This document describes how to find the resulting amplitude when two waves with the same wavelength, frequency and direction interfere. It provides an equation that calculates the resulting amplitude based on the phase difference between the two waves. It then works through an example problem, finding that the resulting amplitude of two interfering waves - one described by a graph and one by an equation - is 5.543m.

1. FINDING THE RESULTANT AMPLITUDE
FROM INTERFERENCE OF WAVES
TRAVELLING IN THE SAME DIRECTION
by Mine Sher

2. SUMMARY
When two waves that have the same wavelength,
frequency, amplitude and direction are superimposed,
finding the resulting amplitude is fairy simple.
All we need to do is find how many radians out of phase the
two component waves are, and then use the equation,
resulting A = 2Acos(phase/2)

3. PROBLEM
A wave is described in the graph below (wave-a). A second
wave (wave-b) is superimposed on wave-a. Wave-b is
described by the equation: x(1)= 3cos(pi/4). What is the
amplitude of the resulting wave?

4. INFO
From the graph, we can find out what the period of the two
component waves are.
We know the phase of wave-b, but not the phase of wave-a.
We can find the phase of wave-a by finding the phase constant for
wave-a.
Note: for the general equation x(t) = Acos(wt+phi)
x(t) = Acos(wt+phase constant) and x(t) = Acos(phase)
The equation described in slide two can be used to find amplitude
once the difference in phase is calculated.

5. SOLVE
First, we see the period to be 12 seconds.
We can use 2pi/T to find w=pi/6

6. We can make an incomplete equation with the information
so far.
x(t)=3cos((pi/6)+phase constant)
We can plug in a known point on the graph to calculate the
phase constant, like the point (1,0)
x(1)=3cos((pi/6)(1)+phase constant)=0
x(1)=cos((pi/6)+phase constant=0
pi/6+phase constant = +/- pi/2
when we solve for the phase constant we get two answers.
Either pi/3 or -2pi/3

7. To check which of the answers apply to our graph, we can
test wether the answer gives us a correct point on our
graph.
Check pi/3 by using the coordinates (4,-3).
x(4)=3cos((pi/6)(4)+pi/3)=-3
yes, it works!
Using -2pi/3 in the equation would give the amplitude at that
time to be 3, not what is on the graph.

8. Now we know the phase of wave-a at x(1) is(pi/6+pi/3), or
(pi/2).
Compared to wave-b’s phase (pi/4), the phases are (pi/4)
out of phase. (phase of wave-a minus phase of wave-b)
Using the equation for finding amplitude of resulting waves,
we find A = A of component waves x 1.848.
In other words, 5.543m.