Some Mathematical Preliminaries

1. Complex Numbers

2. Taylors Series expansion

3. Time average of physical observables

SHO 1
 Complex Numbers
➢ We will extensively use complex numbers
x = −1
2
x = −1 = i
2 2
formulation throughout this course
x2 = ? i = −1 ➢ Their use is NOT mandatory
Imaginary exponents ➢ But the use gives tremendous
Euler’s equation: convenience in classical wave physics

ei = cos + i sin  z=a+ib : Cartesian representation

z = r ei : Polar representation
Imaginary

Complex Plane
z = r (cos  + i sin )
Complex conjugate
z r sin  b
 z* = r (cos  − i sin )
r cos 
 = tan −1 (b / a)
Real
r = a +b ; 2 2

a
SHO 2
 Euler’s equation: ei = cos + i sin 

• The black ball represents the position of the complex number eiφ as it moves through the complex plane.

• The blue ball represents the position of cos(ωt) and the red ball represents the position of isin(ωt).

• Oscillations that can be described by sin(ωt) or cos(ωt) are called harmonic oscillations.

• If you look at the motion of the black ball from above, it moves in a circle but if you look at the motion of the
black ball from the side, it executes harmonic motion.

• The relationship between circular motion and harmonic oscillations is described easily using complex
numbers.

• Sometimes when we observe a harmonic oscillation it is convenient to imagine that we are looking at circular
motion from the side. We can't measure the component of the motion in the imaginary direction, we just
imagine it.
SHO 3
Multiplication by ei is equivalent to rotating counterclockwise
by  in the complex plane
i i
z = Ae i z = Ae e
* − i * − i − i
z =Ae
*
z =Ae e
*

z z * = | z |2 = A 2
z=a+ib
Re z= a, and Im z = b

Re z = Re z*
SHO 4
Phasor = Rotating Arrow + Associated Phase Angle

An Argand diagram is a plot of complex numbers as

points z = x + iy in the complex plane
the x-axis as the real axis and
the y-axis as the imaginary axis

In the plot
the circle represents the complex modulus |z| of z and
the angle represents its complex argument
The complex argument is also called the phase

SHO 5
Taylor series Brook Taylor
Reasonable approximations to
understand the behavior of complex
systems
If a function is defined in an interval at a
point and its derivative of all order exist at
that point then one can expand the function 1685-1731

at that point
1
f ( x) = f ( x0 ) + ( x − x0 ) f ( x0 ) + ( x − x0 ) 2 f " ( x0 ) + ...
'

2!
Goal is to find the value of a function f(x) at the
position x = x0+a, if we are given the information
of the function at x0
SHO 6
Graphical representation of Taylor series approximation

df an d n f
f ( x0 + a ) = f ( x0 ) + a + ... +
dx x0 n! dx n x
Waves and Oscillations,0 W. F. Smith
SHO 7
 Taylor series expansion
Arbitrary potentials can be well approximated by harmonic oscillator
potentials near minima. Consider a potential V(x). Let us assume a
minimum at x = x0 ,
The Taylor series expansion around x = x0,
1
V ( x) = V ( x0 ) + ( x − x0 )V ( x0 ) + ( x − x0 )2V " ( x0 ) + ...
'
2!



f ( n ) ( x0 )
= ( x − x0 ) n
n!
n =0

V ( x0 ) = 0
'
& V ( x0 )  0,
"

Since x0 is a minimum of V(x)

Example: Expansion of sin at 0 = 0
3 5
sin =  − + − ... Try for cos 
3! 5!
For small  : sin  (where  is in radians)
SHO 8
SHO 9
Taylor series at z = a
1
f ( z ) = f (a) + ( x − a) f (a) + ( x − a) 2 f " (a) + ...
'

2!
Examples: Expand in Taylor series at z = π
sin z
1. f ( z ) =
z −
sin z z −  1 ( z −  )3 1 ( z −  )5
Ans : f ( z ) = =− + − + ...
z − z −  3! z −  5! z − 
Expand in Taylor series at z = 1
1 Try using,
2. f ( z ) =
z+2 (1 − x) −1 = 1 − x + x 2 − x 4 + ...
1 1 z − 1 2 ( z − 1) 2 2 ( z − 1) 3
Ans : f ( z ) = − + − + ...
3 9 1! 27 2 ! 27 3!
SHO 10
Before starting...
Lets learn some important notations

