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Note 12
Oscillatory Motion
Sections Covered in the Text: Chapter 14, except 14.7 & 14.8
In this note we investigate the oscillatory motion of an
object. Oscillatory motion is the motion an object
executes as it moves back and forth through a position
of stable equilibrium. We shall examine two examples
of oscillatory motion, a mass on the end of a spring
and the simple pendulum. Both stand as prototypes
for oscillations of all kinds in nature, ranging from the
movement of molecules of water on the surface of a
pond to oscillations of electric and magnetic fields in
electromagnetic radiation. Both examples are also
featured in lab experiments most of you will be doing
at about this time in the course.
Motion of a Particle Attached to a Spring
In Notes 08 and 09 we examined certain aspects of the
physics of a mass on the end of a spring. But that
study did not deal with the continuous motion of the
mass. We consider the physics of that motion here.
Consider an object (a block) modelled as a particle
connected to the end of a spring and in contact with a
horizontal frictionless surface (Figure 12-1).1
a function this force is
F(x) = –kx .
…[12-1]
This function is called Hooke’s Law. Here k is a
constant called the spring constant. The spring force is
zero at the equilibrium position x = 0.
We have seen in other notes that eq[12-1] is an
example of a central or restoring force. It is restoring in
the sense that if the block is displaced to the right of
the equilibrium position, then the force exerted by the
spring on the block points to the left, toward the
equilibrium position. If the block is displaced to the
left of the equilibrium position, then the force exerted
by the spring on the block points to the right, again
toward the equilibrium position. You should be able
to see that if the block is displaced either right or left
of the equilibrium position and then released, then the
block will move back and forth through the equilibrium position under the action of the varying spring
force—that is, the block will execute a vibrational or
oscillatory motion. If the surface on which the block
rests could be made to be frictionless, then no energy
would be converted to thermal energy in the movement. In principle, the block would oscillate back and
forth forever between positions equidistant on either
side of the equilibrium position. This kind of motion
is called simple harmonic motion (abbreviated SHM).
The system executing such motion is called a simple
harmonic oscillator.
Mathematical Representation of SHM
The motion just described in words can be put into
mathematical form. Equating eq[12-1] to ma and
rearranging we can obtain an expression for the
block’s acceleration:
Figure 12-1. A mass on the end of a spring is shown at three
positions in the course of its oscillatory motion.
The spring exerts a force on the block that is
proportional to the spring’s displacement. Written as
a=–
k
x.
m
…[12-2a]
If the object is confined to 1D space then we can drop
the vector notation and write eq[12-2a] as the
function:
a(x) = –
k
x.
m
…[12-2b]
1
This system might just as well be a glider on an airtrack
connected between two identical springs. In this case the effective kvalue of the system is 2k.
Now x is a function of time and a(x) can be written as
the second derivative of x(t) with respect to time. Thus
12-1
Note 12
d 2 x(t)
k
= – x(t) .
2
dt
m
…[12-3]
We shall see in what follows that we can write k/m as
the square of a quantity ω:
ω2 =
k
.
m
…[12-4]
Eq[12-3] then becomes
d 2 x(t)
= –ω 2 x(t ) .
2
dt
…[12-5]
This equation is a second-order differential equation
in x(t). A solution is 2
x(t) = Acos (ωt + φ )
ω=
where
k
,
m
Figure 12-2. Graphs of eq[12-6] showing the meaning of the
period T and amplitude A for arbitrary φ (a) and φ = 0 (b).
…[12-6]
a(t) =
…[12-7]
consistent with eq[12-4]. A is a constant.
ω is called the angular frequency of the vibration. It
has dimension
€ T–1 and units rad.s–1 (or just s–1 because
rad is a dimensionless quantity). ω can also be written
as
ω = 2π f ,
…[12-8]
dv(t) d 2 x(t)
=
= –A ω 2 cos(ωt + φ ) .
