Transcript Simple Harmonic Motion

Simple Harmonic Motion
Harmonic Motion
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Linear Motion- A B one place to another
Harmonic Motion- Repeat over and over
again ex. Swinging, walking in circles,
pendulums, bicycle wheels etc.
Cycle- building block of motion (back and
forth one cycle)
System – all things we are interested in,
exclude things we don’t. Pendulum, system
includes hanger, string and weight.
Oscillator- any system that shows harmonic
motion (heart, planets, pendulum etc.) Earth
has several oscillating systems.
Harmonic Motion
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Light and sound come from oscillations
Sound- oscillation of air(speaker pushes and
pulls on air creating an oscillation of
pressure)
Color- light waves are electromagnetic
oscillations. Faster oscillations make blue,
slower make red.
Technology- Fast electromagnetic oscillation
(cell phones 100 million cycles/sec)
(FM radio 88 to 107 million cylces/sec)
Harmonic Motion
Period- time for 1 cycle
 Frequency- # of cycles per second
Hertz = cycles/sec = Hz
t = 1/f f= 1/t inversely related
Amplitude- size of cycle, can be distance or
angle
Damping- friction eventually slows down all
oscillations and lowers amplitude
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Frequency
The FREQUENCY of a wave is the inverse of the
PERIOD. That means that the frequency is the
#cycles per sec. The commonly used unit is
HERTZ(HZ).
seconds
3.5s
Period  T 
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 2s
cycles 1.75cyc
cycles 1.75cyc
Frequency f 
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 0.5 c  0.5Hz
s
seconds 3.5 sec
1
1
T
f 
f
T
Pendulum Lab-only change 1 variable at a time!
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Part I:
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Part II
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Set up photogate so it
can read as the ball
passes through.
Adjust things as many
times as necessary but
see what variable
(amplitude, # of washers
or length of string has
biggest effect on Period.
Make sure timer is set on
Period (double the time
for 1 full cycle)
Put results in data table
and write conclusion.
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Start at 30 degree
amplitude, record time it
takes to decrease down
to 10 degree amplitude.
Change mass 4 times,
keeping string length
constant.
Change string length 4
times, keeping mass
constant.
Put results in data table
and write conclusion.
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Springs are like Waves and Circles
The amplitude, A, of a wave is the
same as the displacement ,x, of a
spring. Both are in meters.
CREST
Equilibrium Line
Trough
Ts=sec/cycle. Let’s assume that
the wave crosses the equilibrium
line in one second intervals. T
=3.5 seconds/1.75 cycles. T = 2
sec.
Period, T, is the time for one revolution or
in the case of springs the time for ONE
COMPLETE oscillation (One crest and
trough). Oscillations could also be called
vibrations and cycles. In the wave above
we have 1.75 cycles or waves or
vibrations or oscillations.
Simple Harmonic Motion
Back and forth motion that is caused by a force that is directly
proportional to the displacement. The displacement centers
around an equilibrium position.
Fs x
Springs – Hooke’s Law
One of the simplest type
of simple harmonic
motion is called
Hooke's Law. This is
primarily in reference to
SPRINGS.
Fs  x
k  Constantof Proportion
ality
k  Spring Constant(Unit : N/m)
Fs  kx or  kx
The negative sign only
tells us that “F” is what is
called a RESTORING
FORCE, in that it works in
the OPPOSITE direction
of the displacement.
Hooke’s Law
Common formulas which are set equal to
Hooke's law are N.S.L. and weight
Example
A load of 50 N attached to a spring hanging vertically stretches the
spring 5.0 cm. The spring is now placed horizontally on a table
and stretched 11.0 cm. What force is required to stretch the
spring this amount?
Fs  kx
50  k (0.05)
k
1000 N/m
Fs  kx
Fs  (1000)(0.11)
Fs 
110 N
Hooke’s law practice problems
1. What force is necessary to stretch an ideal spring whose force constant is 120. N/m by an amount
of 30. cm?
2. A spring with a force constant of 600. N/m is used on a scale for weighing fish. What is the
mass of a fish that would stretch the spring by 7.5 cm from its normal length?
3. A spring in a pogo-stick is compressed 12 cm when a 40. kg girl stands on the stick. What is
the force constant for the pogo-stick spring?
4. ****An elastic cord is 80. cm long when it is supporting a mass of 10. kg hanging from it at
rest at rest. When an additional 4.0 kg is added, the cord is 82.5 cm long.
HINT: 4 kg stretches the cord 2.5 cm!!
(a) What is the spring constant of the cord?
(b) What is the length of cord when no mass is hanging from it?
HINT once you have the k value, work the equation for spring force backwards!! THINK: How
much does 10.kg STRETCH the cord??
5. A spring is connected to a wall. A mass on a horizontal surface is connected to the spring and
pulled to the right along the surface stretching the spring by 25 cm. If the pulling force exerted on
the mass was 80.N, determine the spring constant of the spring. You then hang an unknown mass
hanging from the spring causes the spring to stretch 15 cm, what is the mass of the unknown?
