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5
Simple Harmonic Motion
1- Oscillations
There are many examples of oscillations in our world. We say that the object oscillates
when it moves back and forth repeatedly about its equilibrium position. In the case of a
particle oscillates on a straight line, we can represent this oscillation by a function which
repeats its value as time (t) progresses.
x(t + T0 ) = x (t)
(eq 1)
x
x axis
x=0
The origin
Fig(1) The motion of a particle on the x-axis.
Let the value of x is X at some time t, as time passes the value of x changes, but then at
later time (t+T0 ) the value of x is X again. The constant T0 is called the period of this
function. The frequency f 0 is the number of repetitions in one second. Then it is clear
that
f0 
1
T0
(eq2)
A famous example of a function of this kind is the sine function.
x(t) = sin(t)
(eq3)
The sine function is plotted in figure 2. It well-known that the period of the sine
function is 2.
1.5
1
Fig(2)
Sin(t)
0.5
The sine
0
0
2
4
6
8
10
12
14
16
18
20
function
-0.5
-1
-1.5
time (t)
sin(t+2) = sin (t)
(eq4)
Most of the real oscillations around us are not simple. We will study an ideal and simple
case. It is called simple harmonic motion.
2- Simple harmonic motion
In this motion the position of an oscillating particle on the x-axis takes the form
x = A sin ( 0 t+ 0)
(eq 5)
where A, 0 and  0 are constants. Equation 5 is shown in figure 3. We also assume that
the origin of the x-axis is the equilibrium position.
6
4
2
0
-5
0
5
10
15
20
25
-2
-4
-6
Fig (3). The position x as a function of time t. Here we take
A =5 cm, 0 = 0.2 rad/s, and  0 =0.25 rad.
Now we want to understand the physical meaning of the three constants. The maximum
value of the sine function is ( 1 ), which means the maximum value of the position x of
the particle is A. It is the maximum displacement from equilibrium. The constant A is
the amplitude of the oscillation.
From equations 1 and 5, we have
A sin[0 (t +T0 )+  0 ] = A sin(0 t+  0 )
sin[0 (t +T0 )+  0 ] = sin(0 t+  0 )
sin[(0 t +  0 )+ 0 T0 ] = sin(0 t+  0 )
Compare the last equation with equation4 we deduce that
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0 T0 =2, or
2
ω0 
2
T0
(eq 6)
And using relation in eq.2 we get
0 = 2  f 0
(eq7)
0 is called the angular frequency.
Let us assume that the position of the particle at t = 0 is x(0) = x 0 . Then by substituting
in equation 5 we get
x 0 = A sin( 0 )
 x0 

A
  sin 1
(eq 8)
The  is called the phase angle (it is not really an angle).
Exercise 1 A particle oscillates in the form:
x(t) = (8103 m) sin[(4  Hz)t],
where x is in meter and t in seconds. (a) Where is the particle at time =0? and t = 0.2 s.
(b) What is the maximum distance far from the origin? (c) What is the period of this
oscillation?
3- Velocity and speed.
For the particle moves in a simple harmonic motion described by equation 5, we can
calculate the velocity (or in fact, the x-component of the velocity) by differentiate
equation 5 with respect to time.
vx = 0 A cos(0 t+  0 ).
(eq 9)
The maximum value of the velocity is when the cosine is 1, and then the maximum
speed is
vmax = 0 A
(eq 10)
The minimum value of the velocity is when the cosine is -1, and vx = -0 A, but the
speed is maximum here also.
We want to write the speed as a function of the position. To do this we square the two
side of equation 9,
vx 2 = 0 2 A2 cos2 (0 t+  0 ).
And using the fact cos2 +sin2  = 1,
vx 2 = 0 2 A2 [1  sin2 (0 t+  0 )].
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vx 2 = 0 2 A2 -0 2 A2 sin2 (0 t+  0 )= 0 2 {A2 – [A sin(0 t+  0 )]2 }
vx 2 = 0 2 A2 -0 2 A2 sin2 (0 t+  0 )= 0 2 (A2 -x2 )
v  ω0 A 2  x 2
Notice that
(eq11)
v 2x  v . It is clear that the particle has its maximum speed at the origin.
Exercise 2
(a)
A particle makes a simple harmonic motion with angular frequency 0 , and
at time =0; x = 0 and vx = v0 . Find the Amplitude of this motion.
(b)
Find the amplitude if x(t=0) =x0 .
4- Acceleration and force.