2
dt
dt
…[12-10]
Note that v(t) and a(t) have the same sinusoidal form
as x(t). Graphs of x(t), v(t) and a(t) for an arbitrary
phase angle φ are shown in Figures 12-3.
where f is the linear frequency. f has units cycles.s –1 or
hertz (abbreviated Hz).
φ is called the phase angle. Among other things, φ
sets the value of x(t) at t = 0.
The time for one complete vibration is called the
period and is denoted T. T has dimension T and unit s.
The period is the inverse of the frequency (T = f –1). The
maximum displacement of the vibration from
equilibrium is called the amplitude and is denoted A.
Figure 12-2 shows a graph of eq[12-6].
Expressions for the velocity and acceleration of the
block can be obtained from successive differentiations
of eq[12-6] with respect to t. The results are:
v(t) =
and
2
dx(t )
= – Aω sin(ωt + φ )
dt
…[12-9]
To show that eq[12-6] is a solution of eq[12-5] you needn’t
prove that it follows mathematically. It is sufficient to show that
eq[12-6] along with its second derivative satisfy eq[12-5] by
substitution.
12-2
Figure 12-3. Displacement, velocity and acceleration graphs
of the vibration of a block on the end of a spring for an
arbitrary phase angle φ.
You should study Figures 12-3 to understand the
relationships between the displacement, velocity and
acceleration. For example, when the displacement is a
Note 12
maximum positive value, the velocity is zero and the
acceleration is a maximum negative value. Can you
correlate these facts with the sketch of the motion in
Figure 12-1? Let us consider a numerical example.
K(t) =
=
1
2
1
2
mv(t)2 ,
1
2
mω 2 A2 sin 2 (ωt + φ ) = kA2 sin 2 (ωt + φ )
…[12-11]
Example Problem 12-1
A Block-Spring System
A block of mass 200. g is connected to a light horizontal spring of force constant 5.00 N.m–1 and is free to
oscillate on a horizontal frictionless surface. (a) If the
block is displaced 5.00 cm from equilibrium and then
released from rest find the period of its motion. (b)
what is the phase angle φ?
Solution:
(a) From eq[12-7] the period of motion of a mass on
the end of a spring is given by
ω=
so that
k
2π
= 2πf =
m
T
T = 2π
€
= 2π
€
using the expression for v(t) from eq[12-9]. The
potential energy of the system (stored in the spring)
has been seen (Note 08) to be given by
1
2
U( x) = kx2 ,
Substituting eqs[12-11] and [12-12] we have
E = K(t) + U (t)
=
1
2
1
2
kA2 sin 2 (ωt + φ ) + kA2 cos 2 (ωt + φ ) .
This expression can be simplified. Collecting terms
and using a trigonometric relation we have
= 1.26 s.
Note that€T is independent of amplitude.
(b) According to the expression for the position of the
block, eq[12-6], the position at time t = 0 is
…[12-12]
using eq[12-6]. The total mechanical energy E of the
system is the sum of the kinetic and potential
energies:
E = K +U .
m
k
0.2(kg)
5.00(N.m−1 )
1
2
U(t) = kA2 cos2 (ωt + φ )
so that
E=
1
2
kA2 [sin2 (ωt + φ ) + cos 2 (ωt + φ)] ,
=
1
2
kA2 ,
…[12-13]
x(0) = 0.05cos(ω0 + φ) = 0.05.
Thus
φ = 0.
The value of the phase angle is set by the position of
the block at t = 0.
Energy Considerations in SHM
It is instructive to study the mass on the end of a
spring from the point of view of energy as well as
from the point of view of kinematics. If the block is
moving with velocity v(t) then it possesses kinetic
energy K(t) given by
since the expression in square brackets is unity. The
total energy is therefore proportional to the square of
the amplitude. Remarkably, E, which is equal to the
sum of two time dependent functions, is itself independent of time.
K and U can be written as functions of either x or t.
They are graphed as functions of t in Figure 12-5a and
of x in Figure 12-5b. Clearly, at the instant when the
kinetic energy is a maximum the potential energy is a
minimum and vice versa. In one period the kinetic
and potential energies go through two oscillations.
