1.
Phys 101 Learning Object LO6
March 7 2015
Standing Sound Waves by Vivian
Tsang 14153143

2.
Let’s Learn the Basics
• Standing waves occur when two sinusoidal waves
with the same wavelength, frequency, and amplitude
travelling in opposite directions cross paths
• This can be illustrated by the following to equations
for the waves:
▫ DR=Asin(kx-wt) *w= fundamental frequency
▫ DL=Asin(kx+wt) *A= amplitude
• This leads to the equation:
▫ D(x, t)=(2Asin(2pi/lambda)x)*(cos(wt))
◦ *k= 2pi/lambda

3.
Nodes=zero displacement
• At a node, sin((2pi/lambda)x)=0
• Therefore, nodes occur at every interval of
lambda/2
node nodenode
Lambda/2

4.
Antinodes=max displacement
• At an antinode, sin((2pi/lambda)x)=+/-1
• Therefore, antinodes occur at every interval of
lambda/2
antinode
antinode
Lambda/2

5.
Nodes and Antinodes
• The distance between nodes and antinodes are
lambda/4 antinode
antinode
node
node
node
Lambda/4

6.
For a string fixed at both ends…
• Solving for the
frequency, we can
see that the
fundamental
frequency occurs
when n=1.
• Allowed frequencies
are also called
harmonic sequences
• Therefore, the
fundamental
frequency is also
called the
fundamental
harmonic

7.
Allowed Frequencies
• For a string fixed
at both ends, the
allowed frequencies
include integer
multiples of the
fundamental
harmonic
•As we increase the
integer n=2, we get
the second
harmonic
•As we increase the
integer n=3, we get
the third harmonic

8.
Harmonics
• As you’ve noticed from the equations, the
“nth” harmonic is just an “n” multiple of the
fundamental harmonic
• Therefore:
▫ If the first harmonic is f1
▫ The second harmonic is f2=2f1
▫ The third harmonic is f3=3f1
▫ The fourth harmonic is f4=4f1
▫ The fifth harmonic is f5=5f1
▫ And so on …..

9.
Practice!
QUESTION: A string
vibrates at 400Hz with
the pattern shown here.
What is the frequency of
the fourth harmonic?
?

10.
Answer
• Since we are given
the frequency of
the third
harmonic:
• 1) divide this
frequency f3 by
three to find the
fundamental
harmonic f1
• 2) multiply f1 by
4 to get the fourth
harmonic f4!

11.
Getting Harder! Standing Sound Waves
• Longitudinal standing sound waves can travel in
the air in a tube or pipe
• At a closed end, there is a node because air
molecules cannot move
• At an open end, there is an antinode because the
air molecules are free to move

12.
Basic Ideas…
• Notice! the allowed
frequencies for
pipes open at both
ends is similar to
our previous
calculations for
allowed
frequencies on a
string
• Notice! the allowed
frequencies for
pipes open at one
end and closed at
one end are only
odd integer
multiples

13.
Practice!
• The fundamental frequency for a clarinet with
one end closed and one end opened is 300Hz.
When both ends of the same pipe are opened,
what is the new fundamental frequency?

14.
Answer!
• First, we solve for
the unknown ratio
of v/L for the
fundamental
frequency fp1of the
pipe with a closed
end
• Next, we
substitute this
ratio into the
equation fp2 to find
the fundamental
frequency of the
pipe when the two
ends are open

15.
Harder Problem!
• An alphorn is a tube opened at
one end and closed at the
other.
• This alphorn is 4m in length.
• QUESTION: By what length
would you have to change the
alphorn’s length if you want to
increase its fundamental
frequency by 200Hz?
• What will be the new
fundamental frequency?

16.
Hints
• This is a harder problem so let’s break it down:
• Given:
▫ V=the speed of sound in air which is about 343m/s
▫ L= 4m
▫ Since the pipe is open at one end and closed at the other, we
would use the formula f=v/4L
▫ Now, you can solve for f1, the fundamental frequency
• Next:
▫ Use the change of frequency given (200Hz) to solve for the
change in length L=v/4 f
• Finally:
▫ Add up the original fundamental frequency and the change
in frequency to find the new fundamental frequency

17.
Solution!

18.
Thanks for watching!