Hooke’s law lab
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Purpose: Using the Springs and Swings apparatus, find
the spring constant of each of 5 springs provided to you
in lab. Label them with color and size.
Procedure: Placing a mass on the bottom hook of the
spring, record how far down the spring moves mg = k∆x
the mass of each mass is 12 g. You may use as many
of the masses as needed.
Record all data in a data table and label the k value for
each spring.
Find the unknown mass: Using the information provided
from your own data, find the mass of one individual
magnet and of the plastic washer provided to your lab
group. You may use any spring you wish, but you might
want to use two or three to confirm your answer.
Good Luck!
Hooke’s Law from a Graphical Point of View
Fs  kx
Suppose we had the following data:
x(m)
Force(N)
0
0
0.1
12
0.2
24
0.3
36
0.4
48
Fs
x
k  Slope of a F vs. x graph
k
Force vs. Displacement
y = 120x + 1E-14
R2 = 1
80
70
0.5
60
0.6
72
Force(Newtons)
60
50
k =120 N/m
40
30
20
10
0
0
0.1
0.2
0.3
0.4
Displacement(Meters)
0.5
0.6
0.7
We have seen F vs. x Before!!!!
Force vs. Displacement
Work or ENERGY = FDx
y = 120x + 1E-14
R2 = 1
80
Since WORK or ENERGY
is the AREA, we must get
some type of energy when
we compress or elongate
the spring. This energy is
the AREA under the line!
70
Force(Newtons)
60
50
40
30
Area = ELASTIC
POTENTIAL ENERGY
20
10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Displacement(Meters)
Since we STORE energy when the spring is compressed and
elongated it classifies itself as a “type” of POTENTIAL ENERGY, Us.
In this case, it is called ELASTIC POTENTIAL ENERGY.
Elastic Potential Energy
The graph of F vs.x for a
spring that is IDEAL in
nature will always
produce a line with a
positive linear slope.
Thus the area under
the line will always be
represented as a
triangle.
NOTE: Keep in mind that this can be applied to WORK or can be conserved
with any other type of energy.
Conservation of Energy in Springs
Example
A slingshot consists of a light leather cup, containing a stone, that
is pulled back against 2 rubber bands. It takes a force of 30 N to
stretch the bands 1.0 cm (a) What is the potential energy stored
in the bands when a 50.0 g stone is placed in the cup and pulled
back 0.20 m from the equilibrium position? (b) With what speed
does it leave the slingshot?
a ) Fs  kx 30  k (0.01) k  3000 N/m
U s  1 kx 2  0.5(k )(.20) 2 
2
b) E B  E A U s  K
U s  1 m v2  1 (0.050)v 2
2
2
v  49 m/s
60 J
SHM and Uniform Circular Motion
Springs and Waves behave
very similar to objects that
move in circles.
The radius of the circle is
symbolic of the
displacement, x, of a spring
or the amplitude, A, of a
wave.
xspring  Awave  rcircle
SHM and Uniform Circular Motion
•The radius of a circle is symbolic of the
amplitude of a wave.
•Energy is conserved as the elastic
potential energy in a spring can be
converted into kinetic energy. Once
again the displacement of a spring is
symbolic of the amplitude of a wave
•Since BOTH algebraic expressions
have the ratio of the Amplitude to the
velocity we can set them equal to each
other.
•This derives the PERIOD of a SPRING.
Example
A 200 g mass is attached to a spring and executes
simple harmonic motion with a period of 0.25 s If the
total energy of the system is 2.0 J, find the (a) force
constant of the spring (b) the amplitude of the motion
m
0.200
Ts  2
 0.25  2
k
k
U s  1 kx2  2  1 kA2
2
2
k
A
126.3 N/m
0.18 m
Pendulums
Pendulums, like springs, oscillate
back and forth exhibiting
simple harmonic behavior.
A shadow projector would show
a pendulum moving in
synchronization with a circle.
Here, the angular amplitude is
equal to the radius of a circle.
Pendulums
Consider the FBD for a pendulum. Here we have the
weight and tension. Even though the weight isn’t at an
angle let’s draw an axis along the tension.
q
q
mgcosq
mgsinq
m gsin q  RestoringForce
m gsin q  kx
Pendulums
s s
q 
R L
s  qL  Am plitude
m gsin q  RestoringForce
m gsin q  kx
m g sin q  kqL
sin q  q , if q  sm all
m g  kl
m l

k g
Tspring
m
 2
k
What is x? It is the
amplitude! In the picture to
the left, it represents the
chord from where it was
released to the bottom of
the swing (equilibrium
position).
Tpendulum
l
 2
g
Example
A visitor to a lighthouse wishes to determine the
height of the tower. She ties a spool of thread
to a small rock to make a simple pendulum,
which she hangs down the center of a spiral
staircase of the tower. The period of oscillation
is 9.40 s. What is the height of the tower?
l
TP  2
 l  height
g
2
2
2
4
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l
T
g
9
.
4
(9.8)
2
P
TP 
l 
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
2
2
g
4
4(3.141592)
L = Height = 21.93 m