Transcript SHM plus problems

Simple Harmonic Motion
NCEA AS 3.4
Text: Chapters 6-8
1
SHM



A periodic or repeating
motion.
Object moves back and forth
over the same path
Egs pendulum, mass on
spring, bobbing up and down
in waves, electrons in A.C
circuits,
2


Equilibrium position
– the undisturbed
position of the
object (mid-point of
motion)
Amplitude –
maximum distance
moved by object
from equilibrium
EQUILIBRIUM
Terms to know…
A
A
3
Definition of SHM

What makes it different to other
forms of periodic motion?
• There is a “restoring force” acting on
the object trying to return it to it’s
equilibrium position
• The size of this restoring force (and
also therefore the acceleration) is
directly proportional to the displacement
from equilibrium
4
Definition of SHM

In other words…
F  ky
where
F  restoring force
k  a constant
y  displacement from equilibrium
5
SHM vs Circular Motion


Another way to think of SHM is as
the component in one direction of
the motion of an object moving in a
circle.
Eg compare the motion of your foot
and your knee when riding a bike
Do pg122 Qu 1-4
6
Reference Circles



This is the circle of
motion associated
with an object
doing SHM.
It’s centre is at
equilibrium
It’s radius is the
amplitude of the
SHM.
m
p
7
Reference Circles


Point p moves
around the circle at
constant speed and
is always
horizontally in line
with the mass on
the spring
The period of the
SHM equals the
period of the
circular motion.
m
p
8
Displacement-time

p
m
y
q
At any time t,
displacement y can
be calculated by:
y  A sin q
q
but since  
t
we write it as :
y  Asin t
9
Displacement


The displacement of an object doing
SHM varies as a sine function.
Eg ink pot pendulum
10
Displacement-time
t=
0
1
2
3
4
5
6
7
8
m
m
m
m
m
m
m
m
m
11
Displacement-time

A graph of displacement-time can be
drawn by using a rotating radius
vector known as a Phasor.
t=2
t=3
t=1
ω
t
t=0
1
2
3
4
5
6
7 8
t=7
t=5
t=6
12
Period and frequency
2

T
  2f
1
T
f

Just like circular
motion, SHM has a
frequency, period and
angular frequency .
13




Note
It is really important to know where
the SHM motion is at t=0s
If t=o at y=0 i.e. equilibrium,
displacement will be a sin graph.
If t=0 at y=+A you will have a ……
displacement graph.
What will it be for y=-A at t=0????
Y=-Acosθ
14
Sam is studying his grandfathers clock.
He measures the length of the
pendulum 1.0m and
notices it ticks at 1.0s intervals. The
mass swings 3.0cm from the middle.
Y=+A @ t=0s
What is the amplitude of the motion?
3.0cm
What is the period of the motion?
2.0s
Calculate the angular frequency
15
2
2
1


 3.1 rad s
T
2.0
Sketch a displacement time graph for
the motion. Label the axis
3.0
t-s
Y-cm
0.5
-3.0
1.0
1.5
2.0
16
Write down the equation for the
displacement of the motion
y=0.030cos3.1t
What is the displacement at t=0.25s?
y=0.030cos(3.1x0.25)
=2.1cm
17
Velocity

vm
vp
p
m
q
For p and m to be
staying in line
horizontally they
must both have the
same vertical
component to their
speeds at any
time.
18
Velocity

Vm will be equal to
the vertical
component of vp.
vm  v p cos q
vm
vp
q
Since vp is constant :
v p  r  A
So we write it as :
vm  Aω cos ωt
19
Velocity-time Graphs


A phasor diagram
can be used to
draw velocity time
graphs
The velocity
phasor

• Has length = A
• Is 90° or /2 rads
ahead of the
displacement phasor
20
Velocity-time Graphs

The velocity-time graph:
t=0
t=1
t=7
ω
t
t=6
1
2
3
4
5
6
7 8
t=5
t=3
t=4
21
Velocity


The maximum speed the object
travels is when vm=vp=A
The object has this maximum speed
when displacement is zero, and zero
speed when displacement is
maximum.
22
Velocity
velocity
t
ω
1
2
3
4
5
6
7 8
displacement
23
Back to the Grandfather clock…………
remember : y=0.030cos3.1t
Write an equation for the velocity of
the pendulum.
v=-Aωsinωt
v=-0.093sin3.1t
What is the maximum speed? 9.3 cms-1
What is the speed at t=0.25s?
v=-0.093sin(3.1x0.25)
=-6.5 cms-1
What is the significance of the
negative sign The pendulum is moving
towards the negative side
24
Acceleration


Point p has
centripetal
acceleration in
towards the centre
of the circle
The object has
acceleration
towards
equilibrium equal
to the vertical
component of ap.
p
m
am
ap
q
25
Acceleration
am  a p sin q
However
q
2
am
ap
Negative sign
indicates
direction
v
ap 
r
2

A 
2

 A
A
So we write it as :
am   A sin t
2
26
Acceleration
Rememberin g that y  A sin t
and am   A sin t
2
then am   y
2
(Compare this to SHM definition )
27
Acceleration-time Graph

Looks like this:
t=6
t=7
t=5
t
t=0
ω
t=1
1
2
3
4
5
6
7 8
t=3
t=2
28
Acceleration


The acceleration phasor is 180° or 
radians ahead of the displacement
phasor.
This is because when the
displacement is at it’s greatest
positive value, the acceleration is
greatest back towards the centre, ie
negative, and vice versa
29
All three….
velocity
t
ω
1
2
3
4
5
6
7 8
acceleration
displacement
30
All three…
y  A sin t
v  A cos t
a   A sin t
2

