Transcript No Slide Title

Chapter 13
Simple Harmonic Motion
In chapter 13 we will study a special type of motion
known as: simple harmonic motion (SHM). It is
defined as the motion in which the position coordinate
x of an object that moves along the x-axis varies with
time t as:
x(t) = Asin(t)
or
x(t) = Acos(t)
The condition for having simple harmonic motion is
that the force F acting on the moving object is:
F = -kx
(13-1)
Periodic motion is defined as a motion that repeats itself in time.
The period T  the time required to complete one repetition Units: s
A special case of periodic motion is known as “harmonic” motion. In
this type of motion the position x of an object moving on the x-axis is
given by:
x(t) = Asin(t)
or
x(t) = Acos(t)
The most general expression for x(t) is:
The parameters A, , and  are constants
(13-2)
x(t) = Asin(t + )
Plots of the sine and cosine functions. Make sure that you can
draw these two functions
/2 = 90º
 = 180º
3/2 = 270º
2= 360º
Note: The period of the sine
and cosine function is 2
(13-3)
Most general form of simple harmonic motion
x(t) = Asin(t + )
A  amplitude
t
  angular frequency
m
O
.
x-axis
  phase
x(t)
sin(a + b) = (sina)(cosb) +(cosa)(sinb)

x = (Acos )sint + (Asin)cost
The coordinate x is a superposition of sint and cost
functions
Special case 1:  = 0  x = A sint
Special case 1:  = /2  x = A cost
(13-4)
The effect of the phase  on is to shift the Asin(t + )
function with respect to the Asin(t) plot This is shown
in the picture below
Red curve: Asin(t)
Blue curve: Asin(t + )
(13-5)
t
m
O
.
x-axis
x(t)
Determination of A and 
These two parameters are determined
from the initial conditions
x(t = 0) and v(t = 0)
dx
x  A sin( t   )
Velocity v 
 A cos( t   )
dt
x(0)  A sin 
eqs.1
v(0)  A cos 
eqs.2
Divide eqs.1 by eqs.2

tan 
Once we have  we can get A from eqs.1
The acceleration

 x(0)
v(0)
x(0)
A
sin 
dv
a 
 -  2 A sin( t   )  -  2 x
dt
(13-6)
x = Asin(t)
v = Acos(t)
Plots of x(t), v(t) and a(t) versus t
for  = 0
a = -2Asin(t)
(13-7)
Consider a point P moving around O on a circular path of
radius R win constant angular velocity . The Cartesian
coordinates of P are: x and y. The angle  = t. From
triangle OAP we have: OA = x = Rcost and AP = Rsint
P
A

The projection of point
P on either the x- or the
y-axis executes
harmonic motion as
point P is moving with
uniform speed on a
circular orbit
(13-8)
m
k
O
Fs
m
A mass m attached to a
x axis
spring (spring constant k)
moves on the floor along the
x axis x-axis without friction
x
k
d2x
Fnet  Fs  kx  ma  a   x , a  2
m
dt
d2x
k
 x
k and m are constants
2
m
dt
k
k

is also a constant
 2 
m
m

2
d x
2



x
2
dt
(13-9)
m
k
x axis
O
Fs
m
x axis
x
d2x
2



x 
2
dt
d2x
2


x0
2
dt
k
2
 =
m
d 2x
2


x0
is the equation of motion. It is called a
2
dt
"differential equation" because it involves derivatives
A function x(t) is called a "solution" of a differential equation if
when it is substituted into the differential equation it "satisfies" it.
In this case the expression:
d 2x
2


x
2
dt
must be equal to zero
(13-10)
m
k
x axis
O
Fs
m
x axis
d2x
2


x0
2
dt
k
2
 =
m
x
Solution candidate: x  A sin( t )

dx
d2x
= A cos( t )  2  - 2 A sin( t ) 
dt
dt
d2x
2
2
2


x


A
sin(

t
)


