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Chapter 23 Simple Harmonic Motion
- 23.1 Introduction: Periodic Motion
- 23.1.1 Simple Harmonic Motion: Quantitative
- 23.2 Simple Harmonic Motion: Analytic
- 23.2.1 General Solution of Simple Harmonic Oscillator Equation
- Example 23.1: Phase and Amplitude
- Example 23.2: Block-Spring System
- 23.3 Energy and the Simple Harmonic Oscillator
- 23.3.1 Simple Pendulum: Force Approach
- 23.3.2 Simple Pendulum: Energy Approach
- 23.4 Worked Examples
- Example 23.3: Rolling Without Slipping Oscillating Cylinder
- Example 23.4: U-Tube
- 23.5 Damped Oscillatory Motion
- 23.5.1 Energy in the Underdamped Oscillator
- 23.6 Forced Damped Oscillator
- 23.6.1 Resonance
- 23.6.2 Mechanical Energy
- Example 23.5: Time-Averaged Mechanical Energy
- 23.6.3 The Time-averaged Power
- 23.6.4 Quality Factor
- 23.7 Small Oscillations
- Example 23.6: Quartic Potential
- Example 23.7: Lennard-Jones 6-12 Potential
- Appendix 23A: Solution to Simple Harmonic Oscillator Equation
- Appendix 23B: Complex Numbers
- Appendix 23C: Solution to the Underdamped Simple Harmonic Oscillator
- Appendix 23D: Solution to the Forced Damped Oscillator Equation....................
Chapter 23 Simple Harmonic Motion
…Indeed it is not in the nature of a simple pendulum to provide equal and
reliable measurements of time, since the wide lateral excursions often
made may be observed to be slower than more narrow ones; however, we
have been led in a different direction by geometry, from which we have
found a means of suspending the pendulum, with which we were
previously unacquainted, and by giving close attention to a line with a
certain curvature, the time of the swing can be chosen equal to some
calculated value and is seen clearly in practice to be in wonderful
agreement with that ratio. As we have checked the lapses of time
measured by these clocks after making repeated land and sea trials, the
effects of motion are seen to have been avoided, so sure and reliable are
the measurements; now it can be seen that both astronomical studies and
the art of navigation will be greatly helped by them…
1
Christian Huygens
23.1 Introduction: Periodic Motion
There are two basic ways to measure time: by duration or periodic motion. Early clocks
measured duration by calibrating the burning of incense or wax, or the flow of water or
sand from a container. Our calendar consists of years determined by the motion of the
sun; months determined by the motion of the moon; days by the rotation of the earth;
hours by the motion of cyclic motion of gear trains; and seconds by the oscillations of
springs or pendulums. In modern times a second is defined by a specific number of
vibrations of radiation, corresponding to the transition between the two hyperfine levels
of the ground state of the cesium 133 atom.
Sundials calibrate the motion of the sun through the sky, including seasonal
corrections. A clock escapement is a device that can transform continuous movement into
discrete movements of a gear train. The early escapements used oscillatory motion to stop
and start the turning of a weight-driven rotating drum. Soon, complicated escapements
were regulated by pendulums, the theory of which was first developed by the physicist
Christian Huygens in the mid 17
th century. The accuracy of clocks was increased and the
size reduced by the discovery of the oscillatory properties of springs by Robert Hooke.
By the middle of the 18
th century, the technology of timekeeping advanced to the point
that William Harrison developed timekeeping devices that were accurate to one second in
a century.
23.1.1 Simple Harmonic Motion: Quantitative
1 Christian Huygens, The Pendulum Clock or The Motion of Pendulums Adapted to Clocks By Geometrical
Demonstrations, tr. Ian Bruce, p. 1.
symbol ω is used for angular speed in circular motion. For uniform circular motion the
angular speed is equal to the angular frequency but for non-uniform motion the angular
speed is not constant. The angular frequency for simple harmonic motion is a constant by
definition.) We therefore have several different mathematical representations for
sinusoidal motion
y ( t ) = A sin( 2 π t / T ) = A sin( 2 π f t ) = A sin( ω 0
t ). ( 23. 1. 5 )
23.2 Simple Harmonic Motion: Analytic
Our first example of a system that demonstrates simple harmonic motion is a spring-
object system on a frictionless surface, shown in Figure 23.
