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Physics 430: Lecture 11
Oscillations
Dale E. Gary
NJIT Physics Department
October 7, 2010
5.1 Hooke’s Law
 As you know, Hooke’s Law (spring law) gives the force due to a spring, in
the form (assuming it is along the x axis):
where x is the displacement relative to the spring’s equilibrium point. The
force can be in either direction—in the +x direction when x is negative, and
in the -x direction when x is positive (at least for the case when k is
positive).
 Since the force is conservative, we can write the force as minus the
gradient of a potential energy
 From what we have learned about potential energy graphs, you can
immediately see that this upward curving parabola indicates that x = 0 is a
stable equilibrium (if k were negative, the parabola would be curved
downward and x = 0 would be an unstable equilibrium).
 However, Hooke’s Law has a general validity for the following reason: for
any general potential energy curve U(x) in the vicinity of an equilibrium
point at x = xo, which we can take to be zero, we can always perform a
Taylor-series expansion about that point:
s

kx
x
Fx -

)
(
.
)
( 2
2
1
kx
x
U 
.
)
0
(
)
0
(
)
0
(
)
( 2
2
1

+


+

+
 x
U
x
U
U
x
U
October 7, 2010
Hooke’s Law-2
 The first term is the constant term, but since potential energy can be
defined with any zero point, the constant term can be ignored (or
considered to be zero).
 The second term is the linear term, but since x = 0 is an equilibrium point,
the slope there is by definition zero, so the first non-zero term is the third
term:
 This says that for sufficiently small displacements from an equilibrium point,
Hooke’s Law is ALWAYS valid for any potential energy function. This
justifies our consideration of this case in detail.
 Of course, Hooke’s Law can relate to any coordinate, not just x. Let’s revisit
the box on a cylinder problem as an illustration.
.
)
( 2
2
1
kx
x
U 
October 7, 2010
Example 5.1: Cube Balanced on a
Cylinder
 Statement of the problem:
 Consider again the cube of Example 4.7 and show that for small angles q the
potential energy takes the Hooke’s Law form U(q) = ½ kq2.
 Solution:
 We found before that the potential energy for the cube in terms of q was
 For small q, we can make the approximations cos q ~ 1 – q2/2; sin q ~ q. Notice
that we kept the TWO leading terms for cos q. Why? With these substitutions,
we have
 Ignoring the constant term, we see that this is in the form of Hooke’s Law, with
a “spring constant” k = mg(r – b).
 Notice that the equilibrium is stable only when k > 0, i.e. when r > b, as we found
after quite a bit more work before.
 Notice also that since is a parabola, the turning points (at least
for small displacements) are equidistant from the equilibrium point.
].
sin
cos
)
[(
)
( q
q
q
q r
b
r
mg
U +
+

.
)
(
)
(
]
)
1
)(
[(
)
( 2
2
1
2
2
2
1
q
q
q
q b
r
mg
b
r
mg
r
b
r
mg
U -
+
+

+
-
+

2
1
2
( )
U x kx

October 7, 2010
5.2 Simple Harmonic Motion
 Let’s now look at all of this from the point of view of the equation of
motion. Consider a cart on a frictionless track attached to a spring with
spring constant k. Since we have
the equation of motion is
where we introduce the constant
which represents the angular frequency with which the cart will oscillate, as
we will see.
 Because Hooke’s Law always applies near equilibrium for any potential
energy, we will find oscillations to be very common, governed by the
general equation of motion such as for a pendulum vs. angle f:
kx
x
Fx -

)
(
x
x
m
k
x
kx
x
m 2

-

-


-
 



m
k


f

f 2
-



x
x = 0
October 7, 2010
Exponential Solutions
 The equation is a second order, linear, homogeneous differential
equation. Therefore, it has two independent solutions, which can be written
 As you can easily check, both of these solutions do satisfy the equation.
However, you should have come to expect two arbitrary constants in the
solution to a second-order differential equation (two constants of
integration), and in fact any linear combination of these two solutions is also
a solution
This is called the superposition principle, which works for any linear system.
 Any solution containing two arbitrary constants is in fact the general
solution to the equation.
 This solution can be written in terms of Sine and Cosine by using Euler’s
equation
 Plugging these into our original solution, we can write
.
)
(
and
)
( t
i
t
i
e
t
x
e
t
x 
 -


