2Wave characteristics and periodic motion test w solutions

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PHYSICS TOPICAL: Wave Characteristics and Periodic Motion Test Time: 21 Minutes* Number of Questions: 16 * 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 Wave Characteristics and Periodic Motion Test Passage I (Questions 1–6) A woman lets herself drop from a bridge above a river Attached firmly to her waist is one end of a strong elastic cord (a “bungee cord”) of negligible mass with an unstretched length of L = 20 m The other end is secured to the bridge at the point from which she drops The height of the bridge above the river is 50 m At any point in time, the variable d represents the woman’s distance from the bridge When the woman hangs motionless in equilibrium, her distance from the bridge is deq = 25 m The tension in the bungee cord obeys Hooke’s law, T = –k(d–L), where k is the force constant equal to 100 N/m The total force on the woman is F = mg + T, and it also obeys Hooke’s law, F = –k(d–deq) However, since the bungee cord only supplies a force when it is stretched, Hooke’s law is only satisfied when d is larger than L When d is smaller than L, the bungee is slack, and there is no tension in it at all As the woman falls from the bridge, the bungee cord begins to stretch as she reaches d = 20 m The tension in the cord decelerates the woman until her velocity reaches zero, and she is momentarily at rest At this point her distance from the bridge is dmax = 40 m Since there is now an upward force on her, she begins to accelerate back towards the bridge The woman undergoes a number of such oscillations before finally coming to rest at d = deq Below is a graph of the tension in the bungee cord as a function of time (Note: The potential energy stored in the bungee cord when it is stretched a distance (d–L) is equal to k(d–L)2/2 g, the acceleration due to gravity, is approximately 10 m/s2.) At a certain point in the woman’s fall, her acceleration is momentarily zero What is the value of d at this point? A B C D 10 20 25 40 m m m m What is the woman’s weight? A 50 kg B 100 kg C 500 N D 1000 N How would the graph in Figure look if there were no frictional effects? A The peaks would have the same amplitude since energy is conserved B The maximum tension would decrease since it does not have to counteract the effects of friction C The oscillations would have the same amplitude, but would also swing to negative values to truly represent periodic motion D It would appear the same since the tension and friction forces are independent What is the weight of the woman if she just touches the river at dmax? (Note: Assume negligible energy loss due to frictional forces.) A B C D 600 N 900 N 1200 N l500 N If both the weight of the woman and the force constant k are doubled, then the force on her at a given distance, d, will: (Note: Assume negligible energy loss due to frictional forces.) Figure 1 Why does the woman NOT hit the bridge on her way back up from dmax? A The force constant k is too small B Momentum is not conserved C The force of gravity prevents her from reaching the bridge D Energy is not conserved due to the frictional force of air resistance KAPLAN A B C D be cut in half double quadruple remain unchanged GO ON TO THE NEXT PAGE MCAT Questions through 11 are NOT based on a descriptive passage A person rubs a wet finger around the rim of a thin crystal glass partially filled with water The sound made by the glass is most closely related to which of the following phenomena? A B C D Doppler effect Interference Resonance Refraction 1 When a spring is compressed to its minimum length and NOT permitted to expand: A potential energy is maximum and kinetic energy is minimum B kinetic energy is maximum and potential energy is minimum C potential energy and kinetic energy are maximum D potential energy and kinetic energy are minimum A mass of kg is suspended from a spring of negligible mass When displaced from its equilibrium position it oscillates with a frequency of Hz What is the spring constant of the spring? A π2 N/m B π2 N/m C 4π2 N/m D 16π2 N/m Which of the following will decrease the period of a pendulum swinging with simple harmonic motion? A B C D Increasing the amplitude of the arc Increasing the mass of the bob Increasing the length of the pendulum Increasing the acceleration due to gravity What is the wavelength of a wave with a period of 0.004 seconds traveling at 32 m/s? A B C D 0.128 m 0.256 m 1.28 m 2.56 m GO ON TO THE NEXT PAGE as developed by Wave Characteristics and Periodic Motion Test Passage II (Questions 12–16) A simple model for studying the vibrations of atoms in a solid is the model of the one-dimensional chain shown in Figure The chain consists of N atoms, each of mass m , and each attached to its two neighboring atoms by springs with spring constant k In the absence of vibrations, each atom is located at its equilibrium position For the chain of Figure 1, equilibrium means that each atom is spaced a distance a from its two neighboring atoms k j=2 j=N m m m k A B C D a An λn L A standing wave of wavelength λ n = 4a exists on a chain