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Unit 2:
Simple Harmonic Motion
(SHM)
THE MOST COMMON FORM OF MOTION
FALL 2015
Objectives:
• Define SHM specifically and give an example.
• Describe the motion of pendulums and spring systems and calculate the length required to produce a given frequency.
• Write and apply Hooke’s Law for objects moving with simple harmonic motion.
• Write and apply formulas for finding the frequency f, period T, w angular frequency, velocity v, or acceleration a in terms of displacement x or time t.
Displacement in SHM
m
x = 0 x = +Ax = -A
x
• Displacement is positive when the position is
to the right of the equilibrium position (x = 0)
and negative when located to the left.
• The maximum displacement is called the
amplitude A.
Periodic Motion
Periodic motion is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time.
Amplitude
A
Period, T, is the time
for one complete
oscillation. (seconds,s)
Period, T, is the time
for one complete
oscillation. (seconds,s)
Frequency, f, is the
number of complete
oscillations per
second. Hertz (s-1)
Frequency, f, is the
number of complete
oscillations per
second. Hertz (s-1)
1f
T
Example 1: The suspended mass makes
30 complete oscillations in 15 s. What is
the period and frequency of the
motion?
x F
15 s0.50 s
30 cylcesT
Period: T = 0.500 sPeriod: T = 0.500 s
1 1
0.500 sf
T
Frequency: f = 2.00 HzFrequency: f = 2.00 Hz
Simple Harmonic Motion,
SHMSimple harmonic motion:
• periodic motion in the absence of friction
(Amplitude is always the same/ Energy is
Conserved)
• produced by a restoring Force that is
directly proportional to the displacement
and oppositely directed (restoring to
equilibrium).
• Displacement does not affect the frequency/
The energy of the system does not affect the
frequency.
7
https://www.youtube.com/watch?v=SZ
541Luq4nE
https://www.youtube.com/watch?
v=VnGkoMoUkgI
Conceptual Example 4. Moving Lights10
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.
Damped Harmonic Motion
13
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.”
14
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).
16
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.
Simple Harmonic Motion and
the Reference Circle
17
Simple harmonic motion, like any motion, can be described
in terms of displacement, velocity, and acceleration.
18DISPLACEMENT
19For any object in simple
harmonic motion, the time
required to complete one
cycle is the period, T (MKS)
Instead of the period, it is more
convenient to speak of the angular
frequency w of the motion, the
frequency being just the number of
cycles of the motion per second.
20
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.w
21FREQUENCY OF VIBRATION
F = ma
tAmtAk www coscos 2
sradinm
k/ww
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
astronaut.
23
m
k
T
w
2
Velocity in SHM
m
x = 0 x = +Ax = -A
v (+)
• Velocity is positive when moving to the rightand negative when moving to the left.
• It is zero at the end points and a maximumat the midpoint in either direction (+ or -).
v (-)
Velocity as Function of
Position.
m
x = 0 x = +Ax = -A
x va
kv A
m
vmax when
x = 0:
2 2kv A x
m 2 2 21 1 1
2 2 2mv kx kA
Check Your Understanding 2 26
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
Example 5: A 2-kg mass hangs at the end
of a spring whose constant is k = 800 N/m.
The mass is displaced a distance of 10 cm
and released. What is the velocity at the
instant the displacement is x = +6 cm?
m+x
½mv2 + ½kx 2 = ½kA2
2 2kv A x
m
2 2800 N/m(0.1 m) (0.06 m)
2 kgv
v = ±1.60 m/sv = ±1.60 m/s
Example 5 (Cont.): What is the maximum
velocity for the previous problem? (A = 10
cm, k = 800 N/m, m = 2 kg.)
m+x
½mv2 + ½kx 2 = ½kA2
800 N/m(0.1 m)
2 kg
kv A
m
v = ± 2.00 m/sv = ± 2.00 m/s
0
The velocity is maximum when x = 0:
Example 3. The Maximum
Speed of a Loudspeaker
Diaphragm
29
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?
30
(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.
(a)
Acceleration in SHM
m
x = 0 x = +Ax = -A
• Acceleration is in the direction of the restoring force. (a is positive when x is negative, and negative when x is positive.)
• Acceleration is a maximum at the end points and it is zero at the center of oscillation.
+x-a
-x+a
F ma kx F ma kx
Acceleration vs.
Displacement
m
x = 0 x = +Ax = -A
xv
a
Given the spring constant, the displacement, and the mass, the acceleration can be found from:
or
Note: Acceleration is always opposite to displacement.
F ma kx F ma kx kx
am
kxa
m
Example 3: A 2-kg mass hangs at the end
of a spring whose constant is k = 400 N/m.
The mass is displaced a distance of 12
cm and released. What is the
acceleration at the instant the
displacement is x = +7 cm?
m+x
(400 N/m)(+0.07 m)
2 kga
a = -14.0 m/s2a = -14.0 m/s2
a
Note: When the displacement is +7 cm(downward), the acceleration is -14.0 m/s2
(upward) independent of motion direction.
kxa
m
kxa
m
Example 4: What is the maximum
acceleration for the 2-kg mass in
the previous problem? (A = 12 cm,
k = 400 N/m)
m+x
The maximum acceleration occurs when the restoring force is a maximum; i.e., when the stretch or compression of the spring is largest.F = ma = -kx xmax = A
400 N( 0.12 m)
2 kg
kAa
m
amax = ± 24.0 m/s2amax = ± 24.0 m/s2Maximum
Acceleration:
Example 5.
The Loudspeaker Revisited—The
Maximum Acceleration
35
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?
36
(b) the maximum acceleration occurs at x = +A and x = –A
(a)
Hooke’s Law
When a spring is stretched, there is a restoringforce that is proportional to the displacement.
F = -kx
The spring constant k is a property of the spring given by:
k = DF
Dx
F
x
m
Ch10. Simple Harmonic Motion and
Elasticity
kxFApplied
kxFApplied
38
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.
39The 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.
Example 1. A Tire Pressure Gauge
40
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?
Conceptual Example 2.
Are Shorter Springs Stiffer Springs?41
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.
Example 2: A 4-kg mass suspended from a
spring produces a displacement of 20 cm.
What is the spring constant?
F20 cm
m
The stretching force is the weight (W = mg) of the 4-kg mass:
F = (4 kg)(9.8 m/s2) = 39.2 N
Now, from Hooke’s law, the force constant k of the spring is:
k = =DF
Dx
39.2 N
0.2 mk = 196 N/mk = 196 N/m
Check Your Understanding 1 43
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
44
2.1c
mkx = ma
2/0.942.0100
1.2smk
mNk /1801001.2
0.942.0