CHAPTER
10 SIMPLE HARMONIC MOTION
AND ELASTICITY
CONCEPTUAL QUESTIONS
4.
REASONING AND SOLUTION Simple harmonic motion is the oscillatory motion
that occurs when a restoring force of the form of Equation 10.2, F  kx , acts on an
object. The force changes continually as the displacement x changes.
A steel ball is dropped onto a concrete floor. Over and over again, it rebounds to its
original height. During the time when the ball is in the air, either falling down or
rebounding up, the only force acting on the ball is its weight, which is constant.
Thus, the motion of the bouncing ball is not simple harmonic motion.
7.
REASONING AND SOLUTION
The time required for a particle in simple
harmonic motion to travel through one complete cycle (the period) is independent of
the amplitude of the motion, even though at larger amplitudes the particle travels
further. This is possible because, at larger amplitudes, the maximum speed of the
particle is greater. Thus, even though the particle must cover larger distances at
larger amplitudes, it does so with greater speeds.
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9.
REASONING AND SOLUTION The elastic potential energy that a spring has by
virtue of being stretched or compressed is given by Equation 10.13:
PE elastic  (1/ 2)kx2 , where x is the amount by which the spring is stretched or
compressed relative to its unstrained length. The amount of stretch or compression
appears squared, so that the elastic potential energy is positive and independent of
the sign of x. Therefore, the amount of elastic potential energy stored in a spring
when it is compressed by one centimeter is the same as when it is stretched by the
same amount.
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13. REASONING AND SOLUTION The playground swing may be treated, to a good
approximation, as a simple pendulum. The period of a simple pendulum is given by
T  2 L / g . This expression for the period depends only on the length of the
pendulum and the acceleration due to gravity; for angles less than 10° the period is
independent of the amplitude of the motion. Therefore, if one person is pulled back
4° from the vertical while another person is pulled back 8° from the vertical, they
will both have the same period. If they are released simultaneously, they will both
come back to the starting points at the same time.
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15. REASONING AND SOLUTION The amount of force F needed to stretch a rod is
given by Equation 10.17: F  Y( L / L0 )A , where A is the cross-sectional area of the
rod, L0 is the original length, L is the change in length, and Y is Young's modulus
of the material.
Since the cylinders are made of the same material, they have the same Young's
modulus. However, the cross-sectional area of the material in the hollow cylinder is
smaller than that of the solid cylinder since most of its cross section is empty. When
identical forces of magnitude F are applied to the right end of each cylinder, each
stretches by an amount L  (FL0 )/(YA) . Since the hollow cylinder has a smaller
cross-sectional area of material compared to the solid cylinder, the hollow cylinder
will stretch the most.
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PROBLEMS
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2.
REASONING The weight of the person causes the spring in the scale to
compress. The amount x of compression, according to Equation 10.1, depends on the
magnitude FApplied of the applied force and the spring constant k.
SOLUTION
a.
Since the applied force is equal to the person’s weight, the spring constant
is
k
FApplied
x

670 N
0.79  10
2
 8.5  104 N / m
m
(10.1)
b.
When another person steps on the scale, it compresses by 0.34 cm. The
weight (or applied force) that this person exerts on the scale is



FApplied  k x  8.5  104 N / m 0.34  102 m  290 N
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6.
REASONING The weight of the block causes the spring to stretch. The amount x
of stretching, according to Equation 10.1, depends on the magnitude FApplied of the
applied force and the spring constant k.
SOLUTION
a. The applied force is equal to the weight of the block. The amount x that the
spring stretches is equal to the length of the stretched spring minus the length of the
unstretched spring. The spring constant is
(10.1)
k
FApplied
x

4.50 N
 30.0 N / m
0.350 m  0.200 m
(10.1)
b.
When a block of unknown weight is attached to the spring, it stretches it
by 0.500 m – 0.200 m. The weight (or applied force) of the block is
FApplied  k x   30.0 N / m  0.500 m  0.200 m   9.00 N
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17.
REASONING AND SOLUTION
a.
Since the object oscillates between  0.080 m , the amplitude of the
motion is 0.080 m .
b. From the graph, the period is T  4.0 s . Therefore, according to Equation 10.4,