Both functions are sinusoidal.
Notice that at x = 0, the potential energy is zero and
the kinetic energy is a maximum (since v is a maximum). At x = ±A, the potential energy is a maximum
and the kinetic energy is zero (since v is zero). At all
12-3
Note 12
positions of x the sum of K and U is constant, just as at
all times t the sum of K and U is constant. Be prepared
to describe a vibrating system such as this one in
words.
The Simple Pendulum
A simple pendulum is essentially an object of mass m
suspended on the end of a string (Figure 12-6).3 For
the simple pendulum to work as intended, gravity
must be present. The system has an equilibrium
position defined when the mass is in a rest position
with the string vertical. In this position the angular
displacement of the string with respect to the vertical
is θ = 0 (rad) or the linear displacement of the mass
relative to the rest position along the arc of its path is s
= 0 (m).
Figure 12-6. A simple pendulum showing the forces on the
mass and their components.
Figure 12-5. Graphs of K and U as functions of time (a) and
of x (b). For convenience, the phase angle in (a) is taken to
be zero.
When the mass is displaced so that the string makes a
non-zero angle with respect to the vertical and
released, the pendulum will swing back and forth. If
friction with the air is negligible, then the mass will
swing continuously between maximum angular
positions θmax and –θmax forever. This kind of motion is
simple harmonic motion. The pendulum system is a
simple harmonic oscillator.
As in the case of a mass on the end of a spring, this
motion can be described mathematically. In the case
of a mass/spring system the driving force is the
spring force. In the case of a simple pendulum the
driving force is the component of the force of gravity,
–mgsinθ, acting tangent to the particle’s path. Applying the second law F = ma we get
–mg sin θ (t) = m
3
d 2 s(t)
,
dt 2
…[12-14]
The story goes that Galileo studied the motion of a simple
pendulum by observing the swinging motion of a chandelier in the
local cathedral. Lacking high technology he counted time in units of
beats of his pulse.
12-4
Note 12
where s(t) is the position of the particle measured
with respect to the equilibrium position along the
particle’s path. Substituting s(t) = Lθ(t), where L is the
length of the string, eq[12-14] can be written
d 2 θ (t)
g
= – sin θ (t) .
2
dt
L
…[12-15]
This is a second order differential equation in θ (t).
Eq[12-15], unlike eq[12-3], is very difficult to solve.
However, if we assume that the angular displacement
θ(t) is small, then we can make the substitution sinθ(t)
≅ θ(t). Then eq[12-15] reduces to
d 2 θ (t)
g
= – θ (t) .
2
dt
L
…[12-16]
Apart from a change in variable and the presence of
different constants, eq[12-16] is the same as eq[12-3].
Furthermore if we put
ω=
g
,
L
…[12-17]
then eq[12-16] becomes of the same form as eq[12-5].
It must therefore have the same form of solution,
namely,
€
θ (t) = θ max cos(ωt + φ ) . …[12-18]
The function θ (t) is sinusoidal as is the function x(t)
(Figure 12-6). The period of the motion is thus given
by
T=
2π
L
= 2π
.
ω
g
The mathematics for a simple pendulum is thus seen
to be the same in essentials as for a mass on the end of
a spring. This underlines a major goal of physics, to
describe various systems with the same tools whenever possible.
A critical difference between the two systems is that
the period of a simple pendulum involves g whereas
the period of the motion of a mass on the end of a
spring does not. A mass on the end of a spring will
function as expected in a spacecraft just as well as on
Earth, but a simple pendulum will not work in a
spacecraft.
Example Problem 12-2
The Simple Pendulum
A simple pendulum (unlike a mass on the end of a
spring) can be used as an instrument to calculate the
local acceleration due to gravity—if the period can be
measured with good accuracy. Assuming the period
of a simple pendulum of length 1.00 m is measured to
be 2.00709 s what is the local value of g?
Solution:
Rearranging eq[12-19] we have for g:
g=
4π 2 L 4π 2 1.00m
=
T2
(2.007s) 2
= 9.80 m.s–2,
to 3 significant figures.
