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}

\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)