2
2

 1.6 rad/s
T
4.0 s
c. Equation 10.11 relates the angular frequency to the spring constant:
  k / m . Solving for k we find
k   2 m  (1.6 rad/s) 2 (0.80 kg)  2.0 N/m
d. At t  1.0 s , the graph shows that the spring has its maximum displacement. At
this location, the object is momentarily at rest, so that its speed is v  0 m/s .
e. The acceleration of the object at t  1.0 s is a maximum, and its magnitude is
a max  A  2  (0.080 m)(1.6 rad/s) 2 = 0.20 m/s 2
18.
REASONING The amplitude of simple harmonic motion is the distance from the
equilibrium position to the point of maximum height. The angular frequency  is related
to the period T of the motion by Equation 10.6. The maximum speed attained by the
person is the product of the amplitude and the angular speed (Equation 10.8).
SOLUTION
a. Since the distance from the equilibrium position to the point of maximum
height is the amplitude A of the motion, we have that A = 45.0 cm = 0.450 m .
(10.1)
b.
motion:
The angular frequency is inversely proportional to the period of the

2
2

 3.31 rad /s
T
1.90 s
(10.6)
c.
The maximum speed vmax attained by the person on the trampoline
depends on the amplitude A and the angular frequency  of the motion:
vmax  A   0.450 m 3.31 rad /s   1.49 m/s
(10.8)
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24.
REASONING AND SOLUTION
a. PE = (1/2) kx2 = (1/2) (425 N/m)(0.470 m)2 = 46.9 J
(10.13)
b.
Assuming that mechanical energy is conserved, KE = PE gives
2(PE)
2(46.9 J)

 55.9 m/s
m
0.0300 kg
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v
SSM REASONING AND SOLUTION If we neglect air resistance, only the
29.
conservative forces of the spring and gravity act on the ball. Therefore, the principle of
conservation of mechanical energy applies.
When the 2.00 kg object is hung on the end of the vertical spring, it stretches the
spring by an amount x, where
x
F mg (2.00 kg)(9.80 m/s 2 )


 0.392 m
k
k
50.0 N/m
This position represents the equilibrium position of the system with the 2.00-kg
object suspended from the spring. The object is then pulled down another 0.200 m and
released from rest ( v0  0 m/s). At this point the spring is stretched by an amount of
0.392 m + 0.200m = 0.592 m . This point represents the zero reference level ( h  0 m)
for the gravitational potential energy.
h = 0 m: The kinetic energy, the gravitational potential energy, and the elastic
potential energy at the point of release are:
(10.1)
KE  12 mv02  12 m(0 m/s)2  0 J
PEgravity  mgh  mg (0 m)  0 J
PE elastic 
1 2 1
2
kx  (50.0 N/m)(0.592 m)  8.76 J
2 0
2
The total mechanical energy E0 at the point of release is the sum of the three
energies above: E0  8.76 J .
h = 0.200 m: When the object has risen a distance of h  0.200 m above the
release point, the spring is stretched by an amount of 0.592 m – 0.200 m = 0.392 m .
Since the total mechanical energy is conserved, its value at this point is still
E  8.76 J . The gravitational and elastic potential energies are:
2
PE gravity  mgh  (2.00 kg)(9.80 m/s )(0.200 m)  3.92 J
PE elastic 
1 2 1
2
kx  (50.0 N/m)(0.392 m)  3.84 J
2
2
Since KE  PEgravity  PEelastic  E ,
KE  E – PEgravity – PEelastic  8.76 J – 3.92 J – 3.84 J = 1.00 J
h = 0.400 m: When the object has risen a distance of h  0.400 m above the
release point, the spring is stretched by an amount of 0.592 m – 0.400 m = 0.192 m . At
this point, the total mechanical energy is still E  8.76 J . The gravitational and elastic
potential energies are:
2
PE gravity  mgh  (2.00 kg)(9.80 m/s )(0.400 m)  7.84 J
PE elastic 
1 2 1
2
kx  (50.0 N/m)(0.192 m)  0.92 J
2
2
The kinetic energy is
KE  E – PEgravity – PEelastic  8.76 J – 7.84 J – 0.92 J= 0 J
The results are summarized in the table below:
h
KE
PE grav
PE elastic
E
0.000 m
0.200 m
0.400 m
0.00 J
1.00 J
0.00 J
0.00 J
3.92 J
7.84 J
8.76 J
3.84 J
0.92 J
8.76 J
8.76 J
8.76 J
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41.
SSM REASONING AND SOLUTION Recall that the relationship between
frequency f and period T is f 1/ T . Then, according to Equations 10.6 and 10.16, the
period of the simple pendulum is given by
L
T  2
g
where L is the length of the pendulum. Solving for g, we obtain
4 2 (1.2 m)
2

