Ch10. Simple Harmonic Motion and Elasticity

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Ch10. Simple Harmonic Motion and Elasticity
The Ideal Spring and Simple Harmonic Motion
The constant k is called the spring constant
A spring that behaves according to
is said to be an ideal spring.
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Example 1.  A Tire Pressure Gauge
In a tire pressure gauge, the air in the tire
pushes against a plunger attached to a spring
when the gauge is pressed against the tire valve.
Suppose the spring constant of the spring is k
320 N/m and the bar indicator of the gauge
extends 2.0 cm when the gauge is pressed against
the tire valve. What force does the air in the
tire apply to the spring?
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Conceptual Example 2.  Are Shorter Springs
Stiffer Springs?
A 10-coil spring that has a spring constant k. If
this spring is cut in half, so there are two
5-coil springs, what is the spring constant of
each of the smaller springs?
Shorter springs are stiffer springs.
Sometimes the spring constant k is referred to as
the stiffness of the spring, because a large
value for k means the spring is stiff, in the
sense that a large force is required to stretch
or compress it.
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HOOKES LAW RESTORING FORCE OF AN IDEAL
SPRING The restoring force of an ideal spring
is
 (10.2) 
where k is the spring constant and x is the
displacement of the spring from its unstrained
length. The minus sign indicates that the
restoring force always points in a direction
opposite to the displacement of the spring.
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When the restoring force has the mathematical
form given by F kx, the type of friction-free
motion illustrated in the figure is designated as
simple harmonic motion.
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The maximum excursion from equilibrium is the
amplitude A of the motion. The shape of this
graph is characteristic of simple harmonic motion
and is called sinusoidal, because it has the
shape of a trigonometric sine or cosine function.
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Check Your Understanding 1
A 0.42-kg block is attached to the end of a
horizontal ideal spring and rests on a
frictionless surface. The block is pulled so that
the spring stretches by 2.1 cm relative to its
unstrained length. When the block is released, it
moves with an acceleration of 9.0 m/s2. What is
the spring constant of the spring?
180 N/m
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2.1cm
kx ma
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The restoring force also leads to simple harmonic
motion when the object is attached to a vertical
spring, just as it does when the spring is
horizontal. When the spring is vertical, however,
the weight of the object causes the spring to
stretch, and the motion occurs with respect to
the equilibrium position of the object on the
stretched spring .
mg kd0, which gives d0 mg/k.
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Simple Harmonic Motion and the Reference Circle
Simple harmonic motion, like any motion, can be
described in terms of displacement, velocity, and
acceleration.
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DISPLACEMENT
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For any object in simple harmonic motion, the
time required to complete one cycle is the period
T
Instead of the period, it is more convenient to
speak of the frequency f of the motion, the
frequency being just the number of cycles of the
motion per second.
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One cycle per second is referred to as one hertz
(Hz). One thousand cycles per second is called
one kilohertz (kHz).
is often called the angular frequency.
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VELOCITY
velocity is not constant, but varies between
maximum and minimum values as time passes
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Example 3.   The Maximum Speed of a Loudspeaker
Diaphragm
The diaphragm of a loudspeaker moves back and
forth in simple harmonic motion to create sound.
The frequency of the motion is f 1.0 kHz and
the amplitude is A 0.20 mm.

• What is the maximum speed of the diaphragm?
• Where in the motion does this maximum speed occur?

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(a)
(b) The speed of the diaphragm is zero when the
diaphragm momentarily comes to rest at either end
of its motion x A and x A. Its maximum
speed occurs midway between these two positions,
or at x 0 m.
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Conceptual Example 4.  Moving Lights
Over the entrance to a restaurant is mounted a
strip of equally spaced light bulbs, as Figure
10.13a illustrates. Starting at the left end,
each bulb turns on in sequence for one-half
second. Thus, a lighted bulb appears to move from
left to right.
Once the apparent motion of a lighted bulb
reaches the right side of the sign, the motion
reverses. The lighted bulb then appears to move
to the left, as part b of the drawing indicates.
Thus, the lighted bulb appears to oscillate back
and forth. Is the apparent motion simple harmonic
motion?
No. Speed is constant.
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ACCELERATION
ac centripetal acceleration a
acceleration of the shadow
(ch8)
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Example 5.  The Loudspeaker RevisitedThe
Maximum Acceleration

• A loudspeaker diaphragm is vibrating at a
  frequency of f 1.0 kHz, and the amplitude of
  the motion is A 0.20 mm.
