Transcript 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, 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?
15 s
T
 0.50 s
30 cylces
x
F
Period: T = 0.500 s
1
1
f  
T 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
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=
DF
Dx
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  ½kx  ½kx
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=
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:
Work  ½kx  ½kx
2
2
8 cm
0
m
2
1
U  ½kx  ½(196 N/m)(0.08 m)
2
U = 0.627 J
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
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
xmax =  A
kA 400 N(  0.12 m)
a

m
2 kg
Maximum
Acceleration:
m
amax = ± 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:
½mvA2 + ½kxA 2 = ½mvB2 + ½kxB 2
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
k
v
A2  x 2
m
mv  kx  kA
2
1
2
2
1
2
vmax when
x = 0:
x = +A
2
v
k
A
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
k
v
A2  x 2
m
800 N/m
v
(0.1 m) 2  (0.06 m) 2
2 kg
v = ±1.60 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
v = ± 2.00 m/s
m
+x
The Reference Circle
The reference circle compares
the circular motion of an object
with its horizontal projection.
x  A cos   wt
x  A cos(2 ft )
x = Horizontal displacement.
A = Amplitude (xmax).
 = Reference angle.
w  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 = wR = wA; w  2f
v = -vT sin  ;
 = wt
v = -w A sin w t
v = -2f A sin 2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  = -ac cos(wt)
v2 w 2 R2
ac  
; ac  w 2 R
R
R
a = -w2A cos(wt)
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2f2x
1
f 
2
a
x
and
T = 1/f
x
T  2
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:
1
f 
2
k
m
m
T  2
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 = -0.2 m
1
f 
2
x=0
k
1

m 2
f = 2.25 Hz
x = +0.2 m
400 N/m
2 kg
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
a =  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 = -2f A sin 2f t
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)
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
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)
t = 0.157 s
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 k’ 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 k’ is
0.025 N m/rad. Find the period.
(Neglect the torsion in the wire)
For Disk:
I = ½mR2
I = ½(0.16 kg)(0.12 m)2
= 0.00115 kg m2
I
0.00115 kg m2
T  2
 2
k'
0.025 N m/rad
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
1
f 
T
Summary (Cont.)
Hooke’s Law: In a spring, there is a restoring
force that is proportional to the displacement.
F  kx
x
The spring constant k is defined by:
m
F
DF
k
Dx
Summary (SHM)
x
a
v
m
x = -A
x=0
F  ma  kx
x = +A
 kx
a
m
Conservation of Energy:
½mvA2 + ½kxA 2 = ½mvB2 + ½kxB 2
Summary (SHM)
1
2
mv  kx  kA
2
1
2
k
v
A2  x 2
m
x  A cos(2 ft )
2
1
2
v0 
2
k
A
m
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