of atoms At a given instant, the atom labeled j = has a displacement of Dn from its equilibrium position Which atom listed below has this same displacement? a j=1 Assume there is a standing wave of wavelength λ n on the chain What is the maximum displacement from equilibrium that an atom can experience during its motion in this wave? k A B C D Figure The relationship between the wavelength of a given standing wave, labeled by n, and its frequency is: f = (k/π2m) sin2(πa/λn), where f is the frequency, λ n is the wavelength of the nth standing wave, and a is the interatomic spacing This formula is called a dispersion relation Because the endpoints of the chain are fixed, the wavelength can only take N values given by λ n = 2L/n, where L is the length of the chain and n = 1, 2, , N The displacement, D n, of the jth atom from its equilibrium position as a function of time is given by the formula: Dn = An sin(2πaj/λn) sin(2πft), Figure shows the graph of the displacement of the fourth atom in a chain of 100 atoms What is the frequency of the wave? 20 displacement (10-12 m) Just as there are characteristic vibrations of an organ pipe, so are there characteristic vibrations of the atoms in the one-dimensional chain Since the chain is fixed at both ends, these vibrations form standing waves with definite frequency and wavelength Because the one-dimensional chain consists of only N atoms, there are only N independent standing waves that can exist on the chain Any arbitrary traveling wave is a linear combination of these The atom labeled j = The atom labeled j = The atom labeled j = No other atom 15 10 - 50 200 400 600 -10 -15 -20 time (10-14 s) Figure × 1012 Hz × 1011 Hz 1.5 × 104 Hz A B C D 200 Hz where An is the amplitude of the nth wave, t is the time, f is the frequency which is related to λ n by the dispersion relation and j = 1, 2, , N GO ON TO THE NEXT PAGE KAPLAN MCAT How does the speed of a wave of wavelength λ n = 3a depend on the interatomic spacing, a? A The speed is directly proportional to the interatomic spacing B The speed is proportional to the cube of the interatomic spacing C The speed is inversely proportional to the interatomic spacing D The speed does not depend on the interatomic spacing In the one-dimensional chain model, what physical aspect of the solid does the spring constant k represent? A B C D The size of the atom The atomic number of the atom The boiling point of the solid The strength of the bond between adjacent atoms END OF TEST as developed by Wave Characteristics and Periodic Motion Test THE ANSWER KEY IS ON THE NEXT PAGE KAPLAN MCAT ANSWER KEY: D B C C C D A D B 10 A 11 12 13 14 15 A B C B A 16 D as developed by Wave Characteristics and Periodic Motion Test EXPLANATIONS Passage (Questions 1–6) D When frictional forces are present, the total mechanical energy (kinetic + potential) of a system is not conserved but is instead gradually dissipated At the top of the bridge, the woman’s initial total mechanical energy is all in the form of gravitational potential energy As the woman falls energy is converted from potential to kinetic as she picks up speed from gravitational acceleration Once the cord starts to stretch, gravitational potential and kinetic energies are converted to the potential energy of the cord that is analogous to the stretching of a spring At the jumper’s maximum distance from the bridge, her total energy is all in the form of potential energy stored in the bungee cord (There may of course still be some gravitational potential energy but we can set the coordinate system such that the lowest position corresponds to zero gravitational potential energy.) As the bungee cord starts to compress and pulls the woman up towards the bridge again, the potential energy of the bungee cord is converted back into kinetic energy and (then to) gravitational potential energy If energy were conserved, the jumper would indeed recover all of the initial energy as gravitational potential energy and return to her initial position, hitting the bridge as she does so However, because of the dissipation of energy by the frictional effects of air resistance, the total amount of energy that can be recovered as gravitational potential energy is less than the initial value and therefore she cannot reach the bridge again Choice A states that the force constant is too small If energy is conserved, the woman will hit the bridge after one bounce no matter what the force constant is: a smaller value of k just means that the cord will stretch farther So choice A is incorrect Choice B states that momentum is not conserved The momentum of a system is conserved if there are no external forces acting on it In this case, the momentum of the woman is not conserved because of the external forces of gravity and the tension force of the cord: as the woman falls, for example, she picks up momentum as she accelerates downward The statement in choice B is correct, but it has nothing to with the reason why she does not hit the bridge Choice C attributes her inability to return to her initial position to the force of gravity As we have seen, however, if energy is conserved she would