PHYSICS TOPICAL: Sound Test 1

Transcription

1 PHYSICS TOPICAL: Sound Test 1 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.

2 MCAT DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptie 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 descriptie 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 proided below for your use with the questions. PERIODIC TABLE OF THE ELEMENTS 1 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc (98) 44 Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La * Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra Ac Rf (261) 105 Ha (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 Ce Pr Nd Pm (145) 62 Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa (231) 92 U 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. 2 as deeloped by

3 Sound Test 1 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 wae is proportional to both the square of the amplitude and the square of the frequency. The perceied olume of a sound is indicated by the sound leel, measured in decibels: β = 10 log(i/i 0 ), where I is the intensity of the sound, and I 0 is a reference intensity, taken to be 10 l2 W/m 2. The characteristics of the sound of a note are determined by the relatie intensity of the harmonic frequencies. For pipes open at one end and closed at the other end, the harmonics are determined by the formula: nλ = 4L, where n = 1, 3, 5, For pipes open at both ends, the harmonics are determined by the formula: nλ = 2L, where n = 1, 2, 3, In both cases, λ is the waelength of the harmonic, and L is the length of the pipe. The fundamental frequency determines the note and corresponds to the waelength where n = 1. 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. 1. Consider an organ pipe closed at one end. What is the ratio of the frequency of the n = 3 harmonic to the fundamental frequency? A. 1:4 B. 1:3 C. 2:1 D. 3:1 2. How many waelengths are there in the organ pipe shown below, and what type of pipe is it? 3. The length of an organ pipe open at both ends is doubled, thereby doubling the waelength of each harmonic it produces. What happens to the intensity? (Note: Assume that the amplitude of each harmonic is kept constant.) A. It is reduced to 1/4 the original alue. B. It remains the same. C. It is doubled. D. It increases for some harmonics and decreases for others. 4. An organ produces a G note with an intensity of 10 6 W/m 2. What is the sound leel of this G note? A. 6 db B. 18 db C. 60 db D. 120 db 5. 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. 6. Pipe 1 is closed at one end, and Pipe 2 is open at both ends. If they each produce the same fundamental waelength, then Pipe 1 must be: A. one half as long as Pipe 2. B. equal in length to Pipe 2. C. two times as long as Pipe 2. D. four times as long as Pipe 2. A. 2, closed at one end B. 2, open at both ends C. 4, closed at one end D. 4, open at both ends GO ON TO THE NEXT PAGE. KAPLAN 3

4 MCAT 7. 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. 4 as deeloped by

5 Sound Test 1 Passage II (Questions 8 12) When a source of sound is moing toward a stationary detector, the frequency of sound perceied by the detector is not the same as that emitted by the source. In order to see this quantitatiely, consider the time T elapsed between the emission of a successie pair of spherical wae fronts (see Figure 1). A source moing with speed s emits the first wae front and then traels a distance s T, where it emits the second wae front. At this point, the first wae front has already traeled a distance T, where = 340 m/s is the speed of sound. Therefore, the waelength detected along the direction of motion is gien by the difference: λ = T s T. Using the fact that λ f = λf = and T = l/f, we find that the shifted frequency f perceied by the detector is: f = f/( - s ). t st 0 s Figure 2 8. A roller skater carrying a portable stereo skates at constant speed past an obserer at rest. Which of the following accurately represents how the frequency perceied by the obserer changes with time? T s A. C. st λ' frequency frequency B. time D. time Figure 1 The preious equation does not describe situations where s. When the source traels faster than the speed of sound, a shock wae is produced by the spherical wae fronts. Figure 2 shows the spherical wae fronts produced by such a source at equally spaced positions oer an arbitrary time t. During this time, the source traels a distance s t, and the first wae front traels a distance t. Howeer, in this case, the source emits each new wae after traeling beyond the front of the preiously emitted wae. The wae 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 wae. frequency time frequency time 9. A police car moing toward a stationary pedestrian at a speed of 10 m/s operates its siren. If the pedestrian perceies the frequency of the siren to be 1030 Hz, what is the frequency emitted by the siren? A. 10 Hz B. 100 Hz C Hz D. 10,000 Hz GO ON TO THE NEXT PAGE. KAPLAN 5

