Download Simple Harmonic Motion

Simple Harmonic
Motion
Sinusoidal Curve and Circular Motion
A mass is oscillating on a spring
Position in
equal time intervals:
This type of motion, is
called Simple Harmonic
Motion. The system is
trying to place the mass
at its static equilibrium
position
Repetitive Motion
A is the equivalent to x
the displacement from
equilibrium and is also
the radius of the circle
Repetitive motion
can be modelled
as circular motion
What Can Algebra Tell Us?
Circle
Spring
Fa  Fs
mac  kx
4 2 r
m 2  kx
T
4 2 r
m 2  kr
T
2
4

m
2
T 
T
k
4 2 m
k
 2
m
k
This tells us that the period
only depends upon the mass
and the spring constant, not on
the distance the spring is
pulled back. In English, the
time for the mass to move
back and forth is ALWAYS the
same even if the amplitude
decreases.
What Can Algebra Tell Us?
Circle
Spring
Fa  Fs
mac  kx
v2
m  kx
r
v2
m  kr
r
kr 2
2
v 
m
We could has also come
up with the equations by
setting the Kinetic Energy
equal to the Spring
Potential Energy and
solving for v
k
vr
m
This tells us that the velocity
depends upon the inverse of
the mass, the spring constant,
and the distance the spring is
pulled back. In English, the
velocity of the spring depends
on how far it was pulled pack
and the spring constant as
well as the bigger the mass
the slower the velocity of the
spring.
What Can Algebra Tell Us?
Circle
Spring
Fa  Fs
mac  kx
mac  kr
ac 
kr
m
This tells us that the spring
experiences no acceleration
when r=0, this is, at the relaxed
position. But we have to use
logic to determine which way
the acceleration is pointing.
Calculus Can Tell us More
Circle
From Force by Spring
F  Fs
ma  kx
d 2x
m 2  kx
dt
d 2x
k
 x
2
dt
m
Therefore x is a function
of time, that when given
a time value, t, I can tell
you the position x, where
the mass is.
The force by the spring is
called a Restoring Force
But, the question is: What
function’s second derivative is
equal to this value?
 k 
x  t   sin 
t 
 m 
Circular Motion
Yes, But what if I didn’t Know Calculus?
Vertical position versus time:
Period T
Period T
Sinusoidal motion
Displacement (cm)
Time (s)
Period T
Sine function: mathematically
y
y=cos(x)
y=sin(x)
1
π/2
π 3π/2 2π 5π/2
-1
2π
3π 7π/2 4π 9π/2
5π
x
Sine function: employed for
oscillations
y
1
y=sin(x)
π/2
π 3π/2 2π 5π/2
3π 7π/2 4π 9π/2
x
5π
-1
Displacement y (m)
A
y= A sin(ωt)
T/2
-A
T
2T
Time t (s)
Sine function: employed for oscillations
Amplitude A is the maximum
1. Maximum displacement A
distance from equilibrium
2. ωT = 2π
2

T
3. Initial condition y(t=0)
Starting from equilibrium:
Starting from A:
Starting from -A:
y=A sin(ωt)
y=A cos(ωt)
y=-A cos(ωt)

 2
 2 
y  A sin t   A sin 
t   A sin 

m
 T 
2


k

Note:
Angular frequency in rad/s


 k 

t  A sin 
t  Therefore
m





k

m
Position, Velocity, Acceleration
 2
x  t   A sin t   A sin 
 T
v  t   A cos t   A
 k 

t   A sin 
t 
m



2
 2
cos 
T
 T
 k 
k

t  A
cos 
t 
m

 m 
4 2
 2
a  t    A sin t    A 2 sin 
T
 T
2
 2
x  t   A cos t   A cos 
 T
 k 

t   A cos 
t 

 m 
2
 2
v  t    A sin t    A
sin 
T
 T
 k 
k

t   A
sin 
t 
m
m



4 2
 2
a  t    A cos t    A 2 cos 
T
 T
2
 k 
k

t    A sin 
t 
m  m 

 k 
k

t    A cos 
t 
m

 m 
Example 1 - determine y(t)
y(cm)
30
5
15
10
t(s)
Period?
T=4 s
Sine/cosine?
Sine
Amplitude?
15 cm
 2 
y (Where
t )  15
sin after 12
t seconds?
is cm
the mass
4s
 2   
y (12s )  15cm sin  12s   15cm sin  6   0cm
 4s

Example 2 – graph y(t)
Amplitude?
3cm
 2
y (t )  3cm cos 
 2s
y(t=0)?
-3cm

t

When will the mass be at +3cm?
Period?
1s, 3s, 5s, …
2s
When will the mass be at 0?
y (cm)
0.5s, 1.5s, 2.5s, 3.5 s …
3
2
-3
4
6
8
t(s)
Summary
Harmonic oscillations are sinusoidal
 Motion is repeated with a period T
 Motion occurs between a positive and negative
maximum value, named Amplitude
 Position: y=A sin(ωt) or y=A cos(ωt)
 Velocity: v=Aωcos(ωt) or v=-Aωsin(ωt)
 Acceleration: a=-Aω2sin(ωt) or a=-Aω2cos(ωt)