hit the bridge in spite of gravity Gravity is a conservative force and, unlike friction, does not dissipate energy C This question actually requires no calculation We know from Newton’s second law that if the acceleration is zero then the net force acting on the object must be zero, since the two are related by F net = ma In this particular case, then, the upward pull on the woman by the bungee cord must equal the downward pull of gravity This is the same point where the woman would just hang motionless in equilibrium Therefore, the point where her acceleration is zero is equal to d = deq = 25 m as given in the passage Choice B may be tempting because 20 m is the length of the unstretched cord This, however, is not the equilibrium length because the weight of the woman will stretch out the cord Choice D corresponds to dmax; at this point the woman’s instantaneous velocity is zero, but not her acceleration C The easiest way to this problem is to use the information given about the woman’s equilibrium position At the point where d = deq, the downward force of gravity (i.e her weight) is equal to the upward tension force of the bungee cord as given by Hooke’s law: Mg = k(deq – L) where k is the force constant of the cord and (deq – L) represents the extension of the cord from its unstretched position Note that we have omitted the negative sign in front of Hooke’s law since it is the magnitude of the forces that are equal Also, the right hand side is obtained by substituting deq for d in the equation for tension A later equation given in the passage, F = –k(d – deq), is not the one we want to use here because that equation gives the net force on the woman in places where d > L, not simply the force of the bungee cord itself Anyway, substituting in numbers into the equation above, we have: Mg = 100 N/m × (25 m – 20 m) = 500 N Be careful! The question asks for the woman’s weight and not her mass Therefore choice C, and not choice A, is correct KAPLAN MCAT A If there were no frictional effects present, energy would be conserved, and so after each period the woman would return to the same position Since tension is directly proportional to the displacement (at least for d > L), the oscillations in the tension would also maintain the same amplitude One can also look at it from the perspective of potential energy: the points of maximum tension correspond to those points where the woman’s gravitational potential energy and kinetic energy have been converted into the potential energy of the bungee cord Since energy is conserved in the absence of dissipative mechanisms, the potential energy of the cord, and hence the tension, would reach the same maximum value each time All the other answer choices are accompanied by a seemingly plausible explanation as to why they might be correct Choice B is incorrect because if anything, the maximum tension would increase since the cord may be stretched farther at maximum amplitude Choice C is incorrect because unlike the case of a normal spring, the tension of a bungee cord cannot pull in the opposite direction (downward) This is stated in the second paragraph of the passage: tension is supplied only when the cord is stretched, and the minimum value of the tension force is zero which applies whenever d ≤ L (A spring is different because it supplies a restoring force not only when it is stretched but also when compressed.) The tension would therefore never take on a negative value; this is true regardless of whether air resistance is present or not Finally, choice D is incorrect because even though tension and friction are separate forces, the two are related in that friction can alter the maximum stretching distance (and hence potential energy) of the cord and that would affect tension B Energy conservation is once again the most efficient way to approach the problem, which is why we need to make the assumption that frictional effects are negligible Before the jump the woman is standing on the bridge and has a gravitational potential energy of Mgh, where h is the height of the bridge above the river (We have chosen to set the zero for potential energy to be at the river level.) There is no kinetic energy and no potential energy of the cord At the point of maximum displacement, the kinetic energy will again be zero, but she will possess potential energy In general, this potential energy will both be gravitational and in the cord, since she may still be at a certain height above the river at that point: PEmax = k (dmax – L)2 + Mg (h – dmax) where the first term is the potential energy of the cord and the second is the gravitational potential energy dmax h h – dmax If the woman is to just touch the river at dmax, though, it must be the case that dmax = h: all of the initial gravitational potential energy is converted into potential energy of the cord The condition to be satisfied is thus: Mgh = k (h – L)2 where the left hand side is the initial gravitational potential energy Substituting in numbers: × 100 N/m × (50 m – 20 m)2 50 × 30 Mg = = 900 N 50 Mg × 50 m = 10 as developed by Wave Characteristics and Periodic Motion Test B The force on the woman is