Standing waves occur when two waves of equal amplitude, wavelength, and frequency travel in opposite directions and superimpose. The result is a wave with a position-dependent amplitude described by A(x)=2Asin(kx), where k is the wavenumber. Nodes occur at integer multiples of half wavelengths, where the amplitude is zero. Anti-nodes occur at odd integer multiples of quarter wavelengths, where the amplitude is maximum (2A). The full equation for a standing wave is D(x,t)=2Asin(kx)cos(ωt), relating amplitude to position and time using sines and cosines respectively.

1. Standing Waves
By Rick Grootes

2.  We will be viewing all of these problems and
equations with a phase constant (ϕ0) of zero for
both waves.
 Remember, this could not be the case for many
examples, as not all waves are as perfect as we
would hope for them to be.
KEY POINT

3.  They are two harmonic waves of equal amplitude, wavelength, and
frequency moving in the opposite direction of one another
 Mathematically, we see these two waves as:
D1(x,t)=Asin(kx-ωt)
and:
D2(x,t)=Asin(kx+ωt)
where:
D(x,t)=D1(x,t)+D2(x,t)
=A[sin(kx-ωt)+sin(kx+ωt)]
 The sum of these two waves gives rise to a simple trigonometric identity
similar to: sin(a-b)+sin(a+b)=2sin(a)cos(b)
 In this case: D(x,t)=2Asin(kx)cos(ωt)
What are standing waves?

4.  Using the formula we found in the previous slide we can
define a position-dependent amplitude A(x):
 A(x)=2Asin(kx)=2Asin(2π*x/λ)
we already know from our formula sheet that:
k=(2π/λ)
 Using the above definition, we can change around our
formula to:
 D(x,t)=A(x)cos(ωt)
Taking it apart!
D(x,t)=2Asin(kx)cos(ωt)

5.  Please find the amplitude and wavelength of the
following examples:
1) A(x)=(80.0cm)sin(8.00x)
2) A(x)=(35.0cm)sin(1.57079x)
3) A(x)=(5624.0cm)sin(0.0981748x)
Please find answers on next slide, but don’t skip!
Lets do a little tester!

6. 1) 2A=80.0cm A = 40.0cm
2π/λ=8.00 λ = π/4 = 0.7854 m
2) 2A=35.0cm A = 17.5cm
2π/λ=1.57079 λ = 4.00 m
3)2A=5624.0cm A = 2812.0 cm
2π/λ=0.0981748 λ = 64.0 m
Answers to Slide 5

7.  These waves are different because the x (position)
is part of the sine function, whereas the t (time) is
part of the cosine function.
 They have nodes and anti-nodes:
 Nodes: certain points on the string that have zero
amplitude and remain at rest at all times
 Anti-nodes: points that move with the maximum
possible amplitude of 2A.
How are these waves different?

8. At the nodes of a standing wave, A(x)=0.
Therefore, they occur when:
sin(2π*x/λ)=0
(2π*x/λ)=mπ where m=±1,±2,…
x=m(λ/2) where m=±1,±2,…
x=0,±(λ/2),±(λ),±(3λ/2),±(2λ),…
How do we find the location of
nodes and anti-nodes:

9.  We already know that anti-nodes occur when
A(x)=±2A. This happens when:
sin(2π*x/λ)=±1
(2π*x/λ)=(m+½)*π where m=0,±1,±2,…
x=(m+½)*(λ/2) where m=0,±1,±2,…
x=±(λ/4), ±(3λ/4), ±(5λ/4),…
Now for Anti-nodes:

10.  When we think about it, two consecutive anti-nodes
will always be half a wavelength apart.
 Similarly, an adjacent node and anti-node will
always be a quarter wavelength apart.
For simplicity

11.  We all know from our trusty formula sheet that:
 Period(T) and frequency(f)
where T=(1/f)
and that angular frequency (ω):
ω=(2π/T)=(2π*f)
Therefore, in full, our equation looks something like:
D(x,t)=2Asin(2π*x/λ)cos(2π*t/T)
We can’t forget about cosine

12.  John and Cindy took a trip to Scotland and wanted
to see some of the greatest cliffs in the world. As
they peered over the edge, they saw huge waves
crashing against the cliff and bouncing off to give
even larger waves to give what seemed to look like
standing waves. With their knowledge of physics,
they created an equation for these huge waves and
thankfully, recorded it all on their handy-dandy
water-proof paper! Go to the next slide to see all
their findings.
Lets take a trip to the rocky coast of
Scotland(Info Part 1)

13. Part 1)
They found that the amplitude of the waves before they
formed a standing wave were 20.0m high a piece.
a) What was the amplitude of the standing wave?
They also found that once the waves merged, they had
nodes at the cliffs (x=0), 15.0m away from the cliffs and
30.0m away from the cliffs.
b) What was the wavelength of the standing
wave?
Collected Data Part 1

14. Part 2)
Then, they wanted to rearrange the data they
collected into the equation:
A(x)=2Asin(kx)
Data Collected Part 2

15.  While they took note of the amplitude and enjoyed the
sights, they also pulled out their stop watches and
calculated how many anti-nodes they saw in 8 seconds.
In those 8 seconds, the saw exactly 24 anti-nodes.
a) what is the frequency of the waves?
b) what is the period of the waves?
c)use all of the information gathered to
formulate an answer to the equation:
D(x,t)=2Asin(kx)cos(ωt)
(Info for Part 3)

16. See next page for solutions to
Part1, Part 2 and Part 3!
THE END