x
x 2

-



.
)
( 2
1
t
i
t
i
e
C
e
C
t
x 
 -
+

).
sin(
)
cos( t
i
t
e t
i






    ).
sin(
)
cos(
)
sin(
)
cos(
)
( 2
1
2
1
2
1 t
B
t
B
t
C
C
i
t
C
C
t
x 


 +

-
+
+

October 7, 2010
Sine and Cosine Solutions
 Thus, the solutions and
are equivalent so long as
 An important issue is that x(t), being a actual position coordinate, has to be
real. In general the first solution looks as if it is complex, while the second
solution looks as if it is real. However, this depends on whether the
coefficients are real or not. Both C1 and C2 can be complex, but if both their
sum and difference C1 + C2 and C1 – C2 are real then B1 and B2 are real and
in fact both versions are completely equivalent. This solution is the definition
of simple harmonic motion (SHM).
 As usual, we determine these constants from the initial conditions.
 If I start the motion by pulling the cart aside (position x(0) = xo) and releasing
it (v(0) = 0) then only the cosine term survives and
 If the cart starts at x(0) = 0 by giving it a kick (velocity v(0) = vo) then only the
sine term survives and
,
)
( 2
1
t
i
t
i
e
C
e
C
t
x 
 -
+

).
sin(
)
( o
t
v
t
x 


),
sin(
)
cos(
)
( 2
1 t
B
t
B
t
x 
 +

).
(
and
, 2
1
2
2
1
1 C
C
i
B
C
C
B -

+

).
cos(
)
( o t
x
t
x 

October 7, 2010
Phase Shifted Cosine Solution
 The two “pure” solutions just described are shown in the figures below.
 The general solution, if I both pull the cart to the side and give it a push
(both xo and vo non-zero), the motion is harder to visualize, but you may
expect it to be the same type of motion, with some phase shift.
 We can demonstrate that for the general case by defining another constant
where A is the hypotenuse of a right triangle whose sides are
B1 and B2. Obviously there is also an associated angle d such
that and . Thus
,
2
2
2
1 B
B
A +

x
t
xo
).
cos(
)
( o t
x
t
x 

x
slope vo
t
).
sin(
)
( o
t
v
t
x 


vo/
xo
A
B1
B2
d
B1 = A cos d B2 = A sin d
).
cos(
)]
sin(
sin
)
cos(
[cos
)
sin(
)
cos(
)
( 2
1
d


d

d


-

+

+

t
A
t
t
A
t
B
t
B
t
x
slope vo
Note that you have already solved this same
problem using energy, in problem 4.28.
October 7, 2010
Solution as Real Part of ei(t-d)
 Notice that in there are still two constants, A and d, that
are to be determined by initial conditions.
 But notice that we can rewrite this using Euler’s equation as
, where Re[] denotes the real part of what is
in the square brackets. As a practical matter, then, we can write the solution
in terms of a complex exponential, and after any mathematical
manipulations, simply throw away any complex part and keep the real part of
the solution as the answer to the physical problem at hand.
 Graphically, if you will recall our discussion from Lecture 5 on
complex exponentials, we can represent the solution as
the x (real) component, or projection, of the complex
number represented by a point in the complex plane
given by .
 You should train yourself to be able to go easily
between the phase shifted cosine and Re[] complex representations.
),
cos(
)
( d
 -
 t
A
t
x
 
)
(
Re
)
cos(
)
( d

d
 -

-
 t
i
Ae
t
A
t
x
d
t-d
C = Ae-id
x
y
A
Ceit =Aei(t-d)
x(t)=Acos(t-d)
)
( d
 -
t
i
Ae
October 7, 2010
Example 5.2: A Bottle in a Bucket
 Statement of the problem:
 A bottle is floating upright in a large bucket of water as shown in the figure. In
equilibrium it is submerged to a depth do below the surface of the water. Show
that if it is pushed down to a depth d and released, it will execute harmonic
motion, and find the frequency of its oscillations. If do = 20 cm, what is the
period of the oscillations?
 Solution:
 The forces on the bottle are mg downward, and the buoyancy force upward. Do
you recall how to determine the buoyancy force?
 The volume of displaced water is Ado, where A
is the cross-sectional area of the bottle. The mass
of the displaced volume is thus rAdo, and the weight
is rgAdo. At this depth, we have
 The equation of motion, then, is
where we have used d = do + x (i.e. x is measured from the
equilibrium position).
The buoyance force is equal to the
weight of the displaced water.
Archimedes’Principle
d
),
( o x
d
gA
mg
x
m +
-
 r