…[12-19]
€
12-5
Note 12
To Be Mastered
•
•
Definitions: angular frequency, linear frequency, period, phase angle
General expression for the position of a simple harmonic oscillator:
x(t) = Acos(ωt + φ )
•
•
•
Physics of: mass on the end of a spring
Physics of: the simple pendulum
Difference between the period of a mass on the end of a spring and the period of a simple pendulum
(memorized):
T = 2π
•
m
k
T = 2π
L
g
Sketches of K, U and E for both systems
€
€
Typical Quiz/Test/Exam Questions
1.
Define the following and give their units:
(a) angular displacement
(b) angular velocity
(c) angular frequency
2.
A simple pendulum is studied at some position on the surface of the earth. If the length of the pendulum is
increased by a factor of 4 then the period of the pendulum
(a) decreases by a factor of 2
(b) descreases by a factor of 4
(c) increases by a factor of 2
(d) remains unchanged.
3.
The period of a simple pendulum of a fixed length L is measured at the surface of the Earth and at the top of a
high mountain. At the surface of the Earth the period of the pendulum is To and the acceleration due to
gravity is go. If at the top of the mountain the period is 2To then what is the acceleration due to gravity there in
terms of go?
4.
The position of a block on the end of a spring (as shown in Figure 12-1) is given by the function
x(t) = 2.0 cos(3.0t + π / 2) meter.
The constant of the spring is 9.0 N.m –1. Assume that friction between block and surface is negligible. Answer
the following questions:
(a) What is the maximum displacement of the block (in m) from the equilibrium position?
(b) What is the angular frequency (in rad.s–1) of the motion?
(c) What is the mass of the block (in kg)?
(d) What was the position of the block at a time t = 0?
12-6
Note 12
5.
A simple pendulum consists of a bob of mass m on the end of a string of length l suspended from a stationary
support (see figure). When released from near position A at time t = 0, the bob swings through B to near C
and back to A repetitively. Assume that friction of the air is negligible. Answer the following questions.
support
θ
l
m
A
B
C
(a) What kind of motion does the bob execute and why?
(b) Sketch one complete cycle of the angular position of the pendulum starting from t = 0. Show clearly on the
sketch the period and amplitude.
(c) Sketch the kinetic and potential energy of the bob over one complete cycle starting from t = 0.
(d) What is the angular acceleration of the bob at position B?
6.
A glider of mass 400.0 g is placed on a horizontal airtrack and connected between two springs whose opposite
ends are connected to block supports that cannot move (see the figure). Each spring has a k value of 1.17
N.m –1. The equilibrium position of the glider (x = 0) is as shown. The glider is displaced 5.0 cm from
equilibrium and then released. Friction is negligible. Answer the following questions.
block
block
glider
airtrack
x=0
(a) Calculate the maximum speed of the glider in m.s–1.
(b) Calculate the speed of the glider in m.s –1 when it is 3.0 cm from equilibrium.
(c) Derive an expression for the acceleration a of the glider in terms of its instantaneous position x.
(d) What is the acceleration of the glider in m.s–2 when it passes through x = 0?
Reminder: The effective k-value of this two spring system is 2k.
12-7
Note 12
7.
A block is connected to the end of a spring and supported on a horizontal frictionless surface. The block is
released at a clock time of t = 0 s from a position x = +A m (see the figure). Answer the following questions.
(a) Write down an expression for the force exerted on the block by the spring.
(b) What kind of motion does the block execute and why?
(c) What is the total work done on the block by the spring on one complete cycle of the motion?
(d) What is the acceleration of the block at the equilibrium position?
8.
A glider released from the upper end of a curved air track moves back and forth through the lowest point on
the track with simple harmonic motion (see the figure). Answer the following questions:
glider
C
A
B
(a) Draw a free body diagram of the glider at positions A, B and C.
(b) What is the total work done by the gravitational field of the Earth on one cycle of the glider from A back to
A?
12-8