SHO 11
Simple Harmonic Oscillations

SHO 12
Simple example of SHM

F = −k x + c x 2

SHO 13
 Simple Pendulum
❖ A point mass suspended from a string or rod of negligible mass
Fnet = −m g sin 
A restoring force acts in the direction l
opposite the displacement from the
equilibrium position.
m l  = −m g sin
  230 & sin   
g g

+  =0   =  max sin t
l l
Amplitude Phase

g 2
=  T = = 2
l
l  g
Note: Small angle approximation is valid till ~ 0.4 radians (= 230)

SHO 14
 Spring-Mass system
Hooke’s Law:
F = −k x
Equation of SHM
m x + k x = 0 friction-free motion
k
x + x = 0  x + 0 2 x = 0
m
Solution of the above equation is
x(t ) = A cos (0 t + )
A and  are arbitrary and can be
determined by the initial conditions
x(t ) = A0 cos (0 t )
❖ At any instant time, one can calculate where
the mass would be
❖ One can describe the motion which is oscillating
in position has period ω0

k 2 m
0 =  T = = 2
m 0 k
SHO 15
Simple example of SHM

F = −k x + c x 2

SHO 16
 Importance of SHO

Know about
❖ The stable equilibrium
❖ Slightly away from stable equilibrium

Harmonic oscillator potential

1 2
V ( x) = k x
2

Lennard-Jones potential
   12   6 
V (r ) = 4    −   
 r   r 
 

SHO 17
 More examples of Free Oscillations
Physical Pendulum
Torque has a sign, depending on the sense of
rotation induced by torque. (point P is cm )

 = I = −( M g sin  )d

 + M g d  = 0
I
The solution is
 (t ) =  max sin( t +  ) max can be find by initial condition

2 M gd
= = I = I cm +  mi ri
2
T I

SHO 18
 Electrical version of Harmonic Oscillator

Initially, C is charged or a current is

induced in L. After that Current and Voltage
Q
oscillate harmonically.
di  Q
VL (t ) = − L = − L Q; and VC =
dt C
 Q
− LQ =
C
The solution is Q(t ) = Qmax cos( t +  ) Qmax can be find by initial
condition
I (t ) = I max cos( t +  + 90 0 )

V (t ) = Vmax cos( t +  ) =
1
LC
SHO 19
 Torsional pendulum

 Where,
I = Moment of Inertia
θ = Angular displacement
 = Rigidity modulus
SHO 20
 Energy of the simple harmonic oscillator
x = A cos(0 t +  ) Since,  = k / m
1 1 1
K .E = m v = m 0 A sin (0 t +  ) = k A2 sin 2 (0 t +  )
2 2 2 2

2 2 2
1 2 1 1
P.E = k x = m0 A cos (0 t +  ) = k A cos (0 t +  )
2 2 2 2 2

2 2 2
1 1
E = K .E + P.E = k A = m 02 A2
2

2 2
2
1 p 1 2
E= kA = 2
+ kx
2 2m 2
2 2
p x
2
+ 2 =1
kA A
p2 x2
+ 2 =1
m 0 A
2 2 2
A SHO 21
Phase space
1-particles:
3 position and 3 momentum coordinates, so
total 6 dimensional space
1D---Phase Line
2D---Phase space and so on

p2 x2
+ 2 =1
m 0 A
2 2 2
A

SHO 22
 Finding Time Average of a physical observable

T is significantly larger than the oscillation time period

1 T /2
 Q = limT → 
2
[Q(t )]2 dt
T −T / 2

1
sin 2 (0t +  ) =
2
𝐸 1 1
Average Energy = = 𝑚𝑣෤ 𝑣෤ = 𝑘𝑥෤ 𝑥෤ ∗
∗
2 4 4

SHO 23
In case of a sinusoidal wave, the RMS value is easy to calculate. If we define Ip to be
the amplitude, then:

where t is time and ω is the angular frequency (ω = 2π/T, where T is the period of
the wave). Since Ip is a positive constant:

Using a trigonometric identity:

but since the interval is a whole number of complete cycles (per definition of
RMS), the sin terms will cancel out, leaving:

= 0.707 Ip

SHO 24
 Differential equation for the SHO

g

+  =0 x + 0 2 x = 0
l
Above equation is 2nd order ordinary homogenous linear differential equation