If t=0 is not equilibrium, then all the
phasors shift their starting positions. The
graphs start in different positions and the
sin/cosine function may change.
Do pg 124 Qu 13-22
31
Back to the Grandfather clock…………again
remember : y=0.030cos3.1t and
v=-0.093sin3.1t
Now write an equation for the acceleration
of the pendulum. a=-A ω2cos ωt
a=-0.29cos3.1t
What is the maximum acceleration? 29 cms-2
What is the acceleration at t=0.25s?
a=-0.288cos(3.1x0.25)
=-21 cms-2
What is the significance of the
negative sign The pendulum is accelerating
towards the negative side 32
What is the acceleration at t=0.75s?
a=-0.288cos(3.1x0.75)
=-0.288 x -0.6847
= 20 cms-2
Notice how the answer
is now positive
That means
acceleration is towards
the positive side
33
Tides
• The tide rises and falls with SHM with a
period of approx. 12 hours. If the
difference between high and low tide is
6.0m, how long does it take to fall 1.0m
from high tide?
34
1.0m

q
3.0m
2.0m
6.0m
2
0
q  cos
 48
3
1
In12 hours circle movesthrough 360
0
0
48
x
12

1
.
6
hours
0
360
35
What is the maximum speed of the tide?
2
v m  A 

2

x3.0
12 x60 x60
1
 0.44mms
What is the speed of the tide when it is
1.0m out from high tide?
36
t=0 @ y=+A
vm   A sin t
2
0

x3.0 x sin 48
12 x60 x60
1
  0.32mms
37
Mass on Spring
F  ky

Where k=the spring
constant measured in Nm-1
and y = the extension or
compression of the spring
y
m
-y
38
Maths….
F  ky
F  ma
k
2
a   y   y
m
k
 
m
k

m
2
39
Maths continued…
2

T
k 2

m T
m
T  2
k


So the larger the
mass, the longer
the period of
oscillation
The stiffer the
spring, the shorter
the period of
oscillation
40
Springs

If more than one
spring is used
then k=k1+k2
because the
restoring force
from both
springs add
k1
together.
k1
k2
m
k2
m
Do pg 134 Qu 1-4
41
Pendulums

q
L
L
A pendulum is only
SHM if q is small,
so that y is
approximately a
straight line.
y
42
Pendulums
Ftension

q

Fresultant
Fweight
The forces can be
resolved into
components.
The component
acting in the
direction of the
motion is the
restoring force.
q
43
Maths

See page 131 in your book to see how
this formula is derived…
L
T  2
g
44
What length of pendulum is required to keep
the same time as a clock?
You want a period of one second.
g=9.81ms-2
L
L
T  2
 1  2
g
9.81
L  0.248m
 25 cm slightly under
Build a pendulum and test it.
45
Pendulums


Period of oscillation is independent of
mass and amplitude, so pendulums
are often used as timing devices. Eg
grandfather clocks
So period of oscillation really only
depends on how long the pendulum
is, as g is fairly constant on earth.
46
Do pg 136 Qu 5-7
Torsional Pendulums



Combination of
rotational motion
and SHM
t proportional to a
Acceleration is
given by:
a   q
2
Where  = 2/T
Do pg 137 Qu 8,9 (hard!)
47
Energy in SHM


Energy must be supplied to start
SHM
During each cycle the energy
changes between kinetic energy and
potential energy
48
Energy in SHM
Etotal
Ekinetic
Epotential
t
49
Energy in SHM

For all objects undergoing SHM:
• Potential energy is maximum and
kinetic energy is zero at maximum
diplacement.
• Potential energy is zero and kinetic
energy is maximum at equilibrium.
50
Energy in SHM
E
Etotal
Ekinetic
Epotential
y
-A
+A
51
Energy

Reminders of
energy equations:
Do pg 146 Qu 1-5
1 2
E p  ky
2
E p  mgh
1 2
Ek  mv
2
Substituti ng v  A
1
2 2
E  m A
2
52
Damping



In practice, energy is lost due to
friction during each oscillation.
This reduces the total amount of
energy and therefore reduces the
maximum displacement that the
object reaches. ie amplitude
decreases.
NOTE: Damping does NOT alter the
period.
53
Max A
Damping
Decay
Envelope
t
54
Damping



Damping is measured as a
percentage loss per cycle.
Eg. An object with amplitude 1m and
20% damping:
1m, 0.8m, 0.64m, 0.512m, 0.41m
(Each value is 80% of previous one)
55
Damping

Damping can be heavy or light
depending on the situation:
• Eg car suspension, balance arms, clock
pendulums,
t
t
56
Forced Oscillations



If an object is free to move , a
driving force can be applied to make
it oscillate.
The frequency of the oscillation
equals the frequency of the driving
force.
The amplitude can vary, depending
on what frequency is driving it.
57
Forced Oscillations
A
Natural
frequency
Driving
Frequency
58
Resonance



If the driving frequency equals the
objects natural frequency, the
amplitude gets very large.
This is called Resonance
Resonance can be useful (eg musical
instruments, tuning radio circuits) or
problematic (eg building or bridge
resonance in winds or earthquakes)
Do pg 147 Qu 6-10
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