given by the equation F = mg + T, where T is equal to –k(d–L) when d > L The force on the woman under this condition can therefore be written as: F = Mg – k (d – L) (d > L) If both the weigh and the force constant are doubled, then, the new force would be F’ = 2Mg – 2k (d – L) = (Mg – k (d – L)) =2F Hence the force is doubled We also need to consider the case where d ≤ L When this is true, tension is equal to zero, and thus the force on the woman is simply the gravitational force: F = mg Under such conditions, if the weight is doubled the force is again doubled So choice B is correct Independent Questions (Questions 7–11) C This question describes a commonly known phenomenon — the “singing” of a crystal glass when rubbed with a wet finger — and asks you to identify the underlying physical principle responsible for it Resonance is the correct answer When a force is exerted at a frequency that is identical to one of the normal mode or natural frequencies of the system, causing the oscillations in the system to increase in amplitude drastically, the system is said to be in resonance The oscillations may be in the form of mechanical oscillations, for example, as in the case of the collapse of the Tacoma Narrows Bridge, or in the form of sound waves, as in the vibrations of the crystal glass here Choice A, Doppler effect, refers to a shift in the frequency of the sound waves perceived because of the relative motion between source and observer Since we not have a moving source or observer here this is incorrect Choice B, interference, arises when two waves interact, either reinforcing or canceling each other We not have two sources of waves here and thus this choice is incorrect Choice D, refraction, is the bending of the direction in which a wave travels when it enters into a different medium This occurs because the speed of a wave is dependent on the medium We have two different media here — crystal and air— but invoking refraction does not address the issue of why sound wave arises in the first place, so this too is incorrect D k , where m m is the mass attached to the spring To relate the angular frequency, measured in radians, to the more familiar frequency f measured in Hz or inverse seconds, we use the relation ω = 2πf Putting the two together and rearranging to solve for k, one obtains: A fundamental relationship exists between the spring constant k and the angular frequency ω: ω = 2πf = k m k = 2πf m k = 4π2f2m Substituting in the numbers given in the passage: k = 4π2(2 s-1)2(1 kg) = 16π2 kg s2 = 16π2 N/m where the change in units in the last step could be made since a newton is a kg•m/s2 KAPLAN 11 MCAT D The swinging of a pendulum can be considered as simple harmonic motion, provided that the angle of swing is not too large The restoring force (provided by gravity) can be modeled as proportional to the displacement, just as in the case of a spring The period of the pendulum is the time it takes to complete one full oscillation and the general formula for the period is (again assuming small amplitude): L g T = 2π where g is the gravitational acceleration and L the length of the pendulum arm From this expression, we see that if the acceleration due to gravity were increased (by transporting it to a more massive planet, for example), the period would decrease Choices A and B involve changing parameters that have no effect on the period (If the amplitude of the arc were increased sufficiently, then our modeling of the motion of the pendulum would no longer be applicable, but the question specifically states that we are in the realm of simple harmonic motion, in which case the period is independent of the (generally small) amplitude.) Choice C, increasing the length of the pendulum, would increase rather than decrease the period 10 A A very important and basic equation on waves is v = fλ, where v is the speed, f is the frequency, and λ the wavelength of the wave In this question, however, we are not given the frequency but the period of the wave The period is the inverse of frequency: T = So our equation becomes: f v= λ T or, solving for wavelength: λ = vT = 32 m/s × 0.004 s = 0.128 m The equations above should be familiar to you, but even had one forgotten them, dimensional analysis would still be helpful here We are after wavelength which is measured in units of length (meters) We are given a quantity with units of meters per second, and another one with units of seconds Multiplying the two would give us a quantity in meters 11 A As discussed in the explanations to the questions of Passage I, an oscillating system has two kinds of energy: kinetic and potential Kinetic energy is energy associated with motion Since its value is equal to (1/2) mv2, it can never be negative A stationary object will have a kinetic energy of zero which is the minimum possible value Here the spring is not permitted to expand—in other words, it is not allowed to move Its kinetic energy is therefore zero This enables us to eliminate choices B and C Conservation of energy should tell us the rest: The total mechanical energy, which is the sum of kinetic and potential energies, remains constant in