Download Chapter 14 - - Simple Harmonic Motion

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:
• Write and apply Hooke’s Law for objects moving
with simple harmonic motion.
• Write and apply formulas for
finding the frequency f, period T,
velocity v, or acceleration a in
terms of displacement x or time t.
• 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.
1
f 
T
Amplitude
A
Period,
Period
Period,T,
T,isisthe
thetime
time
for
forone
onecomplete
complete
oscillation.
oscillation.(seconds,s)
(seconds,s)
Frequency,
Frequency
Frequency,f,f,isisthe
the
number
numberof
ofcomplete
complete
oscillations
oscillationsper
per
-1
second.
second.Hertz
Hertz(s(s-1))
Example 1: The suspended mass makes 30
complete oscillations in 15 s. What is the
period and frequency of the motion?
15 s
T
 0.50 s
30 cylces
x
F
Period:
Period: TT == 0.500
0.500 ss
1
1
f  
T 0.500 s
Frequency:
Frequency: ff == 2.00
2.00 Hz
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
AA restoring
restoring force,
force, F,
F, acts
acts in
in
the
the direction
direction opposite
opposite the
the
displacement
displacement of
of the
the
oscillating
oscillating body.
body.
FF == -kx
-kx
Hooke’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=
F
x
Work Done in Stretching a Spring
Work done ON the spring is positive;
work BY spring is negative.
x
From Hooke’s law the force F is:
F (x) = kx
F
F
m
To stretch spring from
x1 to x2 , work is:
Work  ½ kx  ½ 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 Hooke’s law, the force
constant k of the spring is:
k=
F
x
=

0.2 m
kk == 196
196 N/m
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:
Work  ½ kx  ½ kx
2
2
8 cm
0
m
2
1
U  ½ kx  ½(196 N/m)(0.08 m)
2
UU == 0.627
0.627 JJ
F
2
Displacement in SHM
x
m
x = -A
x=0
x = +A
• 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.
Velocity in SHM
v (- )
v (+)
m
x = -A
x=0
x = +A
• Velocity is positive when moving to the right
and negative when moving to the left.
• 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
• Acceleration is in the direction of the
restoring force. (a is positive when x is
negative, and negative when x is positive.)
F  ma  kx
• 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  ma  kx or
 kx
a
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?
 kx
a
m
(400 N/m)(+0.07 m)
a
2 kg
22
aa == -14.0
m/s
-14.0 m/s
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
xmax =  A
 kA 400 N(  0.12 m)
a

2 kg
m
Maximum
Acceleration:
m
22
aamax
=
±
24.0
m/s
max = ± 24.0 m/s
+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:
22 + ½kx 22 = ½mv 22 + ½kx 22
½mv
½mvAA + ½kxAA = ½mvBB + ½kxBB
Energy of a Vibrating System:
A
x
a
v
m
x = -A
x=0
B
x = +A
• At points A and B, the velocity is zero and the
acceleration is a maximum. The total energy is:
U + K = ½kA2 x =  A and v = 0.
• At any other point: U + K = ½mv2 + ½kx2
Velocity as Function of Position.
x
a
v
m
x = -A
1
2
x=0
mv  kx  kA
2
1
2
2
1
2
vmax when
x = 0:
x = +A
v
2
v
k
A
m
k
A2  x 2
m
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?
½mv2 + ½kx 2 = ½kA2
v
k
m
A2  x 2
800 N/m
v
(0.1 m) 2  (0.06 m) 2
2 kg
vv == ±1.60
±1.60 m/s
m/s
m
+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
½mv2 + ½kx 2 = ½kA2
v
k
800 N/m
A
(0.1 m)
m
2 kg
vv == ±± 2.00
2.00 m/s
m/s
m
+x
The Reference Circle
The reference circle compares
the circular motion of an object
with its horizontal projection.
x  Acos    t
x  A cos(2 ft )
x = Horizontal displacement.
A = Amplitude (xmax).
 = Reference angle.
2f
Velocity in SHM
The velocity (v) of an
oscillating body at any
instant is the horizontal
component of its
tangential velocity (vT).
vT = R = A; 2f
v = -vT sin  ;
 = t
v = - A sin t
vv == -2f
-2f AA sin
sin 2f
2f tt
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  = -ac cos(t)
v2  2 R2
; ac   2 R
ac  
R
R
a = -cos(t)
R=A
a  4 2 f 2 A cos(2 ft )
a  4 f x
2
2
The Period and Frequency as a
Function of a and x.
For any body undergoing simple harmonic motion:
Since a = -4f2x
1
f 
2
a
x
and
T = 1/f
x
T  2
a
The
The frequency
frequency and
and the
the period
period can
can be
be found
found ifif the
the
displacement
displacement and
and acceleration
acceleration are
are known.
known. Note
Note
that
that the
the signs
signs of
of aa and
and xx will
will always
always be
be opposite.
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:
kx
1
f 
2
k
m
m
T  2
k
The
The frequency
frequency ff and
and the
the period
period TT can
can be
be found
found ifif
the
the spring
spring constant
constant kk and
and mass
mass m
m of
of the
the vibrating
vibrating
body
body are
are known.
known. Use
Use consistent
consistent SI
SI units.
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 = -0.2 m
1
f 
2
x=0
k
1

