Ulaby problem 1.8

  • Ulaby
  • Electromagnetics
  • FE Exam
  • PE Exam
  • Wave math

posted on 31 Aug 2016

Ulaby Problem 1.8

Problem Statement 1

Two waves on a string are given by the following functions:

\[\begin{aligned} y_1(x,t) &= 4 \cos{(20t - 30x)} \\ y_2(x,t) &= -4 \cos{(20t + 30x)} \\\end{aligned}\]

Where \(x\) is in centimeters. The waves are said to interfere constructively when their superposition \(\left| y_s \right| = \left| y_1 + y_2 \right|\) is a maximum, and they interfere destructive when \(\left| y_s \right|\) is a minimum.

  • (Part a) What are the directions of propogation of waves \(y_1(x,t)\) and \(y_2(x,t)\)?

  • (Part b) At \(t = (\pi/50)\) s, at what location \(x\) do the two waves interfere constructively, and what is the corresponding value of \(\left| y_s \right|\)?

  • (Part c) At \(t = (\pi/50)\) s, at what location \(x\) do the two waves interfere destructively, and what is the corresponding value of \(\left| y_s \right|\)?

Analysis

General form of wave \(y(x,t) = A \cos{\left( \frac{2 \pi t}{T} - \frac{2 \pi x}{\lambda} + \phi_o \right)}\)

Where:

\[\begin{aligned} A &= \text{amplitude of the wave} \\ T &= \text{time period} \\ \lambda &= \text{spatial wavelength} \\ \phi_0 &= \text{reference phase (which is constant in both time and space)}\end{aligned}\]

The angle \(\phi(x,t)\) is the phase of the wave:

\[\phi(x,t) = \left( \frac{2 \pi t}{T} - \frac{2 \pi x}{\lambda} + \phi_o \right)\]

Let’s rewrite the phase to separate the \(t\) and \(x\) terms for clarity

\[\phi(x,t) = \left( \frac{2 \pi}{T} t - \frac{2 \pi}{\lambda} x + \phi_o \right)\]

Since \(f=\frac{1}{T}\), The angular velocity \(\omega\) is

\[\omega = \frac{2 \pi}{T} = 2 \pi f\]

The phase constant \(\beta\) is defined as

\[\beta = \frac{2 \pi}{\lambda}\]

Unless a reference phase (\(\phi_0\)) is given, we can set it to zero. Now we can write a more compact form for the phase of the wave:

\[\phi(x,t) = \left( \omega t - \beta x \right)\]

Inspect the signs of \(t\) and \(x\). If:

  • one of the signs is positive, and the other is negative, then the wave is traveling in the positive \(x\)-direction

  • the signs are the same, then the wave is traveling in the negative \(x\)-direction.

Now let’s write a simplified wave equation and do some coefficient matching with \(y_1\) and \(y_2\):

\[\begin{aligned} y(x,t) &= A \cos{(\omega t - \beta x)} \\ y_1(x,t) &= 4 \cos{(20t - 30x)} \\ y_2(x,t) &= -4 \cos{(20t + 30x)}\end{aligned}\]

Now superimpose the two waves. First we will set the phase angle of each wave to \(x\) and \(y\) to simplify the algebra as we use a trigonometric identity from Appendix C for our superposition:

\[\begin{aligned} x &= (20t - 30x) \\ y &= (20t + 30x)\end{aligned}\]

Now add the waves together

\[\begin{aligned} y_s &= 4 \cos{x} + (-4 \cos{y}) \\ &= 4 \left[ \cos{x} - \cos{y} \right] \\ &= 4 \left[ -2 \sin{\frac{x+y}{2}} \sin{\frac{x-y}{2}} \right] \\ &= 4 \left[ -2 \left( \sin{\frac{(20t - 30x)+(20t + 30x)}{2}} \right) \left( \sin{\frac{(20t - 30x)-(20t + 30x)}{2}} \right) \right] \\ &= 4 \left[ -2 \left( \sin{\frac{20t - 30x+20t + 30x}{2}} \right) \left( \sin{\frac{20t - 30x-20t - 30x)}{2}} \right) \right] \\ &= 4 \left[ -2 \left( \sin{\frac{40t}{2}} \right) \left( \sin{\frac{-60x)}{2}} \right) \right] \\ &= 4 \left[ -2 \left( \sin{20t} \right) \left( \sin{-30x} \right) \right] \\ &= -8 \left( \sin{20t} \right) \left( \sin{-30x} \right) \\ &= -8 \left( \sin{20t} \right) \left( -\sin{30x} \right) \\ &= 8 \left( \sin{20t} \right) \left( \sin{30x} \right) \\\end{aligned}\]

Now substitute \(t=\frac{\pi}{50}\):

\[\begin{aligned} y_s &= 8 \left( \sin{ \left( \frac{ 20 \pi}{50} \right)} \right) \left( \sin{30x} \right) \\ &= 7.60845 \sin{(30x)}\end{aligned}\]

A sine function reaches its maximum with its phase angle, \(\phi(x,t)\), is equal to \(n \pi + 90^{\circ}\), and reaches its minimum when \(\phi(x,t)\) is equal to \(n \pi\).

Solution to Part (a)

  • Since \(x\) and \(t\) in wave function \(y_1\) have different signs, we conclude that \(y_1\) is traveling in the positive \(x\)-direction.

  • Since \(x\) and \(t\) in wave function \(y_2\) have the same signs, we conclude that \(y_2\) is traveling in the negative \(x\)-direction.

Solution to Part (b)

Set \(\phi(x_1,t) = n \pi + 90^{\circ}\) and solve for \(x\) to determine where waves interfere constructively:

\[\begin{aligned} \phi(x_1,t) &= n \pi + \frac{\pi}{2} \\ 30x &= n \pi + \frac{\pi}{2} \\ x &= \frac{n \pi}{30} + \frac{\pi}{2 \times 30} \\ x &= \boxed{ \frac{n \pi}{30} + \frac{\pi}{60} \text{ (cm) } (n = 0,1,2,...)} \\\end{aligned}\]

Solution to Part (c)

Set \(\phi(x_1,t) = n \pi\) and solve for \(x\) to determine where waves interfere destructively:

\[\begin{aligned} \phi(x_1,t) &= n \pi \\ 30x &= n \pi \\ x &= \boxed{\frac{n \pi}{30} \text{ (cm) } (n = 0,1,2,...)}\end{aligned}\]

Emacs 24.5.1 (Org mode 8.2.10)