Figure 23. 2 Spring-object system
The object is attached to one end of a spring. The other end of the spring is attached to a
wall at the left in Figure 23.2. Assume that the object undergoes one-dimensional motion.
The spring has a spring constant k and equilibrium length l eq
. Choose the origin at the
equilibrium position and choose the positive x - direction to the right in the Figure 23.2.
In the figure, x > 0 corresponds to an extended spring, and x < 0 to a compressed spring.
Define x ( t )to be the position of the object with respect to the equilibrium position. The
force acting on the spring is a linear restoring force, x
F = − k x (Figure 23.3). The initial
conditions are as follows. The spring is initially stretched a distance l 0
and given some
initial speed v 0
to the right away from the equilibrium position. The initial position of the
stretched spring from the equilibrium position (our choice of origin) is x 0
= ( l 0
− l eq
and its initial x - component of the velocity is v x , 0
= v 0
Figure 23. 3 Free-body force diagram for spring-object system
Newton’s Second law in the x - direction becomes
2
2
d x
k x m
dt
This equation of motion, Eq. ( 23. 2. 1 ), is called the simple harmonic oscillator equation
(SHO). Because the spring force depends on the distance x , the acceleration is not
constant. Eq. ( 23. 2. 1 ) is a second order linear differential equation, in which the second
derivative of the dependent variable is proportional to the negative of the dependent
variable,
2
2
d x k
x
dt m
In this case, the constant of proportionality is k / m ,
Eq. ( 23. 2. 2 ) can be solved from energy considerations or other advanced techniques but
instead we shall first guess the solution and then verify that the guess satisfies the SHO
differential equation (see Appendix 22.3.A for a derivation of the solution).
We are looking for a position function x ( t ) such that the second time derivative position
function is proportional to the negative of the position function. Since the sine and cosine
functions both satisfy this property, we make a preliminary ansatz (educated guess) that
our position function is given by
x ( t ) = A cos(( 2 π / T ) t ) = A cos( ω 0
t ) , ( 23. 2. 3 )
where ω 0
is the angular frequency (as of yet, undetermined).
We shall now find the condition that the angular frequency ω 0
must satisfy in order to
insure that the function in Eq. ( 23. 2. 3 ) solves the simple harmonic oscillator equation, Eq.
( 23. 2. 1 ). The first and second derivatives of the position function are given by
d
2 x
dt
2
k
m
B sin
k
m
t
= − ω 0
2 x , ( 23. 2. 11 )
where ω 0
= k / m. The x - component of the velocity associated with Eq. ( 23. 2. 10 ) is
v x
( t ) =
dx
dt
k
m
B cos
k
m
t
The proposed solution in Eq. ( 23. 2. 10 ) has initial conditions 0
x ≡ x ( t = 0 ) = 0 and
v x , 0
≡ v x
( t = 0 ) = ( k / m ) B , thus B = v x , 0
/ k / m. This solution describes an object that
is initially at the equilibrium position but has an initial non-zero x - component of the
velocity, v x , 0
23.2.1 General Solution of Simple Harmonic Oscillator Equation
Suppose 1
x ( t )and 2
x ( t )are both solutions of the simple harmonic oscillator equation,
d
2
dt
2
x 1
( t ) = −
k
m
x 1
( t )
d
2
dt
2
x 2
( t ) = −
k
m
x 2
( t ).