m 2
x = +0.2 m
400 N/m
2 kg
ff == 2.25
2.25 Hz
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 =  A
a  4 f x  4 (2.25 Hz) (0.2 m)
2
2
2
22
aa ==  40
m/s
40 m/s
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
vv == -2f
-2f AA sin
sin 2f
2f tt
x = -0.2 m x = 0
x = +0.2 m
v  2 (2.25 Hz)(0.2 m) sin  2 (2.25 Hz)(2.69 s) 
(Note:  in rads)
v  2 (2.25 Hz)(0.2 m)(0.324)
vv == -0.916
-0.916 m/s
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
v
m
x  A cos(2 ft )
x = -0.2 m x = 0
x 0.12 m
cos(2 ft )  
;
A 0.20 m
2 ft  2.214 rad;
x = +0.2 m
(2 ft )  cos 1 (0.60)
2.214 rad
t
2 (2.25 Hz)
tt == 0.157
0.157 ss
The Simple Pendulum
The period of a simple
pendulum is given by:
L
T  2
g
L
For small angles 
1
f 
2
g
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  2
g
2
2 L
;
T  4
g
(2 s) 2 (9.8 m/s 2 )
L
2
4
L
T 2g
L=
2
4
L = 0.993 m
The Torsion Pendulum
The period T of a torsion
pendulum is given by:
I
T  2
k'
Where
Where k’k’ isis aa torsion
torsion constant
constant that
that depends
depends on
on
the
the material
material from
from which
which the
the rod
rod isis made;
made; II isis
the
the rotational
rotational inertia
inertia of
of the
the vibrating
vibrating system.
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 k’ is
0.025 N m/rad. Find the period.
(Neglect the torsion in the wire)
For Disk:
Disk
I = ½mR2
I = ½(0.16 kg)(0.12 m)2
= 0.00115 kg m2
I
0.00115 kg m 2
 2
T  2
k'
0.025 N m/rad
TT == 1.35
1.35 ss
Note: Period is independent of angular displacement.
Summary
Simple
Simple harmonic
harmonic motion
motion (SHM)
(SHM) isis that
that motion
motion in
in
which
which aa body
body moves
moves back
back and
and forth
forth over
over aa fixed
fixed
path,
path, returning
returning to
to each
each position
position and
and velocity
velocity
after
after aa definite
definite interval
interval of
of time.
time.
The
The frequency
frequency (rev/s)
(rev/s) isis the
the
reciprocal
reciprocal of
of the
the period
period (time
(time
for
for one
one revolution).
revolution).
x
m
F
1
f 
T
Summary (Cont.)
Hooke’s
Hooke’s Law:
Law: In
In aa spring,
spring, there
there isis aa restoring
restoring
force
force that
that isis proportional
proportional to
to the
the displacement.
displacement.
F  kx
x
The spring constant k is defined by:
m
F
F
k
x
Summary (SHM)
x
a
v
m
x = -A
x=0
F  ma  kx
x = +A
 kx
a
m
Conservation of Energy:
22 + ½kx 22 = ½mv 22 + ½kx 22
½mv
½mvAA + ½kxAA = ½mvBB + ½kxBB
Summary (SHM)
1
2
v
mv  kx  kA
2
k
A2  x 2
m
k
A
m
2
1
2
x  A cos(2 ft )
2
1
2
v0 
a  4 f x
2
v  2 fA sin(2 ft )
2
Summary: Period and
Frequency for Vibrating
Spring.
a
v
x
m
x = -A
x=0
x = +A
1
f 
2
a
x
x
T  2
a
1
f 
2
k
m
m
T  2
k
Summary: Simple Pendulum
and Torsion Pendulum
1
f 
2
g
L
I
T  2
k'
L
L
T  2
g
CONCLUSION: Chapter 14
Simple Harmonic Motion