• What is the maximum acceleration of the
  diaphragm, and
• where does this maximum acceleration occur?

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(a)
(b) the maximum acceleration occurs at x A
and x A
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FREQUENCY OF VIBRATION
F ma
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Example 6.  A Body Mass Measurement Device
Astronauts who spend long periods of time in
orbit periodically measure their body masses as
part of their health-maintenance programs. On
earth, it is simple to measure body weight W with
a scale and convert it to mass m using the
acceleration due to gravity, since W mg.
However, this procedure does not work in orbit,
because both the scale and the astronaut are in
free-fall and cannot press against each other.
Instead, astronauts use a body mass measurement
device. This device consists of a spring-mounted
chair in which the astronaut sits. The chair is
then started oscillating in simple harmonic
motion. The period of the motion is measured
electronically and is automatically converted
into a value of the astronauts mass, after the
mass of the chair is taken into account. The
spring used in one such device has a spring
constant of 606 N/m, and the mass of the chair is
12.0 kg. The measured oscillation period is 2.41
s. Find the mass of the astronaut.
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Check Your Understanding 2
The drawing shows plots of the displacement x
versus the time t for three objects undergoing
simple harmonic motion. Which object, I, II, or
III, has the greatest maximum velocity?
II
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Energy and Simple Harmonic Motion
A spring also has potential energy when the
spring is stretched or compressed, which we refer
to as elastic potential energy. Because of
elastic potential energy, a stretched or
compressed spring can do work on an object that
is attached to the spring.
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DEFINITION OF ELASTIC POTENTIAL ENERGY The
elastic potential energy PEelastic is the energy
that a spring has by virtue of being stretched or
compressed. For an ideal spring that has a spring
constant k and is stretched or compressed by an
amount x relative to its unstrained length, the
elastic potential energy is
SI Unit of Elastic Potential Energy joule (J)
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Example 7.  An Object on a Horizontal Spring
An object of mass m 0.200 kg that is vibrating
on a horizontal frictionless table. The spring
has a spring constant k 545 N/m. It is
stretched initially to x0 4.50 cm and then
released from rest (see part A of the drawing).
Determine the final translational speed vf of the
object when the final displacement of the spring
is (a) xf 2.25 cm and (b) xf 0 cm.
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(a) Since x0 0.0450 m and xf 0.0225 m,
(b) When x0 0.0450 m and xf 0 m,
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Conceptual Example 8.  Changing the Mass of a
Simple Harmonic Oscillator
A box of mass m attached to a spring that has a
force constant k. The box rests on a horizontal,
frictionless surface. The spring is initially
stretched to x A and then released from rest.
The box then executes simple harmonic motion that
is characterized by a maximum speed vmax, an
amplitude A, and an angular frequency w.
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When the box is passing through the point where
the spring is unstrained (x 0 m), a second box
of the same mass m and speed vmax is attached to
it, as in part b of the drawing. Discuss what
happens to (a) the maximum speed, (b) the
amplitude, and (c) the angular frequency of the
subsequent simple harmonic motion.
Elastic Potential Energy A2
(a) The maximum speed of the two-box system
remains the same as that of the one-box system.
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Example 9.  A Falling Ball on a Vertical Spring
A 0.20-kg ball is attached to a vertical spring.
The spring constant of the spring is 28 N/m. The
ball, supported initially so that the spring is
neither stretched nor compressed, is released
from rest. In the absence of air resistance, how
far does the ball fall before being brought to a
momentary stop by the spring?
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xf h0, h0 2mg/k.
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Check Your Understanding 3
A block is attached to the end of a horizontal
ideal spring and rests on a frictionless surface.
The block is pulled so that the spring stretches
relative to its unstrained length. In each of the
following three cases, the spring is stretched
initially by the same amount, but the block is
given different initial speeds. Rank the
amplitudes of the resulting simple harmonic
motion in decreasing order (largest first). (a)
The block is released from rest. (b) The block is
given an initial speed v0. (c) The block is given
an initial speed v0/2.
(b), (c), (a)
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The Pendulum
A simple pendulum consists of a particle of mass
m, attached to a frictionless pivot P by a cable
of length L and negligible mass.
