Transcript Section 4.5 – Simple Harmonic Motion; Damped Motion; Combining Waves

Section 4.5 – Simple Harmonic
Motion; Damped Motion;
Combining Waves
Equilibrium is the location where the object is at rest
A
C
B
equilibrium
The amplitude of vibration is the
distance from the equilibrium position to
its point of greatest displacement (A or
C).
The period of a vibrating object is the
time required to complete one vibration that is, the time required to go from point
A through B to C and back to A.
Simple harmonic motion is a
special kind of vibrational motion
in which the acceleration a of the
object is directly proportional to
the negative of its displacement d
from its rest position. That is,
a = -kd, k > 0.
Theorem Simple Harmonic Motion
An object that moves on a coordinate axis
so that its distance d from the origin at
time t is given by either
d  a cost
or
d  a sin t
where a and  > 0 are constants, moves with
simple harmonic motion. The motion has
amplitude a and period 2  .
The frequency f of an object in simple
harmonic motion is the number of
oscillations per unit of time. Thus,
 1
f 
=
2 T
 0
Suppose that the distance d (in centimeters)
an object travels in time t (in seconds)
satisfies the equation
d  15sin 4t
(a) Describe the motion of the object.
Simple harmonic
(b) What is the maximum displacement
from its resting position?
A   15  15 centimeters
Suppose that the distance d (in centimeters)
an object travels in time t (in seconds)
satisfies the equation
d  15sin 4t
(c) What is the time required for one
oscillation?
2 
Period  T 
 seconds
4 2
(d) What is the frequency?
1 2
frequency  f   oscillations per second
T 
The simple harmonic motion on the previous
slides assumes that there is no resistance to
the motion…no friction. The motion of such
systems will continue indefinitely
In most physical systems, this is not the case.
Friction or resistive forces remove energy
from the system, dampening its motion.
Theorem Damped Motion
The displacement d of an oscillating object
from its at rest position at time t is given by
2


b
bt 2 m
2
d  ae
cos   2 t 
4m 

where b is a damping factor (damping
coefficient) and m is the mass of the
oscillating object.
Suppose that a simple pendulum with a bob of
mass 10 grams and a damping factor of .8
grams/sec is pulled 20 centimeters from its at rest
position and released. The period of the pendulum
without the damping effect is 4 seconds.
20 cm
a. Find the equation that describes the position of
the pendulum bob.
b. Using a graphing utility, graph the function
found in part (a).
c. Determine the maximum displacement of the
bob after the first oscillation.
d. What happens to the displacement of the bob as
time increases without bound?
a. Find the equation that describes the position of
the pendulum bob.
m = 10, a = 20, b = 0.8 The period under simple
harmonic motion is 4 seconds.
4

2

2
4
2


b
bt / 2 m
2


d  ae
cos  
t
2 

4m 


    2 0.82 


2 d  20e 0.8t / 2 (10) cos   
t
2
  2  410 


  2 0.64 
d  20e 0.8t / 2 (10) cos

t
 4
400 


b. Using a graphing utility, graph the function
found in part (a).
25
20
15
10
5
-10
-15
-20
-25
23.8
22.1
20.4
18.7
17
15.3
13.6
11.9
10.2
8.5
6.8
5.1
3.4
1.7
-5
0
0
c. Determine the maximum displacement of the
bob after the first oscillation.
25
About 17 cm
20
15
10
5
-10
-15
-20
-25
23.8
22.1
20.4
18.7
17
15.3
13.6
11.9
10.2
8.5
6.8
5.1
3.4
1.7
-5
0
0
d. What happens to the displacement of the bob as
time increases without bound?
Notice that the dampened motion formula contains
the factor
e
0.8t / 20
What happens to this factor as t grows without
bound, t -> ∞?
e
0.8 t / 20
0
Read pages 305-307 on Combining Waves
To graph a function h(x) = f(x) + g(x):
1. Graph f(x) and g(x)
2. Select several x values from the domains of the
f(x) and g(x)
3. Calculate f(x) and g(x) for the values of x
selected in 2.
4. The value of h(x) for a given value of x will be
the sum of f(x) and g(x).
Example: If we pick x = 2 and discover that
for some given f(x) and g(x) we have
f(2) = 4, and g(2) = 7,
then the value of h(2) = 4 + 7 = 11,
and the point (2,11) appears on the graph of
h(x).