6 MCAT 10. A bat flies toward a stationary wall with speed 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 ( + b )/.) A. f b - B. f + b b 12. The sine of the angle θ in Figure 2 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 acuum C. f + b D. f + b b 11. Which of the following graphs represents a plot of the frequency f of a sound wae perceied by a stationary obserer ersus the speed s of the source? A. f' C. f' s s B. f' D. f' s s GO ON TO THE NEXT PAGE. 6 as deeloped by

7 Sound Test 1 Questions 13 through 18 are NOT based on a descriptie passage. 13. A ibrating guitar string produces sound that traels through the air to the human ear. Why is the waelength of the sound traeling through the air NOT the same as the waelength of the wae traeling on the string? A. The speed of the sound wae is not the same as the speed of the wae traeling on the guitar string. B. The frequency of the sound wae is not the same as the frequency of the wae traeling on the guitar string. C. The amplitude of the sound wae is not the same as the amplitude of the wae traeling on the guitar string. D. Sound waes are longitudinal waes, and the wae traeling on the guitar string is transerse. 14. When a sound wae passes from medium 1 haing density ρ l to medium 2 haing density ρ 2, its elocity increases by 30%. What is the ratio of the waelength in medium 2 to that in medium 1? A. 0.3 B. 0.8 C. 1.0 D A sound wae of frequency 300 Hz traels 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 wae is increased until the air in the pipe resonates again, what is its new frequency? A. 450 Hz B. 600 Hz C. 750 Hz D. 900 Hz 18. A car and a train are traeling 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 drier 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 15. How much more intense is a 30 db sound than a 10 db sound? A. 3 times more intense B. 20 times more intense C. 100 times more intense D times more intense 16. A sound wae traeling in air has a waelength 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 7

8 MCAT ANSWER KEY: 1. D 6. A 11. A 16. B 2. B 7. C 12. B 17. D 3. A 8. B 13. A 18. B 4. C 9. C 14. D 5. A 10. D 15. C 8 as deeloped by

9 Sound Test 1 EXPLANATIONS Passage I (Questions 1 7) 1. 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 waelength where n = 1. The other mode of interest corresponds to the waelength where n = 3. The question stem states that the pipe is closed at one end; the releant formula determining the harmonics is therefore nλ = 4L, or rearranging to sole for the waelength λ: λ = 4L n The waelength of the n = 3 harmonic is therefore 1/3 that of the fundamental. For this question, howeer, it is the ratio of the frequencies, not of the waelength, that we are interested in. Frequency is inersely proportional to the waelength ia the familiar relation: f = /λ., the elocity of the sound wae, is the same for the two modes. The frequency of the third harmonic must therefore be three times the frequency of the fundamental, since 3 is the inerse of 1/3. 2. 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 waelengths equialent 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 wae 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. Alternatiely, one may examine the formulas gien in the passage closely if the pipe is closed at one end, then, from the equation gien in the passage, the length of the pipe L would be related to the waelength by: L = n 4 λ, where n = 1, 3, 5, i.e. L = 1 4 λ, 3 4 λ, 5 4 λ, etc. There is thus no way that L can contain 2λ (= 8 λ) exactly. A pipe closed at one end cannot gie rise to the situation 4 depicted in the diagram; the pipe will hae to be open at both ends. 3. A In the question stem, we are gien both the change in waelength 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 gien any relationship between the intensity and either the waelength or length of the pipe directly. What we are gien 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 = ka 2 f 2 where I is the intensity, A the amplitude of the wae, f its frequency, and k some proportionality constant. The question stem states that all the releant amplitudes remain constant, and so we need only concern ourseles with the effects of changing the frequency. All the waelengths are doubled, and so the frequencies, inersely proportional to the waelengths, will be haled. The intensity, dependent on the square of the frequency, will therefore be ( 1 2 )2 = 1 that of before C The sound leel β, measured in decibels (db), is gien by the formula: β = 10 log( I I 0 ) KAPLAN 9

