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# Chapter 14 - Simple Harmonic Motion

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```Chapter 14 - Simple
Harmonic Motion
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
2007
Photo by Mark Tippens
A TRAMPOLINE exerts a restoring force on the
jumper that is directly proportional to the average
force required to displace the mat. Such restoring
forces provide the driving forces necessary for
objects that oscillate with simple harmonic motion.
Objectives: After finishing this
unit, you should be able to:
вЂў Write and apply HookeвЂ™s Law for objects moving
with simple harmonic motion.
вЂў Write and apply formulas for
finding the frequency f, period T,
velocity v, or acceleration a in
terms of displacement x or time t.
вЂў Describe the motion of pendulums
and calculate the length required
to produce a given frequency.
Periodic Motion
Simple 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.
f пЂЅ
1
T
Amplitude
A
Period, T, is the time
for one complete
oscillation. (seconds,s)
Frequency, f, is the
number of complete
oscillations per
second. Hertz (s-1)
Example 1: The suspended mass makes 30
complete oscillations in 15 s. What is the
period and frequency of the motion?
T пЂЅ
15 s
пЂЅ 0.50 s
30 cylces
x
F
Period: T = 0.500 s
f пЂЅ
1
T
пЂЅ
1
0.500 s
Frequency: f = 2.00 Hz
Simple Harmonic Motion, SHM
Simple harmonic motion is periodic motion in
the absence of friction and produced by a
restoring force that is directly proportional to
the displacement and oppositely directed.
x
F
A restoring force, F, acts in
the direction opposite the
displacement of the
oscillating body.
F = -kx
HookeвЂ™s Law
When a spring is stretched, there is a restoring
force that is proportional to the displacement.
F = -kx
x
m
F
The spring constant k is a
property of the spring given by:
k=
DF
Dx
Work Done in Stretching a Spring
Work done ON the spring is positive;
work BY spring is negative.
x
From HookeвЂ™s law the force F is:
F (x) = kx
F
F
m
To stretch spring from
x1 to x2 , work is:
W ork пЂЅ ВЅ kx пЂ­ ВЅ kx
2
2
x1
x2
2
1
(Review module on work)
Example 2: A 4-kg mass suspended from a
spring produces a displacement of 20 cm.
What is the spring constant?
The stretching force is the weight
(W = mg) of the 4-kg mass: 20 cm
F = (4 kg)(9.8 m/s2) = 39.2 N
F
m
Now, from HookeвЂ™s law, the force
constant k of the spring is:
k=
DF
Dx
=
39.2 N
0.2 m
k = 196 N/m
Example 2(cont.: The mass m is now stretched
a distance of 8 cm and held. What is the
potential energy? (k = 196 N/m)
The potential energy is equal to
the work done in stretching the
spring:
8 cm
0
m
W ork пЂЅ ВЅ kx пЂ­ ВЅ kx
2
2
2
1
U пЂЅ ВЅ kx пЂЅ ВЅ (196 N /m)(0.08 m)
2
U = 0.627 J
F
2
Displacement in SHM
x
m
x = -A
x=0
x = +A
вЂў Displacement is positive when the position is
to the right of the equilibrium position (x = 0)
and negative when located to the left.
вЂў The maximum displacement is called the
amplitude A.
Velocity in SHM
v (-)
v (+)
m
x = -A
x=0
x = +A
вЂў Velocity is positive when moving to the right
and negative when moving to the left.
вЂў It is zero at the end points and a maximum
at the midpoint in either direction (+ or -).
Acceleration in SHM
+a
-x
+x
-a
m
x = -A
x=0
x = +A
вЂў Acceleration is in the direction of the
restoring force. (a is positive when x is
negative, and negative when x is positive.)
F пЂЅ m a пЂЅ пЂ­ kx
вЂў Acceleration is a maximum at the end points
and it is zero at the center of oscillation.
Acceleration vs. Displacement
a
v
x
m
x = -A
x=0
x = +A
Given the spring constant, the displacement, and
the mass, the acceleration can be found from:
F пЂЅ m a пЂЅ пЂ­ kx
or
a пЂЅ
пЂ­ kx
m
Note: Acceleration is always opposite to displacement.
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?
a пЂЅ
пЂ­ kx
m
aпЂЅ
пЂ­ (400 N /m )(+0.07 m )
2 kg
a = -14.0 m/s2
a
m
Note: When the displacement is +7 cm
(downward), the acceleration is -14.0 m/s2
(upward) independent of motion direction.