A sin( t )  0
2
dt
Indeed the proposed function x  A sin( t )
d2x
makes 2   2 x
dt
equal to zero and thus it is a solution.
(13-11)
m
k
Summary
x axis
O
Fs
m
x axis
x
The spring-mass system obeys the
following equation:
d2x
k
2
 x  0 ,  
 2 f
2
m
dt
The most general solution of the equation of motion is:
x = Asin(t + )
Note 1: The angular frequency  ( and thus the frequency f , and the
period T) depend on m and k only
Note 2: The amplitude A and the phase  on the other hand are
determined from the initial conditions:
x(t = 0) and
v(t=0)
(13-12)
m
k
x axis
O
Fs
Energy of the simple
harmonic oscillator
m
x axis
x
x  A sin( t   )
v  A cos( t   )
k
2
 
m
kx 2
kA2
Total energy E  U  K , U 

sin 2 ( t   )
2
2
mv 2
mA2 2
kA2
2
K 

cos ( t   ) 
cos 2 ( t   )
2
2
2
kA2
E  U  K 
[sin 2 ( t   )  cos 2 ( t   )]
2
sin 2   cos 2   1
 [sin 2 ( t   )  cos 2 ( t   )]  1
kA2
 E 
2
Energy is constant
(13-13)
X
(13-14)
Why all this fuss about the simple
harmonic oscillator?
E
Consider an object of energy E
(purple horizontal line) moving in
a potential U plotted in the figure
using the blue solid line
Motion in the x-axis is allowed in the regions where U  E
These regions are color coded green on the x-axis. Motion is forbidden
elsewhere (color coded red)
If we have only small departures from the equilibrium positions x01 and
x02 we can approximate U by the potential of a simple harmonic
oscillator (the blue dashed line in the figure)
(13-15)
2
df
2 d f
f ( x)  f ( xo )  ( x  xo ) ( xo )  ( x  xo )
( xo )  ...
2
dx
dx
E
We expand the function U(x) around
one of the minima (x01 in this
example) using a Taylor series
( x  xo1 ) dU
( x  xo1 )2 d 2U
U ( x)  U ( xo1 ) 
( xo1 ) 
( xo1 )  ...
2
1! dx
2! dx
( x  xo1 )2
dU
d 2U
( xo1 )  0 , 2 ( xo1 )  k Set U ( xo1 )  0  U ( x) 
k
dx
dx
2
If we choose the origin to be at xo1  U(x) = kx2/2
This approximation for U is indeed the potential of a simple harmonic
oscillator (dotted blue line)
(13-16)
Simple pendulum: A mass m suspended from a string of
length  that moves under the influence of gravity
s  
ds
d
Velocity v 

dt
dt
Arc length
s
dv
d 2
Tangential acceleration a t =
=
From Newton's second law
2
dt
dt
we have: F  mat where Ft is the net force along the arc.
d 2
d 2
g
Ft  - mg sin   m
 -mg sin  
  si n 
2
2
dt
dt
(13-17)
Simple Pendulum
(13-18)
d 2
g
  sin 
2
dt
This differential equation is too
difficult to solve. For this reason we
consider a simpler version making the
“small angle approximation”
Assume:  <<1 ( is in radians)
Taylor expansion
of the function sin around   0 : sin    If  << 1
d 2
g
 
2
dt
then
sin  
3

5
 ...
3!
5!
The differential equation becomes:
This simpler version we can handle.
(13-19)
Simple pendulum, small angle
approximation:   1
d 2
g
 
2
dt
g
If we call
  2 the equation
above takes the form:
2
d 2
d
x
2
2




Compare
this
equation
with:



x
2
2
dt
dt
whose solution is: x  A sin( t   ) The two equations have
identical form and therefore they have the same solutions: 
   o sin( t   )
where  
g
The period T  2
g
Summary
(13-20)
The simple pendulum
(Small angle approximation:)
 << 1
  o sin(t   )
Angular frequency  
g
Period
T  2
g
Note 1: the period T does not depend on the mass m
Note 2: The pendulum can be used as a clock
gT 2
For example if we set T = 1sec we get 
 0.248 m
2
4
A pendulum of that length is a practical proposition
In the small angle approximation we assumed that  << 1
and used the approximation: sin   We are now going to
decide what is a “small” angle i.e. up to what angle  is the
approximation reasonably accurate?
 (degrees)
(radians)
sin
5
0.087
0.087
10
0.174
0.174
15
0.262
0.259 (1% off)
20
0.349
0.342 (2% off)
Conclusion: If we keep  < 10 ° we make less that 1 % error
(13-21)
O
Energy of a simple pendulum
   o sin( t   )
Arc length s  
Potential energy U  mgh
h  OB  OA
A
U=0
s = 
C
B
OB  , From triangle OAC: OA  cos  
h  (1  cos  ) 
U  mg (1  cos  )
In the small angle approximation   1
cos   1 
2
2