10 MCAT where I 0 = W/m 2. The intensity I in this case is 10 6 W/m 2, and so I logarithmic function that you should know is that log(10 x ) = x. Here, therefore, we hae: I 0 = 10 6 ( 12) = One of the properties of the β = 10 log(10 6 ) = 10 6 = A Each musical note corresponds to a distinct waelength 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 een though the intensity of a sound wae 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 een when the intensities are the same; conersely, beats can be absent een 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. 6. A Let us define L 1 as the length of pipe 1, and L 2 the length of pipe 2. Pipe 1 is closed at one end, and its allowed waelengths are gien by nλ = 4L 1, where n = 1, 3, 5, Pipe 2 is open at both ends, and so the allowed waelengths are gien by nλ = 2L 2, where n = 1, 2, 3, The fundamental waelengths for pipes 1 and 2 are hence 4L 1 and 2L 2 respectiely. These two waelengths are actually the same, as stated in the passage. Hence: Pipe 1 (closed at one end) 4L 1 = 2L 2 L 1 = 1 2 L 2 node antinode l = 4L 1 L1 Pipe 2 (open at both ends) antinode l = 2L 2 antinode L 2 7. C One way to approach this problem is to examine each answer choice indiidually to determine whether it proides 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 waelength of the fundamental frequency (n = 1) is proportional to the length of the pipe: λ 1 = 4L or 2L depending on which pipe we are talking about. Furthermore, since frequency is inersely proportional to the waelength, the fundamental frequency is inersely proportional to the length of the pipe: f 1 = 4L or. Therefore, the pitch pipe, which is said to be much shorter than the organ pipe, will hae a much higher 2L 10 as deeloped by

11 Sound Test 1 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, the waelength is λ = 4L n or 2L n ; the frequency is therefore f = λ = n n or. The frequency is therefore proportional to the 4L 2L number of the harmonic. So a higher harmonic of the pitch pipe will hae 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 reerses the situation gien 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 hae 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 hae 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, howeer, the difference between the organ pipe and pitch frequencies must be that of the organ pipe fundamental. This, howeer, contradicts what we just said about the two tones haing to be close in frequency to gie rise to beats in the first place. So choice D is incorrect, and choice C is the answer. Passage II (Questions 8 12) 8. B We hae a sound source that is moing past an obserer and are asked to predict how the perceied frequency changes. Instead of turning to the formula immediately, let us see how far we can go by qualitatie reasoning. When a sound is emitted from a source that is approaching the obserer, the wae crests are more bunched up and arrie at the obserer more frequently. The obserer consequently hears a higher frequency than if there were no motion. On the other hand, if the source is moing away, each successiely emitted wae crest takes a longer time interal to reach the obserer, and so the perceied frequency of the wae 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 moing 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 perceied frequency gradually increasing then gradually decreasing. This is howeer wrong. The perceied 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 perspectie. 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 s has been reersed in the formula. Choice B is therefore correct. Be sure to distinguish between olume and the frequency the olume will increase and then decrease gradually, but the frequency perceied does not. 9. C This is a straightforward application of the Doppler formula. The one slightly tricky part is that the frequency gien in the stem is the perceied frequency, f. We can first of all eliminate choice D since the source is approaching the obserer: the emitted frequency must be lower than the perceied one. You may in fact choose choice C simply by ealuating the magnitude of the numbers: the source speed, which is the elocity of the police car, is s = 10 m/s. This is a relatiely small percentage of the speed of sound, = 340 m/s as gien in the passage. It is unreasonable, therefore, to expect the perceied frequency to change by factors of 10 and 100, as suggested by choices B and A respectiely. The minor adjustment gien in choice C seems much more reasonable. The actual set-up is: f 340 m/s f = 1030 Hz = (340 m/s 10 m/s) = f f = Hz 34 KAPLAN 11