+x
Example 4: What is the maximum acceleration
for the 2-kg mass in the previous problem? (A
= 12 cm, k = 400 N/m)
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
aпЂЅ
пЂ­ kA
пЂЅ
xmax = п‚± A
пЂ­ 400 N ( п‚± 0.12 m )
m
Maximum
Acceleration:
m
2 kg
amax = В± 24.0 m/s2
+x
Conservation of Energy
The total mechanical energy (U + K) of a
vibrating system is constant; i.e., it is the
same at any point in the oscillating path.
a
v
x
m
x = -A
x=0
x = +A
For any two points A and B, we may write:
ВЅmvA2 + ВЅkxA 2 = ВЅmvB2 + ВЅkxB 2
Energy of a Vibrating System:
A
x
a
v
m
x = -A
x=0
B
x = +A
вЂў At points A and B, the velocity is zero and the
acceleration is a maximum. The total energy is:
U + K = ВЅkA2 x = п‚± A and v = 0.
вЂў At any other point: U + K = ВЅmv2 + ВЅkx2
Velocity as Function of Position.
x
a
v
m
x = -A
1
2
mv пЂ«
2
1
2
x=0
kx пЂЅ
2
1
2
kA
vmax when
x = 0:
x = +A
k
vпЂЅ
2
m
v пЂЅ
k
m
A
A пЂ­x
2
2
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?
ВЅmv2 + ВЅkx
vпЂЅ
k
2
= ВЅkA2
A пЂ­x
2
2
m
vпЂЅ
800 N /m
m
(0.1 m ) пЂ­ (0.06 m )
2
2 kg
v = В±1.60 m/s
2
+x
Example 5 (Cont.): What is the maximum
velocity for the previous problem? (A = 10
cm, k = 800 N/m, m = 2 kg.)
The velocity is maximum when x = 0:
0
ВЅmv2 + ВЅkx 2 = ВЅkA2
vпЂЅ
k
m
AпЂЅ
800 N /m
(0.1 m )
2 kg
v = В± 2.00 m/s
m
+x
The Reference Circle
The reference circle compares
the circular motion of an object
with its horizontal projection.
x пЂЅ A cos пЃ±
пЃ± пЂЅ wt
x пЂЅ A cos(2пЃ° ft )
x = Horizontal displacement.
A = Amplitude (xmax).
пЃ± = Reference angle.
w пЂЅ 2пЃ°f
Velocity in SHM
The velocity (v) of an
oscillating body at any
instant is the horizontal
component of its
tangential velocity (vT).
vT = wR = wA; w пЂЅ 2пЃ°f
v = -vT sin пЃ± ;
пЃ± = wt
v = -w A sin w t
v = -2пЃ°f A sin 2пЃ°f t
Acceleration Reference Circle
The acceleration (a) of an
oscillating body at any instant is
the horizontal component of its
centripetal acceleration (ac).
a = -ac cos пЃ± = -ac cos(wt)
ac пЂЅ
v
w R
2
2
пЂЅ
R
R
R=A
2
; ac пЂЅ w R
2
a = -w2A cos(wt)
a пЂЅ пЂ­ 4пЃ° f A cos(2пЃ° ft )
2
2
a пЂЅ пЂ­ 4пЃ° f x
2
2
The Period and Frequency as a
Function of a and x.
For any body undergoing simple harmonic motion:
Since a = -4пЃ°2f2x
f пЂЅ
1
пЂ­a
2пЃ°
x
and
T пЂЅ 2пЃ°
T = 1/f
пЂ­x
a
The frequency and the period can be found if the
displacement and acceleration are known. Note
that the signs of a and x will always be opposite.
Period and Frequency as a Function
of Mass and Spring Constant.
For a vibrating body with an elastic restoring force:
Recall that F = ma = -kx:
f пЂЅ
1
k
2пЃ°
m
T пЂЅ 2пЃ°
m
k
The frequency f and the period T can be found if
the spring constant k and mass m of the vibrating
body are known. Use consistent SI units.
Example 6: The frictionless system shown
below has a 2-kg mass attached to a spring
(k = 400 N/m). The mass is displaced a
distance of 20 cm to the right and released.