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Example 10.   Keeping Time
Determine the length of a simple pendulum that
will swing back and forth in simple harmonic
motion with a period of 1.00 s.
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Example 11.  Pendulum Motion and Walking
When we walk, our legs alternately swing forward
about the hip joint as a pivot. In this motion
the leg is acting approximately as a physical
pendulum. Treating the leg as a uniform rod of
length D 0.80 m, find the time it takes for the
leg to swing forward.
The desired time is one-half of the period or
0.75 s.
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Damped Harmonic Motion
In the presence of energy dissipation, the
amplitude of oscillation decreases as time
passes, and the motion is no longer simple
harmonic motion. Instead, it is referred to as
damped harmonic motion, the decrease in amplitude
being called damping.
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The smallest degree of damping that completely
eliminates the oscillations is termed critical
damping, and the motion is said to be critically
damped. When the damping exceeds the critical
value, the motion is said to be overdamped. In
contrast, when the damping is less than the
critical level, the motion is said to be
underdamped (curves 2 and 3).
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Conceptual Question 6
REASONING AND SOLUTION A block is attached to a
horizontal spring and slides back and forth in
simple harmonic motion on a frictionless
horizontal surface. A second identical block is
suddenly attached to the first block when the
first block is at one extreme end of the
oscillation cycle.
a. Since the attachment is made at one extreme
end of the oscillation cycle, where the velocity
is zero, the extreme end of the oscillation cycle
will remain at the same point in other words,
the amplitude remains the same.
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b. The angular frequency of an object of mass m
in simple harmonic motion at the end of a spring
of force constant k is given by Equation 10.11
. Since the mass m is doubled
while the force constant k remains the same, the
angular frequency decreases by a factor of .
The vibrational frequency f is related to w by f
w/(2 ) the vibrational frequency f will
also decrease by a factor of
c. The maximum speed of oscillation is given by
Equation 10.8 . Since
the amplitude, A, remains the same and the
angular frequency, w, decreases by a factor of
, the maximum speed of oscillation also
decreases by a factor of .
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Conceptual Question 11
REASONING AND SOLUTION From Equations 10.5 and
10.11, we can deduce that the period of the
simple harmonic motion of an ideal spring is
given by , where m is the
mass at the end of the ideal spring and k is the
spring constant. We can deduce from Equations
10.5 and 10.16 that, for small angles, the
period, T, of a simple pendulum is given by
where L is the length of the
pendulum.
In principle, the motion of a simple pendulum and
an object on an ideal spring can both be used to
provide the period of a clock. However, it is
clear from the expressions for the period given
above that the period of the mass-spring system
depends only on the mass and the spring constant,
while the period of the pendulum depends on the
acceleration due to gravity. Therefore, a
pendulum clock is likely to become more
inaccurate when it is carried to the top of a
high mountain where the value of g will be
smaller than it is at sea level.
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Conceptual Question 12
REASONING AND SOLUTION We can deduce from
Equations 10.5 and 10.16 that, for small angles,
the period, T, of a simple pendulum is given by
where L is the length of the
pendulum. This can be solved for the
acceleration due to gravity to yield
If you were held prisoner in a room and had only
a watch and a pair of shoes with shoelaces of
known length, you could determine whether this
room is on earth or on the moon in the following
way You could use one of the shoelaces and one
of the shoes to make a pendulum. You could then
set the pendulum into oscillation and use the
watch to measure the period of the pendulum. The
acceleration due to gravity could then be
calculated from the expression above. If the
value is close to 9.80 m/s2, then it can be
concluded that the room is on earth. If the
value is close to 1.6 m/s2, then it can be
concluded that the room is on the moon.
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Problem 8
REASONING AND SOLUTION The figure at the right
shows the original situation before the spring is
cut. The weight, W, of the object stretches the
string by an amount x.
Applying F kx to this situation gives W kx
(1)
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The figure at the right shows the situation after
the spring is cut into two segments of equal
length. Let k' represent the spring constant of
each half of the spring after it is cut. Now
the weight, W, of the object stretches each
segment by an amount x'. Applying F kx to
this situation gives
W k'x' k'x' 2k'x'
(2)
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Combining Equations (1) and (2) yields
kx 2k'x' From Conceptual
Example 2, we know that k' 2k so that
kx 2(2k)x' Solving for x'
gives
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Problem 16
REASONING AND SOLUTION From Conceptual Example
2, we know that when the spring is cut in half,
the spring constant for each half is twice as
large as that of the original spring. In this
case, the spring is cut into four shorter
springs. Thus, each of the four shorter springs
with 25 coils has a spring constant of
The angular frequency of simple harmonic motion
is given by Equation 10.11
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Problem 17
REASONING AND SOLUTION
a. Since the object oscillates between , the
amplitude of the motion is 0.08m
b. From the graph, the period is T4.0 s .
