2Wave characteristics and periodic motion test w solutions

PHYSICS 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.


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
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.

2

as developed by


Wave Characteristics and Periodic Motion Test 1
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.)

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.

B.
C.
D.

10
20
25
40

m
m
m
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.
B.
C.
D.

600 N
900 N
1200 N
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.)
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.

KAPLAN

A.

B.
C.
D.

be cut in half.
double.
quadruple.
remain unchanged.

GO ON TO THE NEXT PAGE.

3


MCAT
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.
B.
C.
D.

Doppler effect
Interference
Resonance
Refraction

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.

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
N/m
4

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.
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

1 0 . 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.

4

as developed by


Wave Characteristics and Periodic Motion Test 1
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 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

j=2

j=N

m

m

m

k

A.
B.
C.
D.

a
An

λn
L

1 3 . A standing wave of wavelength λ n = 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
j=1

1 2 . 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 1

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, 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),

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?
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 = 2
The atom labeled j = 4
The atom labeled j = 5
No other atom

15
10
5
0
- 50

200

400

600

-10
-15
-20

time (10-14 s)

Figure 1
2 × 1012 Hz
5 × 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

5


MCAT
1 5 . 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.

1 6 . 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

6

as developed by


Wave Characteristics and Periodic Motion Test 1

THE ANSWER KEY IS ON THE NEXT PAGE

KAPLAN

7


MCAT
ANSWER KEY:
1. D
6. B
2. C
7. C
3. C
8. D
4. A

9. D
5. B
10. A

8

11.
12.
13.
14.
15.

A
B
C
B
A

16. D

as developed by


Wave Characteristics and Periodic Motion Test 1
EXPLANATIONS
Passage 1 (Questions 1–6)
1.

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 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.
2.

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.
3.

C
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.

KAPLAN

9


MCAT
4.

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.
5.

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 =

1
k (dmax – L)2 + Mg (h – dmax)
2

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 =

1
k (h – L)2
2

where the left hand side is the initial gravitational potential energy. Substituting in numbers:
1
× 100 N/m × (50 m – 20 m)2
2
50 × 30 2
Mg =
= 900 N
50

Mg × 50 m =

10

as developed by


Wave Characteristics and Periodic Motion Test 1
6.

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)
= 2 (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)
7.

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 do 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 do 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 do 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.
8.

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
9.

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
1
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 1. The maximum value of the displacement, then, is just An × 1 × 1 = 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 1
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 = 1. We know
that the atom that has the same displacement would be at a position 4a away from the j = 1 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 = 1 + 4 = 5.
A more mathematically involved method involves observing the equation for displacement given in the passage. The
j = 1 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
2

π + 4mπ
2
πj = π (1 + 4m)
j = 1 + 4m
j = 1, 5, 9, …

+ 2mπ =

j = 1 is just the atom itself. j = 5 is the atom that will have the same displacement as the j = 1 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 2 × 10–12 s. The frequency is just the reciprocal of the period:
f=

1
1
=
= 0.5 × 1012 s-1 = 5 × 1011 s-1 = 5 × 1011 Hz
T
2 × 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

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