Breadcrumbs Section. Click here to navigate to respective pages.

Chapter

Chapter

# Appendix A: Waves

DOI link for Appendix A: Waves

Appendix A: Waves book

# Appendix A: Waves

DOI link for Appendix A: Waves

Appendix A: Waves book

## ABSTRACT

A.1 Waves EquationIf we have a stationary wave (non-sinusoidal form), as shown in Fig. A.1.1, as y¢ = f(x¢) at t = 0 (A.1.1)

we can have a traveling wave as y = f(x ± vt) at t = t (A.1.2)where x¢ = x ± vt, ∂ ∂

= x x ¢ 1 and ∂

∂ = ±

x t

v ¢

Also when y = f(x + vt), the wave moves in the left direction. When y = f(x – vt), the wave moves in the right direction. From the space derivative as∂

∂ =

∂ ∂

◊ ∂ ∂

= ∂ ∂

y x

f x

x x

f x¢

¢ ¢

(A.1.3) ∂

∂ =

∂ ∂

∂ ∂

= ∂ ∂ ∂

∂ ◊ ∂ ∂

= ∂ ∂

∂ ∂

= ∂ ∂

y

x x y x

y x x

x x x

f x

f

x

( / )

¢ ¢

¢ ¢ ¢ (A.1.4)and the time derivative as

∂∂ = ∂∂ ◊ ∂∂ = ± ∂∂yt fx xt v fx¢ ¢ ¢ (A.1.5)∂ ∂

= ∂ ∂

∂ ∂

= ∂ ∂ ∂ ∂

◊ ∂ ∂

= ∂ ∂

± ∂ ∂

Ê ËÁ

ˆ ¯˜ ◊± = ∂

t t y t

y t x

x t x

v f x

v v ( / )

¢ ¢

¢ ¢ f

x∂ ¢2 (A.1.6) Equations A.1.4 and A.1.6 yield∂ ∂

= ◊ ∂ ∂

1y

x v

y

t (A.1.7) On the other hand, if we have the wave function of a traveling wave (sinusoidal waveform) as y = A sin k (x ± vt) (A.1.8)the solution to the partial differential equation will be as follows:

Space derivative ∂ ∂

= ± y x

kA k x vtcos ( ) (A.1.9) ∂

∂ = - ±

x k A k x vtsin ( ) (A.1.10)

Time derivative ∂ ∂

= ± ± y t

kvA k x vtcos ( ) (A.1.11)

∂ ∂

= - ± 2

t k v A k x vtsin ( ) (A.1.12) Equations A.1.10 and A.1.12 also yield∂

∂ = ◊

∂ ∂

1y

x v

y

t (A.1.13) On the other hand, a harmonic wave (sinusoidal form) can be written as y A k x vt= ±

sin ( ) (A.1.14)where A and k are the amplitude and the propagation constant of the periodic wave form/character, respectively. Since the difference between sine and cosine functions is a relative translation of 0.5p radian, it is also enough to treat only one of these wave functions. For a sine wave of amplitude, as shown in Fig. A.1.2., at constant time t

Figure A.1.2 Harmonic waves. y A k x vt A k x vt= + + = + +sin ( ) sin ( )l p2 (A.1.15) A sin (kx + kl + kvt) = A sin(kx + kvt + 2p) (A.1.16)Clearly, kl = 2p (A.1.17) At a constant point x, y = A sin k(x + v[t + T]) = A sin k(x + vt) + 2p (A.1.18) y = A sin (kx + kvt + kvT) = A sin (kx + kvt + 2p) (A.1.19)Clearly, KvT = 2p (A.1.20)

Equations A.1.17 and A.1.20 yield v T

= l and v f= l (A.1.21)

If we define the angular frequency w p

= 2 T

, a more compact form of the harmonic wave in Eq. A.1.14 is y A x t T

= ±Ê ËÁ

ˆ ¯˜cos

l (A.1.22) y A kx t= ±

sin ( )w (A.1.23) Furthermore, at a constant point x, the phase angle j is constant or j = k(x ± vt) (A.1.24) dj = 0 = k(dx ± vdt) (A.1.25)dx

dt v= ∓ (A.1.26) From Eq. A.1.14, the general form of the harmonic wave at initial phase j0, as shown in Fig. A.1.3, can be modified as y A k x vt= ± + cos

sin [ ( ) ]f

Figure A.1.3 Harmonic waves showing the initial phase. If we define the initial conditions for harmonic wave at x = 0, t = 0, and y = y0, we get y A

1 0= Ê ËÁ

ˆ ¯˜

y A

using sine function (A.1.28)

The initial phase j0 is generally set equal to zero for simplicity. Let a complex number be defined asz a ib= + (A.1.29)where a = Re(z), b = lm(z), and i2 = −1. z can be shown in the polar plot as seen in Fig. A.1.4.