 2  mg  2
 U  mg 1  1   
2 
2

(13-22)
Energy of a simple pendulum
   o sin( t   )
Arc length s  
(13-23)
ds
Velocity v 
  o cos( t   )
dt
mgl 2 mgl o2
Potential energy U 

sin 2 ( t   )
2
2
mv 2 m 2 o2 2
Kinetic energy K 

cos 2 ( t   )
2
2
2
mgl

g
o
2 
 K
cos 2 ( t   )
2
2
mgl o2
mgl

o
cos 2 ( t   )  sin 2 ( t   )  
E U  K 
2
2
The energy of the pedulum is constant
O

A
r
C
d
mg
The Physical Pendulum is an object of
finite size suspended from a point other
than the center of mass (CM) and is
allowed to oscillate under the influence
of gravity. The torque about point O due
to the gravitational force is:
  mgd From triangle OAC  d  r sin  
  mgr sin  (eqs.1) The torque is also given by the equation:
d 2
  I 2 (eqs.2) We eliminate  between eqs.1 and eqs.2 
dt
d 2
mgr

sin 
2
I
dt
(13-24)
O

r
d 2
mgr

sin 
2
I
dt
In the small angle approximation:   1
sin    
mg
3

5
 ...
3!
5!
d 2
mgr
sin   


2
I
dt
mgr
We define:
  2 The equation of motion becomes:
I
d 2
2



 This equation is an old friend and we can handle it
2
dt
(13-25)
O
(13-26)
r

mg
d 2
mgr
2
2




where

=
I
dt 2
I is the moment of inertia about an axis
that passes through the point of suspension O.
Use the parallel axis theorem: I  I cm  mr
2
mgr
  
I cm  mr 2
2
d 2
2
Compare the equation of the physical pendulum




2
dt
d2x
2
with that of the simple harmonic oscillator



x
2
dt
the two equations have the same form and therefore have
the same solution    o sin( t   )
Overall Summary
k
Fs
m
x axis
Spring-mass system
d 2x
k
2
  ,  
2
dt
m
x  A sin( t   )
Simple pendulum
x

d 2
2



 , 
2
dt
   o sin( t   )
r
g
Physical pendulum
(13-27)
d 2
mgr
2
   ,  
2
dt
I
   o sin( t   )
Damped Harmonic Oscillator
Consider the spring-mass system in which in
addition to the spring force Fs = -kx we have a
damping force
Fd = -bv.
The constant b is
known as the “damping coefficient” In this
example Fd is provided by the liquid in the lower
part of the picture
-kx
-bv
v x
d 2x
Fxnet
(eqs.1)
Fxnet  m 2 (eqs.2)
dt
d 2x
dx
If we eliminate Fxnet form eqs.1 and eqs.2 we get: m 2  kx  b
dt
dt
d 2x
k
b dx
Divide both parts by m 
 x
2
dt
m
m dt
dx
 kx  b
dt
(13-28)
2
(13-29)
d x
k
b dx
 x
2
dt
m
m dt
Solution:
x  Aet sin( t   )
2
b
k
2
2
   o  2 , o 
4m
m
b
Damping factor  
2m
A plot of the solution is given in the figure. The damping force . Results
in an exponential decrease of the amplitude with time. The mechanical
energy in this case is not conserved. Mechanical energy is converted
into heat and escapes.
Driven harmonic oscillator
k
Fs
m
Fosint
x axis
x
Consider the spring-mass system that oscillates with angular frequency
o = (k/m)1/2
If we drive with a force F(t) = Fosint
(where in general   o)
the spring-mass system will oscillate at the driving angular frequency 
with an amplitude A() which depends strongly on the driving
frequency as shown in the figure to the right. The amplitude has a
pronounced maximum when  = o This phenomenon is called
resonance
(13-30)
The effects of resonance can be quite dramatic as these pictures of the
Tacoma Narrows bridge taken in 1940 show. Strong winds drove the
bridge in oscillatory motion whose amplitude was so large that it
destroyed the structure. The bridge was rebuilt with adequate damping
that prevents such a catastrophic failure
(13-31)