Waves and wave equation

A disturbance or variation that transfer energy progressively

from point to point in a medium

In this course we are learning about the Mechanical waves, the

Electromagnetic waves and the Matter waves
 Transversal waves

Transverse wave: Oscillations 900 from axis of travel, the particle

displacement is perpendicular to the wave propagation. The particles
moves up and down about their equilibrium position.

Example: EM waves (X-rays, light, heat, microwaves, radio, etc.) ,

Ocean wave

https://www.phy.olemiss.edu/~perera/animations/waves.html
Longitudinal wave: The particle in a medium oscillates back and forth
about their equilibrium position but it is the disturbance which travels
NOT the individual particles in the medium.

Example: Sound wave, seismic p-waves, molecular motion in

compression and rarefaction, etc.

http://www.geo.mtu.edu/UPSeis/waves.html
 One-Dimensional Traveling Waves in a rope

As the wave passes, the motion of

any actual bit of rope is in the y-
direction, at right angles (transverse)
to the direction of the wave itself,
which is of course along the rope.
Wave keeps its shape
At the same amplitude,
x1 − c t1 = x2 − c t 2 ,
x2 − x1
⇒c= ;
t 2 − t1
wave is moving in the
+ ve direction with velocity c.
(We are neglecting frictional effects—in a real rope, the bump gradually
gets smaller as it moves along.)
 Longitudinal wave in a rod (elastic medium)
How disturbance evolve?

Youngs modulus (modulus of Elasticity) :

stress F / A Y A
Y= = ⇒ F = ξ = k ξ
strain ξ / L  L 
Rod with Young’s modulus Y,
density ρ and area of cross-section A.
𝜉(𝑥) 𝜉(𝑥 + Δ𝑥)

𝑥 𝑥 + Δ𝑥

𝑥 𝑥 + Δ𝑥
If disturbance is created in a rod of density ρ and area of cross-section A, it
propagate as a linear chain of spring-mass system in the continuum limit:

∂2 ξ
At a position x, force equation of slab : F = ρ A∆ x 2
∂t
( It is simply mass times acceleration )
Consider a linear chain of identical spring and masses. The displacement of
jth mass satisfy the equation of motion

k k k
m m m m

ξ j−1 (t ) ξ j (t ) ξ j+1 (t )

m ξj (t ) = −k (ξ j − ξ j +1 ) − k (ξ j − ξ j −1 )

Let Δ𝑥 be the separation between the masses.

Case1: if Δ𝑥 is finite this means the chain is discrete

Case 2: If Δ𝑥 is very small this means the chain is continuous

In the continuum limit ( ):

[ξ j +1 ]
(t ) − ξ j (t ) → [ξ ( x + ∆ x, t ) − ξ ( x, t )] Use Taylor series

 ∂ ξ ( x, t ) ( ∆ x ) 2 ∂ 2 ξ ( x, t ) 
= ξ ( x, t ) + ∆ x + ... − ξ ( x, t )
 ∂x 2 ∂x 2

∂ ξ ( x, t ) ( ∆ x ) 2 ∂ 2 ξ ( x, t )
=∆x +
∂x 2 ∂ x2
 [ξ (t ) − ξ
j j −1 ]
(t ) → [ξ ( x, t ) − ξ ( x − ∆ x, t )]
 ∂ ξ ( x, t ) ( ∆ x ) 2 ∂ 2 ξ ( x, t ) 
= ξ ( x, t ) − ξ ( x, t ) + ∆ x − ...
 ∂x 2 ∂x 2

∂ ξ ( x, t ) ( ∆ x ) 2 ∂ 2 ξ ( x, t )
=∆x −
∂x 2 ∂ x2

Implies,

[ξ j +1 ] [ ]
(t ) − ξ j (t ) − ξ j (t ) − ξ j −1 (t ) =
 ∂ ξ ( x, t ) ( ∆ x ) 2 ∂ 2 ξ ( x, t )   ∂ ξ ( x, t ) ( ∆ x ) 2 ∂ 2 ξ ( x, t ) 
= ∆ x + −∆x − 
 ∂ x 2 ∂ x 2
  ∂ x 2 ∂ x 2