From equation 9 we can calculate the acceleration (ax = dvx /dt)
ax = 0 2 A sin(0 t+  0 )
(eq12)
And by comparing equation 12 with equation 5, we notice that
ax =  02 x (or x = 02 x )
(eq 13)
It says that the acceleration is always opposite to the position. Multiply both side by
the mass of the article m, and using Newton’s second law we get,
Fx = m 02 x
(eq14)
This is very important for simple harmonic motion. This equation tells us that the force
which causes the harmonic motion of the particle is always opposite to the position, and
depends linearly on position x. Such a force is called linear restoring force.
Example 1: A body attached to a spring.
The force by a spring on an attached mass is a good example of the restoring force.
fig(4) A linear simple harmonic oscillator. The surface is frictionless.
For the system shown in figure 4, the force on the block is given by Hook’s law,
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Fx = k x
(eq15)
The constant k is the spring constant. Substitute in equation 14 for the force from
equation 15 we get
k x = m0 2 x
ω0 
 k = m 0 2 ,
k
m
(eq16)
What is the period (T0 ) of the oscillation of a system of a mass m attaches to spring with
a spring constant k. The force of the spring is given by equation 15, then applying
Newton’s second law Fx = m ax , we get,
-k x = m ax
and in simple harmonic motion the acceleration ax is given by equation 13, then
-k x = m0 2 x,
From this it easy to show that 0 
k
, and using equation 6 we get,
m
T0  2
m
k
(eq17)
Example 2 The Simple pendulum.
The so-called simple pendulum consists of a small
plumb bob of mass m swinging at the end of a
light, inextensible string of length L, Figure 5. The
motion is along a circular arc defined by the angle
, as shown. The restoring force is the component
of the weight mg acting in the direction of
increasing  along the path of motion:
Fs = - mg sin .
If we treat the bob as a particle, the differentia l
equation of motion is, therefore,
ms  mg sin 
There is no simple analytical solution of this
differential equation. Now s = L, and, for small
, sin =  to a fair approximation. So, after
canceling the m’s and rearranging terms, we can
Fig(5)
The simple pendulum
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write the differential equation of motion in terms s
as follows:
s   g
g
s   s
L
(eq18)
Compare this to equation 13 ( x = 0 2 x) and replace s with x, we notice that the
angular frequency of the pendulum
0 
g
L
(eq19)
and using relation eq 6 we get the famous period of the pendulum
T0  2
L
g
(for small )
(eq20)
5- Energy consideration.
The total energy in simple harmonic motion is conserved, E = Constant.
V+
1
2
mv2 = constant
(eq 18)
where U is the potential energy. In the case of oscillation a spring-block system in
example 1 the potential energy is
V=
1
2
k x2
(eq 19)
This by use that V = 0 at x = 0. The derivation equation is as follows.
xx
xx
xx
x
 x2 

k
k
x
x
dx
dx
V =   - kx dx =
= 
= k  =

x 0
x 0
 2 0
x 0
1
2
k x2
The total energy of a spring- block system is
E =  k x2 + m v2 = Constant.
(eq 20)
The maximum kinetic energy is when the body moves over the origin (x=0). The speed
at this point is o A as give by eq 10, then the maximum energy Tmax is given by
Tmax =  m o 2 A2
(eq 21)
On the other hand, the maximum potential energy is when the block at the maximum
distance from the origin,
Vmax =  k A2
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(eq 22)
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Table 1. Energy consideration for spring-block system
position
x = -A
x=0
x = +A
any x
Potential energy
 k A2
U=0
U =  k A2
 k x2
Kinetic energy
T=0
 m o 2 A2
0
 m v2
Total energy
 k A2
 m o 2 A2
 k A2
 k x2 + m v2
Exercise. A particle of mass = 2 g oscillates with frequency 60 s -1 . The maximum
potential energy is 100 J. What is the Amplitude of this oscillation?
6 – Damped oscillation.
In previous section, we considered the case where
only a restoring force acts on the body. Energy was
conserved. And the system was described by the
differential equation 14, which I will rewrite it as
𝑚𝑥̈ + 𝑘 𝑥 = 0,
with k is a positive constant. The chosen solution
was (equation 5). Now, we will consider (in addition
to the restoring force) a retarding force which
depends on the velocity v x, of the magnitude (cv x).
The direction of the retarding is always opposite to
the velocity . Applying Newton’s second law,
𝐹𝑥(𝑛𝑒𝑡) = 𝑚 𝑥̈
−𝑘 𝑥 − 𝑐 𝑥̇ = 𝑚 𝑥̈
𝑥̈ +
𝑘
𝑐
𝑥 + 𝑥̇ = 0
𝑚
𝑚
(eq 22)
It is convenient to use the damping factor , defined as
𝑐
𝛾≡
2𝑚
and, as before, 𝜔2 =
𝑘
, the we can rewrite equation 22 as
𝑚
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𝑥̈ + 2𝛾 𝑥̇ + 𝜔0 2
(eq 23)
This is a second-order differential equation with constant coefficients. The general
solution can be written as
𝑥 (𝑡) = 𝐴1 𝑒 −(𝛾−𝑞)𝑡 + 𝐴2 𝑒 −(𝛾+𝑞)𝑡
𝑞 = √𝛾 2 − 𝜔0 2 .
where
(eq 24)
(eq 25)
The student can verify that this is a solution by direct substitution into Equation 24.
