Download Document

General Physics I
Spring 2011
Oscillations
1
Oscillations
• A quantity is said to exhibit oscillations if it varies with time
about an equilibrium or reference value in a repetitive
fashion.
• Oscillations are periodic when the time for one complete
cycle of the oscillations is constant. This repeat time is
called the period (T) of the oscillation.
• Let us assume that for a given oscillation, f cycles are
completed in one second. The number of cycles per
second is called the frequency of the oscillation. Since f
cycles are completed in one second, it follows that one
cycle is completed in 1/f seconds. The time taken to
complete one cycle is the period. Thus,
T=1
f
or
f = 1.
T
The unit of frequency is the hertz (Hz). 1 Hz = 1 cycle/sec.
2
Simple Harmonic Motion
• Consider an object attached to
a spring at one end. The other
end of the spring is fixed. The
object rests on a frictionless
horizontal surface. The object is
in equilibrium when the spring is
not stretched or compressed. If
the object is displaced from the
equilibrium position and
released, it will oscillate about
this position.
• If the displacement of the object
from its equilibrium position is
graphed as a function of time,
one obtains sinusoidal behavior
(sine or cosine function).
f =1/T .
3
Simple Harmonic Motion
Oscillatory motion for which the position is a sinusoidal function
of time is called simple harmonic motion.
4
Restoring Force
• An oscillation must have a
restoring force. This is the net force
that always acts in a direction
toward the equilibrium position,
and thus tends to “restore” the
object back to the equilibrium
position. The oscillation is
sustained because when the object
reaches the equilibrium position
(where the net force is zero), it has
a non-zero velocity and so its
inertia carries it past the
equilibrium position. The restoring
force then slows it down, it stops
momentarily and then moves back
toward the equilibrium position.
Marble in a bowl
Object attached to spring
5
Linear Restoring Force
• The spring force is a linear restoring
force because the force is proportional to
the displacement from the equilibrium
position:
(Fsp )x =−k ∆x.
• The equation above is called Hooke’s
law. Note that the force is in the opposite
direction to the displacement of the free
end of the spring from its equilibrium
position. Thus, the force always points
toward the equilibrium position.
• If we take the equilibrium position to be x
= 0, then ∆x = x. So,
(Fsp )x =−kx.
• A linear restoring is required for simple
harmonic motion.
Slope is –k, where k is the
spring constant.
6
Workbook: Chapter 14, Questions 1, 2
7
Describing Simple Harmonic Motion
• The graph to the right shows the
position (or displacement) x versus
time t for a particle undergoing simple
harmonic motion (SHM). The position
as time progresses is described by
the cosine function:


 2π t 
x = Acos 
.
 T 


In the equation above, A is the
magnitude of the maximum
displacement from equilibrium, which
is the amplitude. When
x = A or –A, the particle is at its
maximum distance from equilibrium
and so the velocity is zero. The
particle changes directions at these
points, which are called turning
points.
8
Describing Simple Harmonic Motion





•
•
•
•
•

x = Acos 2π t .
T 
Note that the argument of the cosine function (the quantity
inside the parentheses) must be in radians.
At t = 0, the argument of the cosine function is zero. The cosine
of zero radians = 1. Thus, x = A. The object starts at the
maximum positive displacement from equilibrium.
At t = T/4 (one-fourth of one period), the argument of the cosine
function is π/2 radians. The cosine of π/2 radians is zero. Thus,
at this time, the object is at the equilibrium position.
At t = T/2 the argument of the cosine function is π radians. The
cosine of π radians is -1. Thus, at this time, the particle is at x =
-A and is about to change directions.
At t = 3T/4, the argument of the cosine function is 3π/2 radians.
The cosine of 3π/4 radians is zero. Thus, the particle is at the
equilibrium position again.
9
Describing Simple Harmonic Motion






x = Acos 2π t .
T 
• At t = T, the argument of the cosine function is 2π radians. The
cosine of 2π radians = 1. Thus, x = A. The object is back to its
starting point. One cycle has been completed.
Velocity
• To find the velocity of the object at any instant, we can find the
slope of the position graph. It is easy to see that the slope is
zero when x = ±A, so the velocity is zero at these times. These
are the turning points as mentioned before. Also, the slope has
maximum magnitude when x = 0, i.e., at the equilibrium. Thus,
the speed of the object is greatest at the equilibrium position.
10
Describing Simple Harmonic Motion
• From the slope of the position
graph, we find that the velocity of
the object undergoing SHM is
given by


 2π t 
vx =−vmax sin 
,
 T 


where
vmax = 2π A.
T
• Like the cosine function, the sine
function will be positive, zero, or
negative at different times. Thus,
the velocity will negative, zero, or
positive at these times. A negative
velocity simply indicates motion in
the negative x direction (leftward).
11
Describing Simple Harmonic Motion
Acceleration
• One can find the acceleration versus time by finding the slope
of the velocity vs. time graph. However, another way to obtain
the acceleration vs. time behavior is to use Newton’s second
law. The restoring force is the only force acting along the
direction of motion and so is the net force in this direction. The
acceleration is given by ax =(Fnet)x/m. But, (Fnet)x = −kx. Thus,
k x.
ax =− m
• We see that the acceleration is proportional to the position but
in the opposite direction to it. Thus, the acceleration vs. time
graph must also be a cosine graph, but an inverted one
because of the minus sign. Using our previous expression for x
as a function of t, we find that




k x =− kA cos  2π t .
ax =− m
m
 T 


12
Describing Simple Harmonic Motion
• We can rewrite the acceleration
as a function of time as


 2π t 
ax =−amax cos 
,
 T 


where
2A
kA
4
π
amax = m = 2 .
T
13
Describing Simple Harmonic Motion
14
A mass attached to a spring oscillates back and forth as
indicated in the position vs. time plot below. At point P, the
mass has
1.
2.
3.
4.
5.
6.
7.
positive velocity and positive acceleration.
positive velocity and negative acceleration.
positive velocity and zero acceleration.
negative velocity and positive acceleration.
negative velocity and negative acceleration.
negative velocity and zero acceleration.
zero velocity but is accelerating (positively or
negatively).
8. zero velocity and zero acceleration.
15
Vertical Springs
• If an object is mounted on a
vertical spring, it will execute
simple harmonic motion in exactly
the same way as an object that is
attached to a horizontal spring.
• The vertical orientation causes
gravity to lower the equilibrium
position. That is all.
16
Workbook: Chapter 14, Question 7
Textbook: Chapter 14, Question 23,
Problem 9
17