Then the sum 1 2
x ( t ) = x ( t ) + x ( t ) of the two solutions is also a solution. To see this,
consider
d
2 x ( t )
dt
2
d
2
dt
2
( x 1
( t ) + x 2
( t )) =
d
2 x 1
( t )
dt
2
d
2 x 2
( t )
dt
2
Using the fact that 1
x ( t )and 2
x ( t )both solve the simple harmonic oscillator equation
( 23. 2. 13 ), we see that
2
2 1 2 1 2
d k k k
x t x t x t x t x t
dt m m m
k
x t
m
Thus the linear combination 1 2
x ( t ) = x ( t ) + x ( t )is also a solution of the SHO equation,
Eq. ( 23. 2. 1 ). Therefore the sum of the sine and cosine solutions is the general solution ,
x ( t ) = C cos( ω 0
t ) + D sin( ω 0
t ) , ( 23. 2. 16 )
where the constant coefficients C and D depend on a given set of initial conditions
0
x ≡ x ( t = 0 ) and v x , 0
≡ v x
( t = 0 ) (^) where 0
x and v x , 0
are constants. For this general
solution, the x^ - component of the velocity of the object at time t is then obtained by
differentiating the position function,
v x
( t ) =
dx
dt
= − ω 0
C sin( ω 0
t ) + ω 0
D cos( ω 0
t ). ( 23. 2. 17 )
To find the constants C and D , substitute t = 0 into the Eqs. ( 23. 2. 16 ) and ( 23. 2. 17 ).
Because cos( 0 ) = 1 and sin( 0 ) = 0 , the initial position at time (^) t = 0 is
x 0
≡ x ( t = 0 ) = C. ( 23. 2. 18 )
The x - component of the velocity at time t = 0 is
v x , 0
= v x
( t = 0 ) = − ω 0
C sin( 0 ) + ω 0
D cos( 0 ) = ω 0
D. ( 23. 2. 19 )
Thus
C = x 0
and D =
v x , 0
ω 0
The position of the object-spring system is then given by
x ( t ) = x 0
cos
k
m
t
v x , 0
k / m
sin
k
m
t
and the x - component of the velocity of the object-spring system is
v x
( t ) = −
k
m
x 0
sin
k
m
t
cos
k
m
t
Although we had previously specified 0
x > 0 and v x , 0
> 0 , Eq. ( 23. 2. 21 ) is seen to be a
valid solution of the SHO equation for any values of 0
x and v x , 0
Example 23.1: Phase and Amplitude
Show that x ( t ) = C cos ω 0
t + D sin ω 0
t = A cos( ω 0
t + φ) , where A = ( C
2
2 )
1 2 > 0 , and
φ = tan
− 1 (− D / C ).
function x ( t ) = A cos( ω 0
t + φ) reaches its maximum value at a later time t = T / 8 than
the function x ( t ) = A cos( ω 0
t ) as shown in Figure 23.4c.
(a)
(b)
(c)
Figure 23. 4 Phase shift of x ( t ) = A cos( ω 0
t + φ) (a) to the left by φ = π / 4 , (b) no shift
φ = 0 , (c) to the right φ = − π / 4
Example 23.2: Block-Spring System
A block of mass m is attached to a spring with spring constant k and is free to slide
along a horizontal frictionless surface. At t = 0 , the block-spring system is stretched an
amount 0
x > 0 from the equilibrium position and is released from rest, v x , 0
= 0. What is
the period of oscillation of the block? What is the velocity of the block when it first
comes back to the equilibrium position?
Solution: The position of the block can be determined from Eq. ( 23. 2. 21 ) by substituting
the initial conditions 0
x > 0 , and v x , 0
= 0 yielding
0
( ) cos
k
x t x t
m
and the x - component of its velocity is given by Eq. ( 23. 2. 22 ),
v x
( t ) = −
k
m
x 0
sin
k
m
t
The angular frequency of oscillation is ω 0
= k / m and the period is given by
Eq. ( 23. 2. 7 ),
T =
2 π
ω 0
= 2 π
m
k
The block first reaches equilibrium when the position function first reaches zero. This
occurs at time 1
t satisfying
1 1
k m T
t t
m k
π π
= = = (^). ( 23. 2. 31 )
The x - component of the velocity at time 1
t is then
v x
( t 1
k
m
x 0
sin
k
m
t 1
k
m
x 0
sin( π / 2 ) = −
k
m
x 0
= − ω 0
x 0
Note that the block is moving in the negative x - direction at time t 1
; the block has moved
from a positive initial position to the equilibrium position (Figure 23.4(b)).
where we used the identity that cos
2 ω 0
t + sin
2 ω 0
t = 1 , and that ω 0
= k / m (Eq.
The mechanical energy in state 2 is equal to the initial potential energy in state 1, so the
mechanical energy is constant. This should come as no surprise; we isolated the object-
spring system so that there is no external work performed on the system and no internal
non-conservative forces doing work.
