Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
Chapter 4
Wave Motion
4.1 Circular Motion
Wave motion
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Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
Types of waves
There are two types of waves. These are transverse waves and longitudinal waves.
A transverse wave is one in which the vibrations of the particles in the wave are at
right angles to the direction in which the energy of the wave is travelling.
Fig.4.3 Transverse wave on a rope
Figure 4.3 shows a transverse wave moving along a rope. The particles of the rope
vibrate up and down, whilst the energy travels at right angles to this, from A to B.
There is no transfer of matter from A to B. Examples of transverse waves include
light waves, surface water waves and all emw.
A longitudinal wave is one in which the direction of the vibrations of the particles
in the wave is along the direction in which the energy of the wave is travelling.
Fig.4.4 longitudinal wave on a stretched spring
Fig.4.4 shows a longitudinal wave moving along a stretched spring. the coils of the
spring vibrate along the length of the spring. Whilst the energy travels along the
same line, from A to B. note that the spring itself does not move from A to B.
examples of longitudinal waves include sound waves
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Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
4.2 Description of wave:
In this section, we will introduce an important wave function which is called
sinusoidal wave shown in figure 4-5. The sinusoidal wave is the simplest example of
a periodic continuous wave. The figure shows a sinusoidal wave traveling to the
right with speed v. Figure 4-6 a shows a snapshot of a sinusoidal wave traveling
through a medium. If we focus on one element at (x=0) we see that elements moves
up and down with displacement y(t) as shown in figure 4-6b. You should notice that
the wave is periodic in both space and time, the space period is represented by  and
the time period is represented by T.
Figure 4-5 A sinusoidal wave traveling to the right.
Figure 4-6 a) A snapshot of a sinusoidal wave traveling through a medium. b) The
position of one element of the medium as a function of time.
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i-
Displacement: the displacement (y) of any element of the medium is
the displacement from its equilibrium position.
ii-
Crest: the point at which the displacement of the element is the highest
Trough: the point at which the displacement of the element is the
lowest
iii-
Amplitude: the maximum value of the displacement
iv-
Wavelength: the distance from one crest to the next is called the
wavelength (). Or generally, it is the distance in which the wave
repeats itself.
v-
Period: the time the wave takes for one complete wave to pass a
particular point is called the period of the wave (T).
vi-
Frequency: the number of waves passing a particular point per second
is called frequency of the wave (f).
The frequency is related to the period by: f 
vii-
1
T
Speed of propagation: the distance the wave travel per second is called
the speed of propagation (v).
v
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
 f
T
Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
4.3Mathematical representation of a wave:
Fig.4-6a
Consider the sinusoidal wave in Figure 4-6a, which shows the position of the
wave at t = 0. Because the wave is sinusoidal, we expect the wave function at
this instant to be expressed as 𝑦(𝑥, 0) = 𝐴 𝑠𝑖𝑛 𝑎𝑥, where A is the amplitude and
a is a constant to be determined. At x = 0, we see that
𝑦(0, 0) = 𝐴 𝑠𝑖𝑛 𝑎(0) = 0,
Consistent with Figure 4-5a. The next value of x for which y is zero is x =λ/2.
Therefore,
𝜆
𝜆
𝑦 ( ,0) = 𝐴𝑠𝑖𝑛 (𝑎 ) = 0
2
2
For this equation to be true, we must have 𝑎𝜆/2 = 𝜋, 𝑜𝑟 𝑎 = 2𝜋/𝜆. Therefore,
the function describing the positions of the elements of the medium through which
the sinusoidal wave is traveling can be written
2𝜋
𝑦(𝑥, 𝑜) = 𝐴𝑠𝑖𝑛(
𝜆
𝑥)
(1)
where the constant A represents the wave amplitude and the constant λ is the
wavelength. Notice that the vertical position of an element of the medium is the
same whenever x is increased by an integral multiple of λ.
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If the wave moves to the right with a speed v, the wave function at some
later time t is
2𝜋
𝑦(𝑥, 𝑡) = 𝐴𝑠𝑖𝑛 [(
𝜆
(𝑥 − 𝑣𝑡)]
(2)
The wave function has the form 𝑓(𝑥 − 𝑣𝑡). If the wave were traveling to the left, the