.
o
gAd
mg r

October 7, 2010
Example 5.2, cont’d
 Solution, cont’d:
 Since the weight of the bottle is the same as the displaced water at equilibrium:
we can replace mg and get
but we can express so we end up with the simple result
 This is the same form as the simple harmonic oscillator, with the same solution.
We identify
 You may recognize this as the same expression as for a pendulum of length l = do.
Note that the result does not depend on the shape or weight of the bottle, or the
density of the water, or anything else. If do = 20 cm, then
,
/ m
gAx
x r
-



.
o
gAd
mg r

,
/
/ o
d
g
m
gA 
r
,
o
x
d
g
x -



,
o
d
g


.
s
9
.
0
2
2 o



g
d




October 7, 2010
Energy Considerations
 We have done these problems considering the equation of motion, i.e. using
Newton’s 2nd Law. We can also consider what happens from an energy point
of view.
 Considering again the problem of a cart on a spring, with
we have a potential energy
 Differentiating x to get the velocity, we find the kinetic energy is
where we have used the fact that
 Then the total energy is What happens is that the kinetic
energy (going as sin2) is 180 degrees out of phase with the potential energy
(which goes as cos2). When U is maximum, T is minimum, and vice versa.
When added together, the cos2 q + sin2 q = 1, so total energy is constant.
.
2
2
1
kA
U
T
E 
+

 
 
2 2 2 2
1 1
2 2
2 2
1
2
sin
sin ,
mx m A t
T kA t
  d
 d
 -
 -
 .
cos2
2
2
1
2
2
1
d
 -

 t
kA
kx
U
.
/
2
m
k


),
cos(
)
( d
 -
 t
A
t
x
October 7, 2010
Problem 5.8: Mass on a Spring
 Statement of the problem (a):
 (a) If a mass m = 0.2 kg is tied to one end of a spring whose force constant k = 80
N/m and whose other end is held fixed, what are the angular frequency , the
frequency f, and the period  of its oscillations?
 Solution (a):
 The angular frequency for a mass on a spring is given by
 The corresponding frequency is
 The period is
 Statement of the problem (b):
 (b) If the initial position and velocity are xo = 0 and vo = 40 m/s, what are the
constants A and d in x(t) = A cos(t - d)?
 Solution (b):
 We have taken the + sign, since the velocity is positive. Thus,
Hz.
2
.
3
2




f
.
s
20
kg
2
.
0
N/m
80 1
-
m
k




s.
31
.
0
1


f

2
0
)
cos(
)
0
( 
d
d 



-
 A
x
m.
2
/
)
sin(
)
sin(
)
0
(
)
0
( o
o
2 





-

 