Therefore, according to Equation 10.4,
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c. Equation 10.11 relates the angular frequency
to the spring constant
. Solving for k we find
d. At t1.0 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 v0 m/s
e. The acceleration of the object at t1.0 s is
a maximum, and its magnitude is
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Problem 21
REASONING The frequency of vibration of the
spring is related to the added mass m by
Equations 10.6 and 10.11
The spring constant can be determined from
Equation 10.1
SOLUTION Since the spring stretches by 0.018 m
when a 2.8-kg object is suspended from its end,
the spring constant is, according to Equation
10.1,
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Solving Equation (1) for m, we find that the mass
required to make the spring vibrate at 3.0 Hz is
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Problem 25
0. 2m
0.392m
0. 2m
0. 2m
2.0kg
point of release
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Problem 25
REASONING AND SOLUTION If we neglect air
resistance, only the 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
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
(v00 m/s).
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At this point the spring is stretched by an
amount of
.This point represents the zero reference
level ( 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
The total mechanical energy E0 at the point of
release is the sum of the three energies above
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h 0.200 m When the object has risen a
distance of above the release
point, the spring is stretched by an amount of
.
Since the total mechanical energy is conserved,
its value at this point is still
. The gravitational and elastic potential
energies are
Since
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h 0.400 m When the object has risen a
distance of above the release
point, the spring is stretched by an amount of
.
At this point, the total mechanical energy is
still . The gravitational
and elastic potential energies are
The kinetic energy is
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The results are summarized in the table below
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Problem 32
f 3.0 HZ
k
m
A 5.0810-2 m
k
m/2
Max speed at halfway of the amplitude.
m
m/2
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REASONING AND SOLUTION
a. Now look at conservation of energy before and
after the split Before split (1/2) mvmax2
(1/2) kA2
Solving for the amplitude A gives
After split If new amplitude is A
(1/2) (m/2)v'2 (1/2) (m/2)(vmax)2 (1/2)
kA'2
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Solving for the amplitude A' gives
Therefore, we find that
Similarly, for the frequency, we can show that
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b. If the block splits at one of the extreme
positions, the amplitude of the SHM would not
change, so it would remain as
5.0810-2 m
The frequency would be
f' f (3.00 Hz) 4.24 Hz
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Problem 38
.
.
t 0.25 s
x
.
Object is resting on the spring. F kx mg
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REASONING AND SOLUTION
Using f 1/T 1/(0.250 s) 4.00 Hz and
also
we can find the ratio k/m 4 2f2 632
N/(kg?m)
With the object resting on the spring, F kx
mg so that,
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When the mass leaves the spring, potential energy
of the spring has been converted to gravitational
energy, i.e.,
(1/2) kx'2 mgh
if h is the height, it can reach
Where x' 0.0500 m 0.0155 m 0.0655 m
Solving for h we get
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Problem 41
REASONING AND SOLUTION Recall that the
relationship between frequency f and period T is
. Then, according to Equations 10.6
and 10.16, the period of the simple pendulum is
given by
where L is the length of the pendulum. Solving
for g and noting that the period is
T  (280 s)/100  2.8 s, we obtain
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Problem 68
REASONING The force F that the spring exerts on
the block just before it is released is equal to
kx, according to Equation 10.2. Here k is the
spring constant and x is the displacement of the
spring from its equilibrium position. Once the
block has been released, it oscillates back and
forth with an angular frequency given by Equation
10.11 as , where m is the
mass of the block. The maximum speed that the
block attains during the oscillatory motion is
vmax A (Equation 10.8). The magnitude of
the maximum acceleration that the block attains
is amax A 2 (Equation 10.10).
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SOLUTION a. The force F exerted on the block by
the spring is
b. The angular frequency of the resulting
oscillatory motion is
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c. The maximum speed vmax is the product of the
amplitude and the angular frequency
d. The magnitude amax of the maximum acceleration
is