quantity 𝑥 − 𝑣𝑡 would be replaced by 𝑥 + 𝑣𝑡.
By definition, the wave travels through a displacement Δx equal to one wavelength λ
in a time interval Δt of one period T. Therefore, the wave speed, wavelength, and
period are related by the expression
𝑣=
𝛥𝑥
𝛥𝑡
𝜆
= 𝑇 = 𝜆𝑓
(3)
Substituting this expression for v into Equation 2 gives
𝑥
𝑡
𝜆
𝑇
𝑦(𝑥, 𝑡) = 𝐴𝑠𝑖𝑛 [2𝜋 ( − )]
(4)
We can express the wave function in a convenient form by defining two other
quantities, the angular wave number k (usually called simply the wave number)
and the angular frequency 𝜔.
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𝑘=
𝜔=
2𝜋
𝑇
2𝜋
(5)
𝜆
= 2𝜋𝑓
(6)
Using these definitions, Equation (4) can be written in the more compact form
𝑦 = 𝐴 sin(𝑘𝑥 − 𝜔𝑡)
(7)
Using Equations 5 and 6, the wave speed v originally given in Equation 3 can be
expressed in the following alternative forms:
𝑣=
𝜔
𝑘
𝑣 = 𝜆𝑓
The wave function given by Equation 7 assumes the vertical position y of an
element of the medium is zero at x = 0 and t = 0. That need not be the case. If it
is not, we generally express the wave function in the form
𝑦 = 𝐴 sin(𝑘𝑥 − 𝜔𝑡 + 𝜙)
(8)
Where ϕ is the phase constant, this constant can be determined from the initial
conditions.
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4.4 Relation between Vibrational Motion and Circular Motion
Fig 4.7
Consider a particle located at point P on the circumference of a circle of radius A, as
in Figure 4.7, with the line OP making an angle ϕ with the x axis at t = 0. We call
this circle a reference circle for comparing simple harmonic motion with uniform
circular motion, and we take the position of P at t = 0 as our reference position. If
the particle moves along the circle with constant angular speed 𝜔 until OP makes an
angle ɵ with the x axis, as in Figure 4.6, then at some time t ˃ 0, the angle between
OP and the x axis is 𝜔𝑡 + 𝜙.As the particle moves along the circle, the projection of
P on the y axis, , moves up and down along the y axis between the limits 𝑦 = ±𝐴 .
From the right triangle OPQ, we see that this y coordinate is
𝑦(𝑡) = 𝐴sin(𝜔𝑡 + 𝜙)
This geometric interpretation shows that the time interval for one complete
revolution of the point P on the reference circle is equal to the period of motion T for
simple harmonic motion between 𝑦 = ±𝐴.
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Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
The phase constant ϕ for simple harmonic motion corresponds to the initial angle
that OP makes with the x axis. The radius A of the reference circle equals the
amplitude of the simple harmonic motion.
Example
A sinusoidal wave traveling in the positive
x- direction has amplitude of 15.0 cm, a
wavelength of 40.0 cm, and a frequency of
8.00 Hz. The vertical position of an
element of the medium at
t = 0 and x = 0 is also 15.0 cm as shown
in Figure.
(A) Find the wave number k, period T,
angular frequency ⍵, and speed v of the
wave.
(B) Determine the phase constant ϕ and
write a general expression for the wave
function.
Solution
Evaluate the wave number from equ.(5)
Evaluate the period of the wave from equ.
1
𝑇=
𝑓
Evaluate the angular frequency of the
wave from equ.(6)
Evaluate the wave speed from equ.(3)
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Evaluate the phase constant ϕ
Substitute A =15.0 cm, y = 15.0 cm, x = 0,
and t = 0 into Equation ( 8):
Write the wave function:
4.6 Propagation of a sinusoidal wave on a taut string:
If we attach an oscillating blade to one side of a taut string as shown in figure
4-8, we can create a transverse wave that propagates along the string. So, each
element of the string such as (P), also oscillates vertically with simple harmonic
motion. The wave equation is described by:
yx, t   Asin kx  t 
Figure 4-8 one method of producing a sinusoidal wave on a string.
We can use this equation to describe the motion of any element of the string.
The speed (vy) of any element of the string is given by:
 y  x, t  
dy  x, t 
 A coskx  t   
dt x cons tan t
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The acceleration (ay) of any element of the string is given by:
ay 
dv y  x, t 
dt
  2 A sin kx  t   
x cons tan t
  2 y  x, t 
Note: The speed (vy) of any element of the string should not be confused with the
speed of propagation of the wave (v).
The Speed of Waves on Strings
The speed of the transverse wave on a taut string depends on the tension (T) in
the string, and the mass of the string per unit length (). The speed of the transverse
wave is given by:
v
T