d
 
v
A
v
A
A
x
v 
2

d  -

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physics430_lecture11.ppt

  • 1. Physics 430: Lecture 11 Oscillations Dale E. Gary NJIT Physics Department
  • 2. October 7, 2010 5.1 Hooke’s Law  As you know, Hooke’s Law (spring law) gives the force due to a spring, in the form (assuming it is along the x axis): where x is the displacement relative to the spring’s equilibrium point. The force can be in either direction—in the +x direction when x is negative, and in the -x direction when x is positive (at least for the case when k is positive).  Since the force is conservative, we can write the force as minus the gradient of a potential energy  From what we have learned about potential energy graphs, you can immediately see that this upward curving parabola indicates that x = 0 is a stable equilibrium (if k were negative, the parabola would be curved downward and x = 0 would be an unstable equilibrium).  However, Hooke’s Law has a general validity for the following reason: for any general potential energy curve U(x) in the vicinity of an equilibrium point at x = xo, which we can take to be zero, we can always perform a Taylor-series expansion about that point: s  kx x Fx -  ) ( . ) ( 2 2 1 kx x U  . ) 0 ( ) 0 ( ) 0 ( ) ( 2 2 1  +   +  +  x U x U U x U
  • 3. October 7, 2010 Hooke’s Law-2  The first term is the constant term, but since potential energy can be defined with any zero point, the constant term can be ignored (or considered to be zero).  The second term is the linear term, but since x = 0 is an equilibrium point, the slope there is by definition zero, so the first non-zero term is the third term:  This says that for sufficiently small displacements from an equilibrium point, Hooke’s Law is ALWAYS valid for any potential energy function. This justifies our consideration of this case in detail.  Of course, Hooke’s Law can relate to any coordinate, not just x. Let’s revisit the box on a cylinder problem as an illustration. . ) ( 2 2 1 kx x U 
  • 4. October 7, 2010 Example 5.1: Cube Balanced on a Cylinder  Statement of the problem:  Consider again the cube of Example 4.7 and show that for small angles q the potential energy takes the Hooke’s Law form U(q) = ½ kq2.  Solution:  We found before that the potential energy for the cube in terms of q was  For small q, we can make the approximations cos q ~ 1 – q2/2; sin q ~ q. Notice that we kept the TWO leading terms for cos q. Why? With these substitutions, we have  Ignoring the constant term, we see that this is in the form of Hooke’s Law, with a “spring constant” k = mg(r – b).  Notice that the equilibrium is stable only when k > 0, i.e. when r > b, as we found after quite a bit more work before.  Notice also that since is a parabola, the turning points (at least for small displacements) are equidistant from the equilibrium point. ]. sin cos ) [( ) ( q q q q r b r mg U + +  . ) ( ) ( ] ) 1 )( [( ) ( 2 2 1 2 2 2 1 q q q q b r mg b r mg r b r mg U - + +  + - +  2 1 2 ( ) U x kx 
  • 5. October 7, 2010 5.2 Simple Harmonic Motion  Let’s now look at all of this from the point of view of the equation of motion. Consider a cart on a frictionless track attached to a spring with spring constant k. Since we have the equation of motion is where we introduce the constant which represents the angular frequency with which the cart will oscillate, as we will see.  Because Hooke’s Law always applies near equilibrium for any potential energy, we will find oscillations to be very common, governed by the general equation of motion such as for a pendulum vs. angle f: kx x Fx -  ) ( x x m k x kx x m 2  -  -   -      m k   f  f 2 -    x x = 0
  • 6. October 7, 2010 Exponential Solutions  The equation is a second order, linear, homogeneous differential equation. Therefore, it has two independent solutions, which can be written  As you can easily check, both of these solutions do satisfy the equation. However, you should have come to expect two arbitrary constants in the solution to a second-order differential equation (two constants of integration), and in fact any linear combination of these two solutions is also a solution This is called the superposition principle, which works for any linear system.  Any solution containing two arbitrary constants is in fact the general solution to the equation.  This solution can be written in terms of Sine and Cosine by using Euler’s equation  Plugging these into our original solution, we can write . ) ( and ) ( t i t i e t x e t x   -   x x 2  -    . ) ( 2 1 t i t i e C e C t x   - +  ). sin( ) cos( t i t e t i           ). sin( ) cos( ) sin( ) cos( ) ( 2 1 2 1 2 1 t B t B t C C i t C C t x     +  - + + 
  • 7. October 7, 2010 Sine and Cosine Solutions  Thus, the solutions and are equivalent so long as  An important issue is that x(t), being a actual position coordinate, has to be real. In general the first solution looks as if it is complex, while the second solution looks as if it is real. However, this depends on whether the coefficients are real or not. Both C1 and C2 can be complex, but if both their sum and difference C1 + C2 and C1 – C2 are real then B1 and B2 are real and in fact both versions are completely equivalent. This solution is the definition of simple harmonic motion (SHM).  As usual, we determine these constants from the initial conditions.  If I start the motion by pulling the cart aside (position x(0) = xo) and releasing it (v(0) = 0) then only the cosine term survives and  If the cart starts at x(0) = 0 by giving it a kick (velocity v(0) = vo) then only the sine term survives and , ) ( 2 1 t i t i e C e C t x   - +  ). sin( ) ( o t v t x    ), sin( ) cos( ) ( 2 1 t B t B t x   +  ). ( and , 2 1 2 2 1 1 C C i B C C B -  +  ). cos( ) ( o t x t x  