What is the frequency of the motion?
x
a
v
m
x=0
x = -0.2 m
f пЂЅ
1
k
2пЃ°
m
пЂЅ
x = +0.2 m
1
400 N /m
2пЃ°
2 kg
f = 2.25 Hz
Example 6 (Cont.): Suppose the 2-kg mass
of the previous problem is displaced 20 cm
and released (k = 400 N/m). What is the
maximum acceleration? (f = 2.25 Hz)
x
a
v
m
x=0
x = -0.2 m
x = +0.2 m
Acceleration is a maximum when x = п‚± A
a пЂЅ пЂ­ 4пЃ° f x пЂЅ пЂ­ 4пЃ° (2.25 H z) ( п‚± 0.2 m)
2
2
2
a = п‚± 40 m/s2
2
Example 6: The 2-kg mass of the previous
example is displaced initially at x = 20 cm
and released. What is the velocity 2.69 s
after release? (Recall that f = 2.25 Hz.)
x
a
v
m
v = -2пЃ°f A sin 2пЃ°f t
x = -0.2 m x = 0
x = +0.2 m
v пЂЅ пЂ­ 2пЃ° (2 .2 5 H z)(0 .2 m ) sin пЃ› 2 пЃ° (2 .2 5 H z)(2 .6 9 s) пЃќ
v пЂЅ пЂ­ 2пЃ° (2.25 H z)(0.2 m )(0.324)
v = -0.916 m/s
The minus sign means it
is moving to the left.
Example 7: At what time will the 2-kg mass
be located 12 cm to the left of x = 0?
(A = 20 cm, f = 2.25 Hz) -0.12 m
x
a
m
x пЂЅ A cos(2пЃ° ft )
x = -0.2 m x = 0
cos(2пЃ° ft ) пЂЅ
x
A
пЂЅ
v
пЂ­ 0.12 m
x = +0.2 m
пЂ­1
(2 пЃ° ft ) пЂЅ cos ( пЂ­ 0.60)
;
0.20 m
tпЂЅ
t = 0.157 s
2 пЃ° (2.25 H z)
The Simple Pendulum
The period of a simple
pendulum is given by:
T пЂЅ 2пЃ°
L
L
g
For small angles пЃ±.
f пЂЅ
1
g
2пЃ°
L
mg
Example 8. What must be the length of a
simple pendulum for a clock which has a period
of two seconds (tick-tock)?
L
T пЂЅ 2пЃ°
L
g
T
2
пЂЅ 4пЃ°
2
L
2
;
L =
g
2
LпЂЅ
T g
4пЃ°
2
2
(2 s) (9.8 m /s )
4пЃ°
2
L = 0.993 m
The Torsion Pendulum
The period T of a torsion
pendulum is given by:
T пЂЅ 2пЃ°
I
k'
Where kвЂ™ is a torsion constant that depends on
the material from which the rod is made; I is
the rotational inertia of the vibrating system.
Example 9: A 160 g solid disk is attached to
the end of a wire, then twisted at 0.8 rad
and released. The torsion constant kвЂ™ is
0.025 N m/rad. Find the period.
(Neglect the torsion in the wire)
For Disk:
I = ВЅmR2
I = ВЅ(0.16 kg)(0.12 m)2
= 0.00115 kg m2
T пЂЅ 2пЃ°
I
k'
пЂЅ 2пЃ°
0.00115 kg m
2
T = 1.35 s
Note: Period is independent of angular displacement.
Summary
Simple harmonic motion (SHM) 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.
The frequency (rev/s) is the
reciprocal of the period (time
for one revolution).
x
m
F
f пЂЅ
1
T
Summary (Cont.)
HookeвЂ™s Law: In a spring, there is a restoring
force that is proportional to the displacement.
F пЂЅ пЂ­ kx
x
The spring constant k is defined by:
m
F
k пЂЅ
DF
Dx
Summary (SHM)
x
a
v
m
x = -A
x=0
F пЂЅ m a пЂЅ пЂ­ kx
x = +A
a пЂЅ
пЂ­ kx
m
Conservation of Energy:
ВЅmvA2 + ВЅkxA 2 = ВЅmvB2 + ВЅkxB 2
Summary (SHM)
1
2
vпЂЅ
mv пЂ«
2
k
1
2
A пЂ­x
2
m
x пЂЅ A cos(2пЃ° ft )
kx пЂЅ
2
2
1
2
kA
v0 пЂЅ
2
k
A
m
a пЂЅ пЂ­ 4пЃ° f x
2
v пЂЅ пЂ­ 2пЃ° fA sin(2пЃ° ft )
2
Summary: Period and
Frequency for Vibrating
Spring.
a
v
x
m
x = -A
f пЂЅ
f пЂЅ
x=0
1
пЂ­a
2пЃ°
x
1
k
2пЃ°
m
T пЂЅ 2пЃ°
x = +A
пЂ­x
a
T пЂЅ 2пЃ°
m
k
Summary: Simple Pendulum
and Torsion Pendulum
f пЂЅ
1
g
2пЃ°
L
T пЂЅ 2пЃ°
L
T пЂЅ 2пЃ°
L
g
I
k'
CONCLUSION: Chapter 14
Simple Harmonic Motion
```
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