∂ 2
ξ ( x, t )
= (∆ x) 2

∂ x2
 Y A Y A
The force in terms of Y F =  ξ =   ∆ξ = k ∆ξ
 L   ∆x
Δ𝜉
Here, is the local strain.
Δ𝑥

Now the force balance equation can be written as,

m ξj (t ) = −k (ξ j − ξ j +1 ) − k (ξ j − ξ j −1 )
[
= k (ξ j +1 − ξ j ) − (ξ j − ξ j −1 ) ]
∂ 2 ξ ( x, t ) Y A 2 ∂ ξ ( x, t )
2
ρ A∆ x = ( ∆ x)
∂t 2
∆x ∂ x2 Y
c =
2
∂ ξ ( x, t ) Y ∂ ξ ( x, t )
2 2
∂ ξ ( x, t ) 1 ∂ ξ ( x, t )
2 2 s
ρ
= ⇒ = 2
∂t 2
ρ ∂x 2
∂x 2
cs ∂t2

The general solution the differential equation is f ( x − c t ) and f (x + c t)

In case of a pipe containing a fluid medium, one can obtain a similar
expression

∂ 2 ξ ( x, t ) B ∂ 2 ξ ( x, t ) ∂ 2 ξ ( x, t ) 1 ∂ 2 ξ ( x, t )
= ⇒ = 2
∂t 2
ρ ∂x 2
∂x 2
cB ∂ t 2
Where B is the bulk modulus of the fluid
B
c =
2

ρ
B

The general solution of the differential equation is

f ( x − c t ) and f ( x + c t )
Wave is moving in +ve x direction Wave is moving in -ve x direction
1. Consider a sound wave travelling in a solid with Young’s modulus
Y =2 ×1011 kg-m-1s-2, and whose mass density is ρ = 2 ×103 kg-m-3.
The wave-solution has the form ξ(x, t) = A cos2(k x - (2π ×102) t),
where x and t are measured in SI units. The wavelength of the given
wave form is

Y 2 ×10 11
c= = = 10 4
m/s,
ρ 2 ×10 3

ω = 4 π ×10 2 rad/s,
ω 4 π ×10 2
ν= = = 200 Hz,
2π 2π
4
c
10
λ= = = 50 m.
ν 200
So, the two aspects of the set of wave equation

1. The function must propagate the disturbance

f (x ± c t)

wave moving to the left

∂ 2 ξ ( x, t ) 1 ∂ 2 ξ ( x, t )
= 2
∂x 2
c ∂t2

2. Different forms of the solution of wave equations

2π
ξ = ξ 0 cos(ω t − k x) or ξ = ξ 0 cos (c t − x) or ξ = ξ 0 ei (ω t − k x )
λ
Transverse wave on a flexible string
y

Linear density 𝜇 under a tension T 0,0

x

∆x T
Consider an element of the string
between x and x + ∆x

For a small displacement y, we can assume that the magnitude of the tension T remain
the same, but acts in different direction at the two ends of the elements of the string.
This give rise to a net vertical force.

The forces on the bit of string (neglecting the tiny force of gravity, air resistance, etc.)
are the tensions T at the two ends. The tension will be uniform in magnitude along the
string, but the string curves if it’s waving, so the two T vectors at opposite ends of the
bit of string do not quite cancel, this is the net force we’re looking for.
Assumption while deriving the wave equation

1. The string is indefinitely long

2. The magnitude of the tension is constant, independent of position

3. The angle of inclination of the displaced string wrt x-axis at any

point is small.

4. An element Δx of the string moved only in transverse direction as a

result of the wave disturbance

5. Neglecting friction of surrounding air

6. Neglecting the effect of stiffness of string

7. Neglecting the effect of gravity

8. We neglect the x-component while equating the forces in two

directions. The x-component is responsible for the transport linear
momentum by a transverse wave travelling on the string
 y
T
𝜃 + Δ𝜃