✓ 2nd Order: because the highest derivative is second order

✓ Ordinary: because the derivatives are only w.r.t. one variable t
✓ Homogenous: because 𝑥 Τ𝜃 or its derivative appear in every term
✓ Linear: because 𝑥 Τ𝜃 or its derivatives appear separately and
linearly in each term
SHO 25
Simple Harmonic Oscillator
Let us find the general solution...
d 2x k
The equation of motion is given by: 2
+ 0 x = 0 where 0 =
2 2

dt m
The above is a second order linear homogeneous differential equation with
constant coefficients

x=e pt
, x = p e p t , x = p 2 e p t ,

p 2 e p t + 02 e p t = 0  p =  i 0

The general solution is given by: The constansts c1 and c2 can be

determined by the initial conditions
x = c1 ei 0t + c2 e −i 0t
SHO 26
 i 0t − i 0 t
The general solution is x(t ) = c1 e + c2 e
At t = 0 x(0) = c1 + c2
i 0t − i 0t
The velocity is given by x (t ) = c1 (i 0 )e + c2 (−i 0 ) e
At t = 0 x (0) = i 0 (c1 − c2 )
1 x (0)  1 x (0) 
c1 =  x(0) +  c2 =  x(0) − 
2 i 0  2 i 0 

1 x (0)  i 0t 1  x (0)  −i 0t

x(t ) =  x(0) +  e +  x(0) − e
2 i 0  2 i 0 
SHO general solution 27
 Three Possible Excitations

t = 0, x = a0 t=0
1. Released from extremity
a0

t = 0, x = 0
2. Impulsed at equilibrium t=0
v
v

t = 0, x = a
t=0
3. Hit at intermediate position

a v
SHO
v 28
 1 x (0)  i 0t 1  x (0)  −i 0t
We have the displacement x(t ) =  x(0) +  e +  x(0) − e
2 i 0  2 i 0 
i 0t
And the velocity equation x (t ) = c1 (i 0 )e + c2 (−i 0 ) e −i 0t

Take the special cases

✓ The mass is pulled to one side and released from rest at t = 0

x(0) = a0 and x (0) = 0 t = 0, x = t=0

a0 a0
1 i 0t 1
x(t ) = a0 e + a0 e−i 0t
2 2
 x(t ) =a0 cos0t

SHO 29
 1 x (0)  i 0t 1  x (0)  −i 0t
We have the displacement x(t ) =  x(0) +  e +  x(0) − e
2 i 0  2 i 0 
i 0t
And the velocity equation x (t ) = c1 (i 0 )e + c2 (−i 0 ) e −i 0t

Take the special cases

✓ The mass is hit and is given a speed v0 at its equilibrium position at t = 0

x(0) = 0 and x (0) = v0

t = 0, x = 0
t=0
1  v0  i 0t 1  v0  −i 0t v
x(t ) =  e + − e v
2  i 0  2  i 0 
v0
 x(t ) =
0
sin 0 t

SHO 30
 1 x (0)  i 0t 1  x (0)  −i 0t
We have the displacement x(t ) =  x(0) +  e +  x(0) − e
2 i 0  2 i 0 
i 0t
And the velocity equation x (t ) = c1 (i 0 )e + c2 (−i 0 ) e −i 0t

Take the special cases

✓ The mass is given a speed v at a displacement a at t = 0
1 v  i 0 t 1  v  −i 0t
x(0) = a and x(0) = v  x(t ) =  a +  e +  a − e
t = 0, x = a
2 i 0  2 i 0  t=0

x(t ) = a cos 0t +

v
sin0t  x = a0 cos(0t +  ) a v
0 v
2
2  v  v  −1
where a0 = a +  
 and  = tan  − 


 0   0a 
SHO 31
 Displacement, Velocity and Acceleration

Why sinusoidal function?

Displacement : x(t ) = A cos(t +  )
The linear operations (differentiation,
dx
integration and addition) applied to a Velocity : v(t ) = = −  A sin(t +  )
sinusoidal functions of a definite dt
period generate other sinusoidal d 2x
Acceleration : a(t ) = = − 2 A cos(t +  )
function of the same period, differing dt 2
at most in amplitude and phase
SHO 32
 Mean position

L O P R
x = −A x=0 x x=+A
v(− A) = 0 (min) v(0) = A  (max), v( x) =   A2 − x 2 , v(+ A) = 0 (min)

a(− A) = − A  (max)
2
a(0) = 0 (min), a( x) = − x
2
a( A) = − A  2 (max)

SHO 33
 Use complex number for solution of SHM
i 0 t
z (t ) = A e x=ASin0t

zA(t ) = A e i
A = Complex amplitude t

Real and imaginary x=Asin(0t+/3)

parts of z(t) satisfy
simple harmonic
equation of motion t
x(t)=Re z(t)
x=Asin(0t+/2)

t
SHO 34