Figure 23. 5 State 3 at equilibrium and in motion
State 3: now the object is at the equilibrium position so the potential energy is zero,
3
U = 0 , and the mechanical energy is in the form of kinetic energy (Figure 23.5).
2
3 3 eq
E = K = m v. ( 23. 3. 6 )
Because the system is closed, mechanical energy is constant,
E
1
= E
3
Therefore the initial stored potential energy is released as kinetic energy,
2 2
0 eq
k x = m v , ( 23. 3. 8 )
and the x - component of velocity at the equilibrium position is given by
v x,eq
k
m
x 0
Note that the plus-minus sign indicates that when the block is at equilibrium, there are
two possible motions: in the positive x - direction or the negative x - direction. If we take
0
x > 0 , then the block starts moving towards the origin, and v x,eq
will be negative the first
time the block moves through the equilibrium position.
We can show more generally that the mechanical energy is constant at all times as
follows. The mechanical energy at an arbitrary time is given by
E = K + U =
mv x
2
k x
2
. ( 23. 3. 10 )
Differentiate Eq. ( 23. 3. 10 )
dE
dt
= mv x
dv x
dt
dx
dt
= v x
m
d
2 x
dt
2
Now substitute the simple harmonic oscillator equation of motion, (Eq. ( 23. 2. 1 ) ) into Eq.
( 23. 3. 11 ) yielding
dE
dt
demonstrating that the mechanical energy is a constant of the motion.
23.3.1 Simple Pendulum: Force Approach
A pendulum consists of an object hanging from the end of a string or rigid rod pivoted
about the point P. The object is pulled to one side and allowed to oscillate. If the object
has negligible size and the string or rod is massless, then the pendulum is called a simple
pendulum. Consider a simple pendulum consisting of a massless string of length l and a
point-like object of mass m attached to one end, called the bob. Suppose the string is
fixed at the other end and is initially pulled out at an angle θ 0
from the vertical and
released from rest (Figure 23.6). Neglect any dissipation due to air resistance or frictional
forces acting at the pivot.
Figure 2 3. 6 Simple pendulum
Let’s choose polar coordinates for the pendulum as shown in Figure 23.7a along with the
free-body force diagram for the suspended object (Figure 23.7b). The angle θ is defined
with respect to the equilibrium position. When θ > 0 , the bob is has moved to the right,
and when θ < 0 , the bob has moved to the left. The object will move in a circular arc
centered at the pivot point. The forces on the object are the tension in the string
d
2 θ
dt
2
g
l
θ. ( 23. 3. 18 )
This equation is similar to the object-spring simple harmonic oscillator differential
equation
d
2 x
dt
2
k
m
x. ( 23. 3. 19 )
By comparison with Eq. ( 23. 2. 6 ) the angular frequency of oscillation for the pendulum is
approximately
ω 0
g
l
with period
T =
2 π
ω 0
2 π
l
g
The solutions to Eq. ( 23. 3. 18 ) can be modeled after Eq. ( 23. 2. 21 ). With the initial
conditions that the pendulum is released from rest,
d θ
dt
( t = 0 ) = 0 , at a small angle
θ( t = 0 ) = θ 0
, the angle the string makes with the vertical as a function of time is given by
θ( t ) = θ 0
cos( ω 0
t ) = θ 0
cos
2 π
T
t
= θ 0
cos
g
l
t
The z - component of the angular velocity of the bob is
ω z
( t ) =
d θ
dt
( t ) = −
g
l
θ 0
sin
g
l
t
Keep in mind that the component of the angular velocity ω z
= d θ / dt changes with time
in an oscillatory manner (sinusoidally in the limit of small oscillations). The angular
frequency ω 0
is a parameter that describes the system. The z - component of the angular
velocity ω z
( t ) , besides being time-dependent, depends on the amplitude of oscillation θ 0
In the limit of small oscillations, ω 0
does not depend on the amplitude of oscillation.