The factor q could be imaginary. That why we have three cases.
Case 1: Overdamping, In this case 𝜔0 2 < 𝛾 2 and q is positive real number. And the
solution is given by equation 24,
𝑥 (𝑡) = 𝐴1 𝑒 −(𝛾−𝑞)𝑡 + 𝐴2 𝑒 −(𝛾+𝑞)𝑡
The integration constants A1 and A2 are
determined from the initial conditions.
Case 2: Critical damping, In this case 𝜔0 2 = 𝛾 2 and q is zero. And the solution is
given by
𝑥 (𝑡) = 𝐴 𝑡 𝑒 −𝛾𝑡 + 𝐵𝑒 −𝛾𝑡
(eq 26)
Case 3: Underdamping, In this case 𝜔0 2 > 𝛾 2 and q is imaginary. And the solution
is given by
𝑥 (𝑡) = 𝑒 −𝛾𝑡 [𝐴 sin(𝜔𝑑 + 𝜙0 )]
(eq 27)
𝑘
𝑐2
𝜔𝑑 = √𝜔0 2 − 𝛾 2 = √ −
𝑚 4𝑚2
(eq 28)
with the real 𝜔𝑑 defined as
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Problems.
(1) A string vibrates harmonically with frequency of 512 Hz. The amplitude of oscillation
of the counterpoint of the string is 2 mm. What are the maximum speed and the
maximum acceleration at that point?
(2)
A piston makes simple harmonic motion with an amplitude of 0.1 m. It passes through
the center of its motion with a speed of 0.5 m/s. What is the period of the oscillation?
(3)
A particle moves in simple harmonic motion with a frequency of 10Hz. Find the
displacement x at any time t for the flowing initial condition:
t=0,
(4)
x=0.25 m,
and v= 0.1 m/s
We can describe the simple harmonic motion
x = B cos (0 t + )
Show that it is equivalent to Equation 5, by funding the relation between the
constants.
(5)
We can describe the simple harmonic motion
x = C sin (0 t ) + D cos (0 t )
Show that it is equivalent to Equation 5, by funding the relation between the constants.
(6)
Derive the energy function for the simple pendulum (similar to equation 20)
(7)
Calculate the average kinetic, potential, and total energy of the harmonic oscillator.
(8)
A string vibrates harmonically with a frequency of 512 Hz. If the amplitude of
oscillation of the center point of the string is 2 mm. (), what are the maximum speed
and the maximum acceleration at that point?
(9)
A piston executes simple harmonic motion with an amplitude of 0.1 m. If it passes
through the center of its motion with a speed of 0.5 m/s, what is the period of
oscillation?
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(10) A particle undergoes simple harmonic motion with a frequency of 10 Hz. Find the
displacement x at any time t for the following initial condition:
t=0
x = 0.25 m
v x = 0.l m/s
(11) A particle undergoing simple harmonic motion has a velocity when the displacement
is x1 and a velocity vx2 when the displacement is x2 . Find the angular frequency and
the amplitude of the motion in terms of the given quantities.
(12) A spring of stiffness k supports a box of mass M in which is placed a block of mass m.
If the system is pulled downward a distance d from the equilibrium position and then
released, find the force of reaction between the block and the bottom of the box as a
function of time. For what value of d does the block just begin to leave the bottom of
the box at the top of the vertical oscillations? Neglect any air resistance
(13) Show that the ratio of two successive maxima in the displacement of a damped
harmonic oscillator is constant. (Note: The maxima do not occur at the points of
contact of the displacement curve with the curve 𝐴𝑒 −𝛾𝑡 )
(14) The frequency f d of a damped harmonic oscillator is 100 Hz, and the ratio of the
amplitude of two successive maxima is one half. What is the undamped frequencyf 0 of
this oscillator?
(15) A metal block of mass m (= 1.0 kg) attached to a spring (k = 100 N/m) slides on a
horizontal surface that has been lubricated with a heavy oil so that the block suffers a
viscous resistance retarding force of magnitude c v x. If three kinds of oils are used in
three successive experiments, with c= 15, 20, and 25 Ns/m. In each experiment the
block is pulled 1 meter from the equilibrium position and then released. Determine in
each case whither the motion is overdamping, critical camping, or underdamping. Find
the value of the constants, and then plot x versus time for each case.
Last modification on 1437-11-02
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