Download Ch10 Simple Harmonic Motion and Elasticity

Ch10. Simple Harmonic Motion
and Elasticity
The Ideal Spring and Simple Harmonic Motion
FApplied  kx
The constant k is called the
spring constant
A spring that behaves
according to FApplied  kx 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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HOOKE’S 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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5
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.”
6
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
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has the shape of a trigonometric sine or cosine function.
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
8
2.1cm
kx = ma
2.1
2
k
 0.42  9.0m / s
100
0.42  9.0
k
100  180 N / m
2.1
9
mg = kd0, which gives d0 = mg/k.
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 .
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Simple Harmonic Motion and the
Reference Circle
Simple harmonic motion, like any motion, can be described
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in terms of displacement, velocity, and acceleration.
DISPLACEMENT
12
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.
(a)What is the maximum
speed of the diaphragm?
(b)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
a  ac cos 
ac
centripetal acceleration
a
acceleration of the shadow
ac  r
2
(ch8)
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Example 5.
The Loudspeaker Revisited—The
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.
(a)What is the maximum
acceleration of the
diaphragm, and
(b)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

 k  A cos t   m  A cos t
k

m
2

 in rad / s 
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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 astronaut’s 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
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astronaut.
2
k


T
m
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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
1 2
PEelastic  kx
2
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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E f  E0
1 2 1 2
1 2 1 2 1
1 2
2
mv f  I f  mgh f  kx f  mv0  m0  mgh0  kx0
2
2
2
2
2
2
1 2 1 2 1 2
mv f  kx f  kx0
2
2
2
k 2
2
vf 
( x0  x f )
m
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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.
k

m
Elastic Potential Energy

A2
amax  A
(a) The maximum speed of the two-box system remains the
same as that of the one-box system.
(b) The amplitude of the two-box system is greater than
that of the one-box system by a factor of 2
(c) The angular frequency of the two-box system is smaller
than that of the one-box system by a factor of 2
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2
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.
1
f 
2
g/L
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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:   k / m . Since the mass m
is doubled while the force constant k remains the same,
the angular frequency decreases by a factor of 2 . The
vibrational frequency f is related to w by f = w/(2 ) ; the
vibrational frequency f will also decrease by a factor of 2
c. The maximum speed of oscillation is given by Equation
10.8: vmax  A . Since the amplitude, A, remains the
same and the angular frequency, w, decreases by a factor
of 2 , the maximum speed of oscillation also decreases by a
factor of 2 .
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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 T  2 m/ k , 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 T  2 L / g 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 T  2 L / g where L is the
length of the pendulum. This can be solved for the
acceleration due to gravity to yield: g  4 2 L / T 2
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
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that the room is on the moon.
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)
kx
W
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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'
k'x'
k'x'
W
(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
x 0.160 m
x'  
 0.040 m
4
4
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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 4  420 N/m 1680 N/m
The angular frequency of simple harmonic motion is given
by Equation 10.11:

k
1680 N/m

 6.0 rad/s
m
46 kg
52
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
T=4.0 s . Therefore, according
to Equation 10.4,
2
2


 1.6 rad/s
T
4.0 s
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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
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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:
1
f 
2
k
m
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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1
f 
2
k
k
m
FApplied
x
1 k
f 
2
4 m
2
m
k
4 f
2
2
mg (2.8 kg)(9.80 m/s2 )


 1.52 103 N/m
x
0.018 m
Solving Equation (1) for m, we find that the mass required
to make the spring vibrate at 3.0 Hz is
1.52  103 N/m
m
2 2 
2
2  4.3 kg
4 f
4 (3.0 Hz)
k
56
Problem 25
0. 2m
0.392m
0. 2m
2.0kg
0. 2m
point of
release
P  Eelastic
57
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
F mg (2.00 kg)(9.80 m/s 2 )
x 

 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
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(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:
KE  mv  m(0 m/s)  0 J
1
2
2
0
1
2
2
PEgravity  mgh  mg (0 m)  0 J
PE elastic 
1 2
kx0
2

1
(50.0 N/m)(0.592 m) 2
2
 8.76 J
The total mechanical energy E0 at the point of release is the
sum of the three energies above: E  8.76 J
0
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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 
Since
1 2
kx
2

1
(50.0 N/m)(0.392 m) 2
2
 3.84 J
KE  PEgravity  PEelastic  E
KE  E – PEgravity – PEelastic  8.76 J – 3.92 J – 3.84 J = 1.00 J
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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
kx
2

1
2
(50.0 N/m)(0.192 m)
2
 0.92 J
The kinetic energy is
KE  E  PE gravity  PEelastic  8.76 J  7.84 J  0.92 J  0 J
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The results are summarized in the table below
h
0.000 m
0.200 m
0.400 m
KE
PE grav
PE elastic
E
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
62
Problem 32
f = 3.0 HZ
k
m
A = 5.08*10-2 m
k
m/2
Max speed at halfway of the amplitude.
m
m/2
63
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
A  v max
m
k
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
A  v max
m
2k
Therefore, we find that
A' = A/ 2 = (5.08 * 10–2 m)/ 2 = 3.59*10-2m
k

m
k
2f 
m
1
f 
2
k
m
Similarly, for the frequency, we can show that
f' = f 2 = (3.00 Hz) 2 = 4.24 Hz
65
b. If the block splits at one of the extreme positions,
the amplitude of the SHM would not change, so it
would remain as 5.08*10-2 m
The frequency would be
f' = f 2 = (3.00 Hz) 2 = 4.24 Hz
66
Problem 38
x
.
.
.
t = 0.25 s
Object is resting on the spring.
F = kx = mg
g
x
 0.015 sec ond
k
m
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REASONING AND SOLUTION
Using f = 1/T = 1/(0.250 s) = 4.00 Hz and also
1
f=
2
k
m
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,
g
x=
= 0.0155 m.
k
m
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When the mass leaves the spring, potential energy of the
spring has been converted to gravitational energy, i.e.,
if h is the height, it can reach (1/2) kx'2 = mgh
Where x' = 0.0500 m + 0.0155 m = 0.0655 m
Solving for h we get
2 
2 


k
x'
(0.0655
m)
  
h   
 632 N/(kg m) 
 0.138 m
2

m  2g 
2(9.80 m/s ) 
69
Problem 41
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
T  2
L
g
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
4 L 4 (1.2 m)
2
g
2 
2  6.0 m/s
T
(2.8 s)
2
2
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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   k / m , 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
F   kx   82.0 N /m  0.120 m   9.84 N
b. The angular frequency  of the resulting oscillatory
motion is
82.0 N /m
k


 10.5 rad /s
m
0.750 kg
72
c. The maximum speed vmax is the product of the
amplitude and the angular frequency:
vmax  A   0.120 m10.5 rad /s   1.26 m/s
d. The magnitude amax of the maximum acceleration is
amax  A   0.120 m 10.5 rad /s   13.2 m /s
2
2
2
73