Waves 4

Waves - 4 By Aditya Abeysinghe

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1

A stationary wave is formed when two waves of equal amplitude
and frequency travel in opposite directions in a medium.
The stationary wave is due to the interference of the two waves.
Consider the diagram below:
Anti
node

Node

Original
wave

Node

Anti
node

Anti
node

Anti
node

Node

Node

Reflected
wave

Node

Original
wave
direction

2

The following characteristics are important to
understand the wave.
1. At an antinode, the two waves have constructive
interference.
2. At a node, the two waves have destructive
interference.
3. Distance between consecutive nodes = λ /2
4. Distance between consecutive antinodes = λ /2
5. Distance from a node to the next antinode = λ /4,
λ being the wave length.


3

A transverse wave is formed when one end of a string is fixed and
other end is moved up and down.
See figure below.

Speed of this vibration is given by,
V = √T/ m , where T is the tension and m is the mass
per unit length.

4

Consider the diagram below.

Antinode
Node

Node
l

If the wave length of the wire is λ , then l = λ /2.
So, λ = 2l . Also, V = fλ . Thus, V = f . 2l or f = V/2l.
However, we know that v = V = √T/ m .
So, f = (1/2l) (√T/ m)
This is the fundamental or lowest note obtained from the string .
Thus, this frequency is commonly denoted by the symbol f0.

5

When the tension and the other quantities that affect
the frequency are arranged, so that multiple
frequencies can be obtained for a string, we get
overtones.
For example, for a string which was plucked in the
middle, the first overtone would be as follows:
Antinode

Node

Antinode

Node

Antinode

Node


Node

6

Since the string was plucked in the middle, you’ll
observe an antinode at that point. However, to
remain symmetric, there will be three completely
closed loops should be formed.
Similarly, when the string is vibrating in the second
overtone, to remain symmetric, five loops should be
formed.
Antinode

Node

Antinode

Node

Antinode

Node

Antinode

Node


Antinode

Node

Node

7

Let’s calculate the individual frequencies for the first
and second overtones. (Length of the string = l)
For the first overtone,
l = 3λ1/2 . Therefore, λ1 = 2l/3 . So, f1 = v/ λ1
Thus, f1 = v/ (2l/3 ) or f1 = (3/2l) (√T/ m)
For the second overtone,
l = 5λ2/2. Therefore, λ1 = 2l/5 . So, f2 = v/ λ2
Thus, f2 = v/ (2l/5 ) or f2 = (5/2l) (√T/ m)
Note that f1 = 3 f0 and f2 = 5 f0 .
Thus, overtones are integer multiples of the
fundamental frequency for a particular wave.


8

Waves in pipes are similar to waves in
strings. The only difference is that pipes are
classified, unlike strings, as closed pipe and
open pipe.
The wave formation at the closed end of a
closed pipe should be a node and at the
opened end it should be an antinode.
In contrast, in an open pipe, both ends are
opened, so antinodes are formed at both
ends.

9

When a closed pipe is vibrating in its fundamental frequency, a wave
pattern as shown below is seen.

Antinode

Node

l

If the length of the pipe is l, then λ1 / 4 = l or λ1 = 4l.
Thus, the fundamental frequency, f0 = V / 4l .
As in the waves of strings, overtones can be obtained for a closed
pipe.However, the basic theory remains the same (A node at the closed
end and an antinode at the open end), while the no. of loops changes.

10

For the first overtone,
l = 3λ1/4 .
Therefore, λ1 = 4l/3.
So, f1 = v/ λ1
Thus, f1 = v/ (4l/3 )
= 4V /3l
Therefore, f1 = 3 f0

l

1st Overtone

For the second
overtone,
l = 5λ2/4.
Therefore,
λ1 = 4l/5.
So, f2 = v/ λ2
Thus, f2 = v/ (4l/5 )
= 5V/4l = 5f0

l

2nd Overtone

11

In an open pipe, both ends experience antinodes.
So, an open pipe vibrating in its fundamental frequency will be as
follows:

Antinode

Node

Antinode

l

If the length of the pipe is l, then λ1 / 2 = l or λ1 = 2l.
Thus, the fundamental frequency, f0 = V / 2l .
Overtones for such an open pipe are as follows:

12

For the first
overtone, l = λ1 .
So, f1 = v/ λ1
Thus, f1 = v/ l
Therefore, f1 = 2 f0

l

1st Overtone

l

2nd Overtone

For the second overtone,
l = 3λ2/2. Therefore, λ1 = 2l/3 . So, f2 =
v/ λ2
Thus, f2 = v/ (2l/3 ) = 3V/2l = 3f0
13

The air at the open end of a pipe is free to move. So, the
vibrations extend a little into the air outside the pipe as shown
below.
Antinode

Node

l
l

Node

e
Antinode

e

Antinode

e
14

As illustrated in the diagram it should be noted that if the end
correction is not negligible,
(1) For a closed pipe the new wavelength would be
λ/4 = l + e, so, λ = 4 (l + e).
(2) For an open pipe the new wavelength would be
λ/2 = l + e + e , since two end-corrections are required . So, λ = 2
(l + 2e).
Experimental data show that e is approximately 0.3d , d being the
diameter of a pipe, regardless whether it is closed end or open
end. So, greater the diameter, greater will be the end correction.


15

Waves 4

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