in medium Since, as we have said, the frequency does not change, and since v = fλ, the velocity is proportional to the wavelength, and a velocity in medium of 1.3 times that of the old one means that the wavelength in medium has to be 1.3 times as great as the old wavelength as well 15 C Sound level is the intensity measured in a logarithmic scale More specifically the sound level β is defined by β =10 I log( ), where I is the intensity of the sound and I0 some reference intensity For this question, we need to express the intensity I0 in terms of the sound level Rearranging the equation above, we have: I β log( ) = I0 10 I = 10(β/10) I0 I = I010(β/10) So a sound of 30 dB has an intensity of I0103 = 1000I0; while a sound of 10 dB has an intensity of I0101 = 10I0 The 30-dB sound is therefore 100 times as intense as the 10-dB one A handy rule of thumb regarding decibels is that for every increase of 10 dB, the sound is 10 times more intense So going from 10 dB to 20 dB we make the intensity 10 times larger; going from 20 dB to 30 dB makes the 20-dB sound in turn 10 times more intense, yielding a net increase in intensity by a factor of 10 × 10 = 100 16 B This question calls for a straightforward application of the formula fλ = v Notice that the velocity is given in meters per second, while the wavelength is given in centimeters The easiest thing to would be to convert the wavelength into meters: f= v 340 m/s = = 400 s–1 = 400 Hz λ 0.85m 17 D For a pipe closed at one end to resonate, there must be an antinode at the open end and a node at the closed end An antinode is a point in space at which a standing wave fluctuates with maximum amplitude, while a node is a point at which the standing wave is always at zero displacement For this to be the case, the length of the pipe must be some odd multiple of a quarter-wavelength: λ L = n , where n = 1, 3, 5, … KAPLAN 13 MCAT node antinode λ = 4L L node antinode λ = (4/3)L etc The frequencies for resonance are therefore given by: 4L where n is some odd integer We are told in the question stem that a frequency of 300 Hz resonates in the fundamental mode, i.e.: v 300 Hz = 4L 3v The next mode of resonance occurs for n = 3, i.e f (n=3) = This is three times the fundamental frequency, and so 4L the next frequency to allow for resonance is 300 × = 900 Hz f= 18 B We have a car and a train traveling in the same direction at the same speed (30 m/s) The source and the observer are thus neither approaching nor moving away from each other The frequency perceived by the car will thus not be any different from the emitted frequency, 1000 Hz 14 as developed by ... Hz B 400 Hz C 440 Hz D 800 Hz END OF TEST KAPLAN MCAT ANSWER KEY: D A B C A B C C A 10 D 11 12 13 14 15 A B A D C 16 B 17 D 18 B as developed by Sound Test EXPLANATIONS Passage I (Questions... 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) GO ON TO THE NEXT PAGE as developed by Sound Test Passage I (Questions 1–7) A church organ is a musical instrument that creates sound by forcing... organ pipe can be tuned to the frequency of the beats GO ON TO THE NEXT PAGE as developed by Sound Test Passage II (Questions 8–12) When a source of sound is moving toward a stationary detector,