The fact that the period is independent of the mass of the object follows algebraically
from the fact that the mass appears on both sides of Newton’s Second Law and hence
cancels. Consider also the argument that is attributed to Galileo: if a pendulum,
consisting of two identical masses joined together, were set to oscillate, the two halves
would not exert forces on each other. So, if the pendulum were split into two pieces, the
pieces would oscillate the same as if they were one piece. This argument can be
extended to simple pendula of arbitrary masses.
23.3.2 Simple Pendulum: Energy Approach
We can use energy methods to find the differential equation describing the time evolution
of the angle θ. When the string is at an angle θ with respect to the vertical, the
gravitational potential energy (relative to a choice of zero potential energy at the bottom
of the swing where θ = 0 as shown in Figure 23.8) is given by
U = mgl ( 1 − cos θ ) ( 23. 3. 24 )
The θ - component of the velocity of the object is given by v θ
= l ( d θ / dt ) so the kinetic
energy is
K =
m v
2
m l
d θ
dt
2
Figure 23. 8 Energy diagram for simple pendulum
The mechanical energy of the system is then
E = K + U =
m l
d θ
dt
2
+ mgl ( 1 − cos θ). ( 23. 3. 26 )
Because we assumed that there is no non-conservative work (i.e. no air resistance or
frictional forces acting at the pivot), the energy is constant, hence
dE
dt
m 2 l
2 d θ
dt
d
2 θ
dt
2
d θ
dt
= ml
2
d θ
dt
d
2 θ
dt
2
g
l
sin θ
K
1
mv 1
2
. ( 23. 3. 34 )
Because the energy is constant, we have that U 0
= K
1
or
mgl
θ 0
2
mv 1
2
. ( 23. 3. 35 )
We can solve for the θ - component of the velocity at the bottom of the arc
v θ , 1
= ± gl θ 0
The two possible solutions correspond to the different directions that the motion of the
bob can have when at the bottom. The z - component of the angular velocity is then
d θ
dt
( t 1
v 1
l
g
l
θ 0
in agreement with our previous calculation.
If we do not make the small angle approximation, we can still use energy techniques to
find the θ - component of the velocity at the bottom of the arc by equating the energies at
the two positions
mgl 1 − cos θ 0
( ) =^
mv 1
2 , ( 23. 3. 38 )
v θ , 1
= ± 2 gl 1 − cos θ 0
( ).^ (^23.^3.^39 )
23.4 Worked Examples
Example 23.3: Rolling Without Slipping Oscillating Cylinder
Attach a solid cylinder of mass M and radius R to a horizontal massless spring with
spring constant k so that it can roll without slipping along a horizontal surface. At time t ,
the center of mass of the cylinder is moving with speed V cm
and the spring is compressed
a distance x from its equilibrium length. What is the period of simple harmonic motion
for the center of mass of the cylinder?
Figure 23.9 Example 23.
Solution: At time t , the energy of the rolling cylinder and spring system is
E =
Mv cm
2
I
cm
d θ
dt
2
kx
2
. ( 23. 4. 1 )
where x is the amount the spring has compressed, I cm
= ( 1 / 2 ) MR
2 , and because it is
rolling without slipping
d θ
dt
V
cm
R
Therefore the energy is
E =
MV
cm
2
MR
2
V
cm
R
2
kx
2
MV
cm
2
kx
2
. ( 23. 4. 3 )
The energy is constant (no non-conservative force is doing work on the system) so
dE
dt
2 MV
cm
dV cm
dt
k 2 x
dx
dt
= V
cm
M
d
2 x
dt
2
Because V cm
is non-zero most of the time, the displacement of the spring satisfies a
simple harmonic oscillator equation
d
2 x
dt
2
2 k
3 M
x = 0. ( 23. 4. 5 )
Hence the period is
T =
2 π
ω 0
= 2 π
3 M
2 k
Example 23.4: U-Tube
A U-tube open at both ends is filled with an incompressible fluid of density ρ. The
cross-sectional area A of the tube is uniform and the total length of the fluid in the tube
is L. A piston is used to depress the height of the liquid column on one side by a distance
0
x , (raising the other side by the same distance) and then is quickly removed (Figure
23.1 0 ). What is the angular frequency of the ensuing simple harmonic motion? Neglect
any resistive forces and at the walls of the U-tube.