2
2  6.0 m/s
T
(2.8 s)
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g
4 2 L
48.
REASONING The shear stress is equal to the magnitude of the shearing force
exerted on the bar divided by the cross sectional area of the bar. The vertical deflection
Y of the right end of the bar is given by Equation 10.18 [ F  S(Y / L0 )A ].
SOLUTION
a. The stress is
F mg (160 kg)(9.80 m/s 2 )


 4.9  106 N/m 2
–4 2
A
A
3.2  10 m
b. The vertical deflection Y of the right end of the bar is
0.10 m
F L0
6
2
–6
Y   

(4.9

10
N/m
)
10
2  6.0 10 m
A  S
8.1 10 N/m
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51. REASONING AND SOLUTION We know that
P = B(V/V0) = 2.6  1010 N/m2)(1.0  10–10 m3)/(1.0  10–6 m3)

P = 2.6  106 N/m2
(10.20)
Since the pressure increases by 1.0  104 N/m2 per meter of depth, the depth is
2.6  10 6 N/m 2
 260 m
2
4 N/m
1.0  10
m
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54.
REASONING Both cylinders experience the same compressive force F. The
magnitude of this force is related to the change in length of each cylinder according to
Equation 10.17: F  Y(L / L0 )A . Each cylinder will decrease in length; the total
decrease in the length of the stack is equal to the sum of the decrease in length of each
cylinder.
SOLUTION The length of the copper cylinder decreases by
Lcopper 
FL0
FL0
(6500 N)(3.0  10 –2 m)


 9.0  10 –5 m
2
11
2
–2
2
YA
Y(r ) (1.1 10 N/m ) (0.25  10 m)
Similarly, the length of the brass decreases by
Lbrass 
(6500 N)(5.0  10 –2 m)
 1.8  10 –4 m
10
2
–2
2
(9.0  10 N/m ) (0.25 10 m)
Therefore, the amount by which the length of the stack decreases is
2.7  10 –4 m .
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74. REASONING AND SOLUTION
a. When the block passes through the position x = 0 m, its velocity is a maximum
and can be found from Equation 10.8: vmax = A.
We can find the angular frequency  from the following reasoning: When the
mass is given a displacement x, one spring is stretched by an amount x, while the other is
compressed by an amount x. The total restoring force on the mass is, therefore,
F = k1x  k2x = (k1 + k2)x
Comparison with Equation 10.2 shows that the two spring system has an effective
spring constant keff = k1 + k2. Thus, from Equation 10.11

keff

m
k1  k2
m
Combining this with Equation 10.8 we obtain
v max  A
k1  k 2
m
 (0.070 m)
650 N/m  450 N/m
 1.3 m/s
3.0 kg
b. The angular frequency of the system is
k1  k 2
650 N/m  450 N/m
 19 rad/s
m
3.0 kg
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