  • 8. October 7, 2010 Phase Shifted Cosine Solution  The two “pure” solutions just described are shown in the figures below.  The general solution, if I both pull the cart to the side and give it a push (both xo and vo non-zero), the motion is harder to visualize, but you may expect it to be the same type of motion, with some phase shift.  We can demonstrate that for the general case by defining another constant where A is the hypotenuse of a right triangle whose sides are B1 and B2. Obviously there is also an associated angle d such that and . Thus , 2 2 2 1 B B A +  x t xo ). cos( ) ( o t x t x   x slope vo t ). sin( ) ( o t v t x    vo/ xo A B1 B2 d B1 = A cos d B2 = A sin d ). cos( )] sin( sin ) cos( [cos ) sin( ) cos( ) ( 2 1 d   d  d   -  +  +  t A t t A t B t B t x slope vo Note that you have already solved this same problem using energy, in problem 4.28.
  • 9. October 7, 2010 Solution as Real Part of ei(t-d)  Notice that in there are still two constants, A and d, that are to be determined by initial conditions.  But notice that we can rewrite this using Euler’s equation as , where Re[] denotes the real part of what is in the square brackets. As a practical matter, then, we can write the solution in terms of a complex exponential, and after any mathematical manipulations, simply throw away any complex part and keep the real part of the solution as the answer to the physical problem at hand.  Graphically, if you will recall our discussion from Lecture 5 on complex exponentials, we can represent the solution as the x (real) component, or projection, of the complex number represented by a point in the complex plane given by .  You should train yourself to be able to go easily between the phase shifted cosine and Re[] complex representations. ), cos( ) ( d  -  t A t x   ) ( Re ) cos( ) ( d  d  -  -  t i Ae t A t x d t-d C = Ae-id x y A Ceit =Aei(t-d) x(t)=Acos(t-d) ) ( d  - t i Ae
  • 10. October 7, 2010 Example 5.2: A Bottle in a Bucket  Statement of the problem:  A bottle is floating upright in a large bucket of water as shown in the figure. In equilibrium it is submerged to a depth do below the surface of the water. Show that if it is pushed down to a depth d and released, it will execute harmonic motion, and find the frequency of its oscillations. If do = 20 cm, what is the period of the oscillations?  Solution:  The forces on the bottle are mg downward, and the buoyancy force upward. Do you recall how to determine the buoyancy force?  The volume of displaced water is Ado, where A is the cross-sectional area of the bottle. The mass of the displaced volume is thus rAdo, and the weight is rgAdo. At this depth, we have  The equation of motion, then, is where we have used d = do + x (i.e. x is measured from the equilibrium position). The buoyance force is equal to the weight of the displaced water. Archimedes’Principle d ), ( o x d gA mg x m + -  r   . o gAd mg r 
  • 11. October 7, 2010 Example 5.2, cont’d  Solution, cont’d:  Since the weight of the bottle is the same as the displaced water at equilibrium: we can replace mg and get but we can express so we end up with the simple result  This is the same form as the simple harmonic oscillator, with the same solution. We identify  You may recognize this as the same expression as for a pendulum of length l = do. Note that the result does not depend on the shape or weight of the bottle, or the density of the water, or anything else. If do = 20 cm, then , / m gAx x r -    . o gAd mg r  , / / o d g m gA  r , o x d g x -    , o d g   . s 9 . 0 2 2 o    g d    
  • 12. October 7, 2010 Energy Considerations  We have done these problems considering the equation of motion, i.e. using Newton’s 2nd Law. We can also consider what happens from an energy point of view.  Considering again the problem of a cart on a spring, with we have a potential energy  Differentiating x to get the velocity, we find the kinetic energy is where we have used the fact that  Then the total energy is What happens is that the kinetic energy (going as sin2) is 180 degrees out of phase with the potential energy (which goes as cos2). When U is maximum, T is minimum, and vice versa. When added together, the cos2 q + sin2 q = 1, so total energy is constant. . 2 2 1 kA U T E  +      2 2 2 2 1 1 2 2 2 2 1 2 sin sin , mx m A t T kA t   d  d  -  -  . cos2 2 2 1 2 2 1 d  -   t kA kx U . / 2 m k   ), cos( ) ( d  -  t A t x
  • 13. October 7, 2010 Problem 5.8: Mass on a Spring  Statement of the problem (a):  (a) If a mass m = 0.2 kg is tied to one end of a spring whose force constant k = 80 N/m and whose other end is held fixed, what are the angular frequency , the frequency f, and the period  of its oscillations?  Solution (a):  The angular frequency for a mass on a spring is given by  The corresponding frequency is  The period is  Statement of the problem (b):  (b) If the initial position and velocity are xo = 0 and vo = 40 m/s, what are the constants A and d in x(t) = A cos(t - d)?  Solution (b):  We have taken the + sign, since the velocity is positive. Thus, Hz. 2 . 3 2     f . s 20 kg 2 . 0 N/m 80 1 - m k     s. 31 . 0 1   f  2 0 ) cos( ) 0 (  d d     -  A x m. 2 / ) sin( ) sin( ) 0 ( ) 0 ( o o 2       -     d   v A v A A x v  2  d  -