the absence of friction If the kinetic energy is at a minimum, as it is in this case, the potential energy must be at a maximum, and vice versa Passage II (Questions 12–16) 12 B In the second paragraph we are told that the characteristic vibrations of the atoms of the chain form standing waves of definite frequency and wavelength Then in the third paragraph we are given an equation which tells us that at time t the jth atom is displaced by an amount An times two sine functions The first sine function has as argument that depends on the wavelength and position (the argument of a function is the quantity that the function is operating on: e.g in the expression sinθ, θ is the argument); the second sine function has as argument something that depends on the frequency and the time The crucial thing to realize is that no matter how complicated the argument looks, the maximum value of the sine function is always The maximum value of the displacement, then, is just An × × = A n Note that since the first sine function depends on the particular atom in question, not all of the atoms necessarily reach a displacement of An 13 C In order to answer this question all we really need is to have a picture of the one-dimensional chain in our mind and be able to visualize what it means to have a standing wave of wavelength 4a on the chain If we freeze time at a given instant, 12 as developed by Wave Characteristics and Periodic Motion Test the wavelength tells us how far we have to travel from a given point until the displacement pattern repeats itself Thus if the wavelength of a wave is 4a then at any given instant we have to go a distance of 4a from a given point to find the exact same displacement In this question, we want to find that atom that has the same displacement as the atom labeled j = We know that the atom that has the same displacement would be at a position 4a away from the j = atom Since the atoms are spaced at a distance a apart, a distance of 4a corresponds to four atoms The atom in question is thus j = + = A more mathematically involved method involves observing the equation for displacement given in the passage The j = atom and the one we are after will differ only in position, and thus in the equation, they would differ only in the argument of the first sine function Our goal is to find that j for which the value of the sine function is the same as that for j = 1, since that would result in the same displacement I.e we want: sin(2πaj/λn) = sin(2πa/λn) Since the sine function has a periodicity of 2π, the two sine functions are equal if their argument differ by some multiple of 2π: 2πaj 2πa = + 2mπ, λn λn m = 0, 1, 2, … Since the wavelength is 4a, we can substitute this as the denominator to obtain: πj π = 2 π + 4mπ πj = π (1 + 4m) j = + 4m j = 1, 5, 9, … + 2mπ = j = is just the atom itself j = is the atom that will have the same displacement as the j = atom, and is the one we are looking for 14 B The key to answering this question correctly lies in the understanding of what the graph is and is not telling us It is a graph of displacement of one atom versus time The distance from crest to crest, then, is the amount of time for the atom to complete one oscillation—i.e it is the period, not the wavelength, of the wave Inspection of the graph tells us that the period is 200 × 10–14 s, or × 10–12 s The frequency is just the reciprocal of the period: f= 1 = = 0.5 × 1012 s-1 = × 1011 s-1 = × 1011 Hz T × 10 –12 s 15 A The speed of a wave is simply the product of wavelength and frequency The question is already telling us that the wavelength is 3a, and so we only need to determine how the frequency is affected In the passage we are given the equation for the frequency Substituting into this equation a value of 3a for the wavelength λ, we get: f2 = (k/π2m)sin2(πa/3a) = (k/π2m)sin2(π/3) Note that the a drops out: the frequency in this case is independent of the interatomic spacing Changing a, then, only affects speed through the wavelength, and since v = (3a) × frequency, the speed is directly proportional to the interatomic spacing 16 D In its usual context the spring constant k represents the stiffness of a spring A spring with a large k is hard to compress and stretch We need now to relate this to a solid What determines how hard it is to get the atoms to vibrate? The stronger the bond between atoms, the more restricted an atom’s movement So we can say that the spring constant is related to the strength of the bond between adjacent atoms KAPLAN 13 ... the solid The strength of the bond between adjacent atoms END OF TEST as developed by Wave Characteristics and Periodic Motion Test THE ANSWER KEY IS ON THE NEXT PAGE KAPLAN MCAT ANSWER KEY: D... mind and be able to visualize what it means to have a standing wave of wavelength 4a on the chain If we freeze time at a given instant, 12 as developed by Wave Characteristics and Periodic Motion. .. C D 0.128 m 0.256 m 1.28 m 2.56 m GO ON TO THE NEXT PAGE as developed by Wave Characteristics and Periodic Motion Test Passage II (Questions 12–16) A simple model for studying the vibrations of

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