Note: The potential energy stored in the bungee cord when it is stretched a distance d–L is equal to kd–L2/2.. If both the weight of the woman and the force constant k are doubled, then
Trang 1PHYSICS TOPICAL:
Wave Characteristics
and Periodic Motion Test 1
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
Trang 2
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
1
H
1.0
2
He
4.0 3
Li
6.9
4
Be
9.0
5
B
10.8
6
C
12.0
7
N
14.0
8
O
16.0
9
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.
Trang 3Passage 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.)
Figure 1
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
2 At a certain point in the woman’s fall, her
acceleration is momentarily zero What is the value of
d at this point?
A 10 m
B 20 m
C 25 m
D 40 m
3 What is the woman’s weight?
A 50 kg
B 100 kg
C 500 N
D 1000 N
4 How would the graph in Figure 1 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
5 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 600 N
B 900 N
C 1200 N
D l500 N
6 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.)
A be cut in half.
B double.
C quadruple.
D remain unchanged.
GO ON TO THE NEXT PAGE.
Trang 4
Questions 7 through 11 are NOT
based on a descriptive passage
7 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 Doppler effect
B Interference
C Resonance
D Refraction
8 A mass of 1 kg is suspended from a spring of
negligible mass When displaced from its equilibrium
position it oscillates with a frequency of 2 Hz What
is the spring constant of the spring?
A π2
4 N / m
B π2 N/m
C 4π2 N/m
D 16π2 N/m
9 Which of the following will decrease the period of a
pendulum swinging with simple harmonic motion?
A Increasing the amplitude of the arc
B Increasing the mass of the bob
C Increasing the length of the pendulum
D Increasing the acceleration due to gravity
1 0 What is the wavelength of a wave with a period of
0.004 seconds traveling at 32 m/s?
A 0.128 m
B 0.256 m
C 1.28 m
D 2.56 m
1 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
GO ON TO THE NEXT PAGE.
Trang 5Passage 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 1 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
a
Figure 1
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 relationship between the wavelength of a given
standing wave, labeled by n, and its frequency is:
f 2 = (k/π2m) sin2(πa/λn),
where f is the frequency, λn 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 λn = 2L/n, where L is the length
of the chain and n = 1, 2, , N The displacement, Dn, 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),
where An is the amplitude of the nth wave, t is the time, f
is the frequency which is related to λn by the dispersion
relation and j = 1, 2, , N.
1 2 Assume there is a standing wave of wavelength λn on the chain What is the maximum displacement from equilibrium that an atom can experience during its motion in this wave?
A a
B An
C λn
D L
1 3 A standing wave of wavelength λn = 4a exists on a
chain of atoms At a given instant, the atom labeled
j = 1 has a displacement of Dn from its equilibrium position Which atom listed below has this same displacement?
A The atom labeled j = 2
B The atom labeled j = 4
C The atom labeled j = 5
D No other atom
1 4 Figure 1 shows the graph of the displacement of the
fourth atom in a chain of 100 atoms What is the frequency of the wave?
- 2 0
- 1 5
- 1 0
- 5 0 5
1 0
1 5
2 0
time (10 -14 s) Figure 1
A 2 × 1012 Hz
B 5 × 1011 Hz
C 1.5 × 104 Hz
D 200 Hz
GO ON TO THE NEXT PAGE.
Trang 6
1 5 How does the speed of a wave of wavelength λn = 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
1 6 In the one-dimensional chain model, what physical
aspect of the solid does the spring constant k
represent?
A The size of the atom
B The atomic number of the atom
C The boiling point of the solid
D The strength of the bond between adjacent atoms
END OF TEST
Trang 7THE ANSWER KEY IS ON THE NEXT PAGE
Trang 8
ANSWER KEY:
Trang 9Passage 1 (Questions 1–6)
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 do 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
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 Fnet = 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
The easiest way to do 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
Trang 10
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
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 = 1
2 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
h
dmax
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 = 1
2 k (h – L)2 where the left hand side is the initial gravitational potential energy Substituting in numbers:
Mg × 50 m = 1
2 × 100 N/m × (50 m – 20 m)2
Mg = 50 × 302
50 = 900 N