In this Learning Object we will look at

how the amplitudes of two fixed positions

on a standing wave change with time.

In this Learning Object we will look at

how the amplitudes of two fixed positions

on a standing wave change with time.

But what are standing waves?

In this Learning Object we will look at

how the amplitudes of two fixed positions

on a standing wave change with time.

But what are standing waves?

A standing wave results when two waves

that have the same frequency and

amplitude travel in opposite directions

along the same medium.

A standing wave is therefore the sum of the two waves travelling in opposite directions, and can be expressed by adding the wave functions of the two component waves.

The equation of the standing wave is thus

D(x,t)=D1(x,t)+D2(x,t)

Using this information, we can track the motion of a standing wave by looking at the motion of its component waves at a fixed position.

Here we see two waves, one blue and one red,

travelling along a string in opposite directions

We will being at t0 when the two waves are directly

on top of one another. The black wave is the sum

of the red and blue waves.

We will pick two points of interest on the

standing (black) wave, labeled POI1and

POI2, and track their motion over a

period of time.

At t1, the two waves have moved slightly

away from each other so that they are

π/2 radians out of phase, and POI1has a

smaller amplitude than it did at t0. This is

due to the interference that results from

the superposition of the two waves. POI2has not changed from 0.

At t2 , the two waves are now π radians

out of phase, and thus completely

cancel one another out.

The resulting black wave is a straight line

with amplitude zero, at both points of

interest.

At t3, the red and blue waves are now

π/2 radians out of phase again, and we

see that at POI1 the resulting wave is now

the inverse of what it was at t1. This is an

important observation that will be

discussed shortly.

Finally, at t4, the waves are once again in

phase and we see that the resulting

black wave has reached its maximal

negative amplitude at our point of

interest. Notice that throughout the

entire time period, POI2has remained

zero.

From our investigation, we can see that the

amplitude of the standing (black) wave at

POI1 varied as a function of time. Its

maximum amplitude remained constant.

Thus, the amplitude of POI1 over time can

be represented by a simple sine function.

However, our POI2 never moved from the zero position. Its amplitude as a function of time was always equal to zero.

We can conclude from our investigation that the amplitude of any point on the standing wave will vary as a sinusoidal function with respect to time.

However, certain points along the wave, such as our POI2, have constant zero amplitude.

We call those points “nodes” and they

occur at every location x=mλ/2, where

m is any integer value.

Our other point, POI1, is an example of

what is called an “antinode”, which is

any point where the standing wave

reaches a maximum amplitude. They

occur at every location x=(m+1/2)(λ/2),

where m is any integer value.

Here is a portion of the original graph of

the standing wave with its nodes and

antinodes labeled:

Just like our two points of interest, every

point on the standing wave (over the

course of one wavelength) has its own

unique maximum amplitude, and it

oscillates from its positive to its negative

maximum amplitude

This is different from travelling waves, in

which every point on the wave oscillates

with the same amplitude

Question 1:

A sinusoidal standing wave has a

wavelength equal to π/2. Where are its

first 4 nodes located?

Answer:

A standing wave has nodes at every point

x=mλ/2. The first node will have the lowest

possible value of m, which is zero. Therefore,

the first node occurs at x=0. Each

consecutive node will have a value of m

one greater than the last. That gives us the

second node at x=(0+1)(π/2)(1/2)= π/4, the

third node at x=(1+1)(π/2)(1/2)= π/2 and

the fourth node at x=(2+1)(π/2)(1/2)= 3π/4

Question 2:

Find the locations of the first 4 antinodes

of the standing wave in the previous

problem.

Answer:

The antinodes of a standing wave are located at every point x=(m+1/2)(λ/2). Therefore, as previously, the first antinode is found at the lowest value of m: x=(0+1/2)(π/2)(1/2)= π/8. The second antinode will be at (0+1+1/2)(π/2)(1/2)=3π/8. The third antinode is found at (1+1+1/2)(π/2)(1/2)= 5π/8 and the fourth antinode at (2+1+1/2)(π/2)(1/2)= 7π/8

Pictures from:

http://www.minelab.com/__files/i/5894/S

ineWave.gif

http://en.wikipedia.org/wiki/Standing_w

ave#mediaviewer/File:Standingwaves.sv

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