12 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 b and emits a frequency f. The wall, then, detects a frequency f as gien by the Doppler formula, with s = b : f f = ( 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 moing obserer. The releant formula for this scenario is gien in the question stem, and so the frequency perceied by the bat, f, is gien by: f = f ( + b) = f ( b ) ( + b) = f ( + b) ( 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 gien in the passage: f f = ( s ) Generally, for this type of question there are only a few points to consider to arrie at the correct graph. First, it is always a good idea to look at the behaior of the cure as the independent ariable (the one plotted on the x-axis, in this case s ) goes to infinity. This particular formula, howeer, is described in the passage as being alid only for s <. In other words, the elocity of the source is always less than the elocity of the sound. As s gets closer and closer to, we see that the denominator gets closer and closer to zero (while remaining positie since s < ). Something diided by a ery small number gies a ery 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 s = 0, the formula boils down to f = f: both the source and the obserer are stationary and so there is no reason for the perceied frequency to be different from the emitted one. The graph in choice A is the only one that shows f haing a nonzero, finite alue at s = B Although seemingly difficult, the answer can be obtained just by examining Figure 2 and using trigonometry, instead of any scientific principle. The angle θ is part of a right triangle, with the hypotenuse being s t. The side opposite to the angle has length t. 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 oer the hypotenuse, is therefore: sin θ = t = = : s t s s This is the ratio of the speed of sound to the speed of the source. Independent Questions 13. A When the string ibrates, it generates perturbations among the air molecules and causes them to ibrate. Specifically, the string ibrations excite compression waes in the air which trael 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 ibrations. Choice B is thus incorrect. 12 as deeloped by

13 Sound Test 1 Since we are told that the waelengths 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 wae is independent of its waelength. Choice D is incorrect because een though the statement is true, it is irreleant here. 14. D Again, it is important to keep in mind that as a wae goes from one medium to another, its speed and waelength may change, but its frequency will remain constant. As the wae in this question moes from medium 1 to medium 2, it is said that its elocity increases by 30%. Conerting this percentage into numbers, we can write that the elocity of the wae in medium 2 is 1.3 times that in medium 1. Since, as we hae said, the frequency does not change, and since = fλ, the elocity is proportional to the waelength, and a elocity in medium 2 of 1.3 times that of the old one means that the waelength in medium 2 has to be 1.3 times as great as the old waelength as well. 15. C Sound leel is the intensity measured in a logarithmic scale. More specifically. the sound leel β is defined by β =10 log( I I 0 ), where I is the intensity of the sound and I 0 some reference intensity. For this question, we need to express the intensity in terms of the sound leel. Rearranging the equation aboe, we hae: log( I ) = β I 0 10 I = 10 I (β/10) 0 I = I 0 10 (β/10) So a sound of 30 db has an intensity of I = 1000I 0 ; while a sound of 10 db has an intensity of I = 10I 0. 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 eery 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 = B This question calls for a straightforward application of the formula fλ =. Notice that the elocity is gien in meters per second, while the waelength is gien in centimeters. The easiest thing to do would be to conert the waelength into meters: f = λ = 340 m/s 0.85m = 400 s 1 = 400 Hz 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 wae fluctuates with maximum amplitude, while a node is a point at which the standing wae is always at zero displacement. For this to be the case, the length of the pipe must be some odd multiple of a quarter-waelength: L = n λ 4, where n = 1, 3, 5, KAPLAN 13

14 MCAT node antinode λ = 4L L node antinode λ = (4/3)L etc. The frequencies for resonance are therefore gien by: f = n 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.: 300 Hz = 4L The next mode of resonance occurs for n = 3, i.e. f (n=3) = 3. This is three times the fundamental frequency, and so 4L the next frequency to allow for resonance is = 900 Hz. 18. B We hae a car and a train traeling in the same direction at the same speed (30 m/s). The source and the obserer are thus neither approaching nor moing away from each other. The frequency perceied by the car will thus not be any different from the emitted frequency, 1000 Hz. 14 as deeloped by