𝜃
Resultant of these two forces
T

𝑥 𝑥 + Δ𝑥 x
Fy = T sinθ x + ∆x
− T sinθ x
Fy = −T sin θ + T sin(θ + ∆θ )
for small θ , Fy ≈ T tanθ x + ∆x
− T tanθ x
for small θ ; Fy ≈ T ∆θ
 ∂ y  ∂ y 
Fy = T   −    dm y = T ∆θ ⇒ ( µ ∆ x) y = T ∆θ
 ∂ x  x + ∆x  ∂ x  x  ∂y 1 ∂θ ∂2 y
tan θ = ⇒ = 2
 ∂ y  ∂2 y  ∂ y   ∂2 y ∂ x cos θ ∂ x ∂ x
2
= T   + ∆ x 2 −    = T∆ x 2
 ∂ x  x ∂ x  ∂ x  x  ∂x
∂2 y ∂2 y
( µ ∆ x) 2 = T ∆θ = T∆x 2
∂2 y ∂2 y ∂t ∂x
dm y = µ ∆ x 2 = T∆ x 2
∂t ∂x 2
1 ∂ y ∂2 y T
⇒ 2 2 = 2 where, c2 =
c ∂t ∂x µ
 1 ∂2 y ∂2 y
Wave Equation ⇒ 2 2 = 2
c ∂t ∂x
At the point x,
• The 1st derivative is the slope of the string.

• The second derivative is the rate of change of the slope: in other

words, how much the string is curved at x.

• And, it’s this curvature that ensures the T’s at the two ends of a bit
of string are pointing along slightly different directions, and
therefore don’t cancel.

• This force, then, gives the mass times acceleration on the right.
 Wave Equation
1 ∂2 y ∂2 y
⇒ 2 2 = 2
c ∂t ∂x

• All traveling waves move at the same speed and the speed is
determined by the tension and the mass per unit length.

• The wave can travel in any direction +x or –x.

• So there are two solution of the above wave equation.

f (x ± c t)
• Then sum of the two is also a solution to the equation: which is
superposition of the two.
 Standing Waves from Traveling Waves

Superpose harmonic traveling waves moving in opposite directions to get

a standing wave:

Standing waves on
a Stretched string
l
⇒ y ( x, t ) = A cos(k x − ω t )
1 ∂2 y ∂2 y
= 2 ⇒ y ( x, t ) = A cos(−k x − ω t )
c ∂t
2 2
∂x
= A cos(k x + ω t )
Boundary conditions At, x = 0, y = 0
At, x = l , y = 0
⇒ y ( x, t ) = A e i (ω t − k x ) + B e i (ω t + k x )

At, x = 0, y = 0 ⇒ A + B = 0
i (ω t − k x ) i (ω t + k x )
⇒ y ( x, t ) = A e − Ae
( )
= A ei ω t e −i k x − ei k x = − Aei ω t 2i sin( kx)

At x = l , y = 0
sin kl = 0, ⇒ kl = n π
iω t  nπ 
y ( x, t ) = − A( 2i ) e sin  x
 l 
Equation of standing wave
 iω t  nπ 
y ( x, t ) = − A(2i ) e sin  x
 l 
(n-1) nodes between boundaries
 This is standing wave

iω t  nπ 
y ( x, t ) = − A(2i ) e sin  x
 l 
Wave equation for transverse wave
n
n wavelength, l = λn
th
On a string (both ends are fixed)
2
∂2 y 1 ∂2 y
or,ν n =
c nc
= = 2 2
λn 2 l ∂x 2
c ∂t
 ωn x 
⇒ y ( x, t ) = An sin   cos ωnt ;
 c 
 
nπ T
where, ωn = = n ω1
L µ
y1 ( x, t ) = A cos(k x − ω t ) ;
y2 ( x, t ) = A cos(−k x − ω t ) ;
y ( x, t ) = y1 + y2 = A cos(k x − ω t ) + A cos(−k x − ω t );
  kx − ω t − (−kx − ω t )   kx − ω t + (−kx − ω t )  
y ( x, t ) = A 2 cos  cos  
  2   2 
= 2 A cos(kx) cos(−ω t ) = 2 A cos(kx) cos(ω t )

y1 ( x, t ) = A sin( k x − ω t ) ;
y2 ( x, t ) = A sin( −k x − ω t ) ;
y ( x, t ) = y1 + y2 = A sin( k x − ω t ) + A sin( −k x − ω t );
  kx − ω t + (−kx − ω t )   kx − ω t − (−kx − ω t )  
y ( x, t ) = A 2 sin   cos  
  2   2 
= 2 A cos(kx) sin( −ω t ) = −2 A cos(kx) sin(ω t )