2
 Angular frequency:  
T
m

T  2
k
Energy of Harmonic Motion

According to the law of conservation
of energy, when the mass of attached
to a spring is released, the total
energy of the system remains
constant. This energy is the sum of
the elastic potential energy and the
kinetic energy of the spring
ET 
1 2 1 2
kx  mv
2
2
Energy Example
A 55 g box is attached to a horizontal spring (force constant 24 N/m).
The spring is then compressed to a position A= 8.6 cm to the left of the
equilibrium position. The box is released and undergoes SHM.
a) What is the speed of the box when it is at x=5.1 cm from the
equilibrium position?
b) What is the maximum speed of the box?
Can you solve these questions two ways?
•Using energy?
•Using position and velocity formulas?
Energy Example
A 55 g box is attached to a horizontal spring (force constant 24 N/m).
The spring is then compressed to a position A= 8.6 cm to the left of the
equilibrium position. The box is released and undergoes SHM.
a) What is the speed of the box when it is at x=5.1 cm from the
equilibrium position?
ETi  ETf
Ep i  EKi  E p f  EK f
1 2
1
1
kxi  0  kx 2f  mv 2f
2
2
2
kxi2  kx2f  mv2f
mv2f  kxi2  kx2f
vf 
kxi2  kx 2f
m
2
2
 N
 N
24
0.086
m

24
0.051
m








m
m


vf 
0.055kg
v f  1.4
m
s
Energy Example
a) What is the speed of the box when it is at x=5.1 cm from the equilibrium
position?
Let the position function be:
 k 

24 
x  A cos 
t   0.086 cos 
t 
m
0.055




Let the velocity function be:
 k 
k
v  A
sin 
t   1.79648sin  20.8893 t 
m
 m 
Set the position function = to 5.1 and solve for the t the time.
 24 
0.051  0.086 cos 
t 
 0.055 
 24 
0.051
 cos 
t 
0.086
 0.055 
24
 0.051 
cos 
t

0.086
0.055


1
t  0.044807
Plug this t value into the velocity function
v  1.79648sin  20.8893  0.044807  
 1.4465
m
s
Therefore the speed is 1.4 m/s
Energy Example
A 55 g box is attached to a horizontal spring (force constant 24 N/m).
The spring is then compressed to a position A= 8.6 cm to the left of the
equilibrium position. The box is released and undergoes SHM.
b) What is the maximum speed of the box?
The maximum speed will occur when it is at the equilibrium position (xf=0)
ETi  ETf
Ep i  EKi  E p f  EK f
1 2
1
1
kxi  0  kx 2f  mv 2f
2
2
2
kxi2  kx2f  mv2f
mv2f  kxi2  kx2f
vf 
kxi2  kx 2f
m
2
2
 N
 N
24
0.086
m

24
0.
m








m
m


vf 
0.055kg
v f  1.8
m
s
Energy Example
b) What is the maximum speed of the box?
The maximum speed will occur when it is at the equilibrium position (xf=0)
Let the position function be:
 k 

24 
x  A cos 
t   0.086 cos 
t 
m
0.055




Let the velocity function be:
 k 
k
v  A
sin 
t   1.79648sin  20.8893 t 
m
 m 
Set the position function = to 0 and solve for the t the time.
 24 
0  0.086cos 
t 
 0.055 
 24 
0  cos 
t 
 0.055 

2

24
t
0.055
t  0.075196
Plug this t value into the velocity function
v  1.79648sin  20.8893  0.075196  
 1.79648
m
s
Therefore the speed is 1.8 m/s
Conservation of Energy
When the spring is in Equilibrium with gravity, the forces are
balanced (the downward force of gravity matches the upward
for of the spring. So: Fg  Fs  mg  kd
Where ‘d’ is the
equilibrium extension of
the spring and ‘g’ is the
acceleration due to
gravity. Therefore :
Fs  kd
Fg  mg
mg
d
k
d
Conservation of Energy
To simplify things. I’m
going
to choose
We
will now
set theUg = 0
at thebouncing.
extensionAt any
spring
equilibrium.
I’m freeis:to
time
the total energy
choose any location for
this.
ETotal  EK  U g  U S
d
x=0
Ug=0
This gives us:
ETotal
1 2
1
2
 mv  mgx  k  x  d 
2
2
+x
Conservation of Energy
mg
d
k
Now, let’s expand out
this equation:
ETotal
ETotal
ETotal
1 2
1
2
 mv  mgx  k  x  d 
2
2
1 2
1 2
1 2
 mv  mgx  kx  kxdd  kd
2
2
2
1 2
1 2
 mg  1 2
 mv  mgx  kx  kx 
  kd
2
2
 k  2
Conservation of Energy
Now, let’s expand out
this equation:
1 2
1 2
1 2
ETotal  mv  mgx  kx  mgx  kd
2
2
2
1 2 1 2 1 2
I’m a constant
ETotal  mv  kx  kd
2
2
2
Therefore, I’m a
constant too!
Java Site for SHM
http://canu.ucalgary.ca/map/content/
shm/springEnergy/simulate/page2.ht
ml