The above equation states that: “The greater the tension in the string, the
greater the speed of propagation of the wave along the string, and the greater the
mass per unit length of the sting, the smaller the speed of the wave”.
Example:
A uniform cord has a mass of m = 0.3 kg and a length of L = 6m. The cord
passes over a pulley and supports a 2 kg object as shown in figure. Find the speed of
a pulse traveling along this cord.
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Solution
The tension in the cord is given by:
T  Mg  2  9.8  19.6 N
The mass per unit length of the string is:

m 0.3

 0.05 kg / m
L
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Therefore, the speed of propagation of the wave along the string is:
v
T

19.6
 19.79 m / s
0.05

Example:
A particular wave propagating along a taut string is given by:
y x, t  0.2 sin 0.5x  8.2t ,
where the displacement y is in meter, x is in meter and t is in second. Find:
 
iiiiiiivvviviiviiiix-


The amplitude (A) of the wave.
The wave number (k).
The wavelength ().
The angular frequency ().
The frequency (f).
The period (T).
The velocity of the wave (v) and the direction of propagation.
The maximum speed of the string elements.
The maximum acceleration of the string elements.
Solution
yx, t   0.2 sin 0.5x  8.2t 
The general form of the wave equation is:
yx, t   Asin kx  t   
Comparing the two equations, we got:
A=0.2 m
k=0.5 rad m-1
=8.2 rad/s
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=zero
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iii-
The amplitude of the wave is A=0.2 m
The wave number k=0.5 rad m-1
iii-
The wavelength is
iv-
The angular frequency is =8.2 rad/s
v-
The frequency is
vi-
The period is T
vii-

f 
2 2

 4  12.566 m
k
0.5
 8.2

 1.305 Hz
2 2
1
1

 0.766 s
f 1.305
The speed of the wave is v  f  12.566  1.305  16.4 m / s
 8.2
or v 

 16.4 m / s
k 0.5

In the given wave equation, the sign in front of  is negative so, the direction of
propagation is in positive x direction.
viii- The speed of the string elements is given by:
 y  x, t  
dy  x, t 
dt x cons tan t
d
0.2 sin 0.5 x  8.2t 
dt
 8.2  0.2 cos0.5 x  8.2t 
 1.64 cos0.5 x  8.2t 

The maximum speed of the string elements is: v y , max  1.64 m / s
ix- The acceleration of the string elements is given by:
ay 
dv y  x, t 
dt

x  cons tan t
d
 1.64 cos0.5 x  8.2t 
dt
 1.64  8.2 sin 0.5 x  8.2t 
 13.448 sin 0.5 x  8.2t 
The maximum acceleration of the string elements is:
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a y , max  13.448 m 2 / s
Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
Solved Problem
Transverse pulses travel with a speed of 200 m/s along a taut copper wire whose
diameter is 1.50 mm. What is the tension in the wire? (The density of copper is 8.92
g/cm3.)
Solution
Solved Problem
Transverse waves travel with a speed of 20.0 m/s in a string under a tension of 6.00
N. What tension is required for a wave speed of 30.0 m/s in the same string?
Solution
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Faculty of Engineering
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The proof of
 The string is under tension T
 Consider one small string element of
length ∆s
 The net force acting in the y direction is
 This uses the small-angle
approximation
Because the element forms part of a circle and
subtends an angle 2θ at the center, ∆s = R(2 θ),
and
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