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WAVES -4

Waves - 4 By Aditya Abeysinghe

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Waves - 4 By Aditya Abeysinghe 2

STATIONARY WAVES AND WAVE LENGTH

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:

Node

Antinode

NodeNodeNode

Antinode

Antinode

Antinode

Original wave

Reflected wave

Original wavedirection

Node


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 /24. Distance between consecutive antinodes =

λ /25. Distance from a node to the next antinode =

λ /4,λ being the wave length.


WAVES IN STRINGSA 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.


FINDING THE FREQUENCY OF VIBRATION

Consider the diagram below.

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.

l

Node Node

Antinode

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FINDING THE OVERTONESWhen 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:


Node NodeNode

Node

Antinode Antinode Antinode


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.

Node Node Node Node Node Node

Antinode Antinode Antinode Antinode Antinode


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.


WAVES IN PIPESWaves 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.

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

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.


l

NodeAntinode


For the first overtone, l = 3λ1/4 . Therefore, λ1 = 4l/3. So, f1 = v/ λ1

Thus, f1 = v/ (4l/3 ) = 4V /3lTherefore, f1 = 3 f0

1st Overtone

l

l

2nd Overtone

For the second overtone,l = 5λ2/4. Therefore, λ1 = 4l/5. So, f2 = v/ λ2

Thus, f2 = v/ (4l/5 ) = 5V/4l = 5f0


OPEN PIPEIn an open pipe, both ends experience antinodes.

So, an open pipe vibrating in its fundamental frequency will be as follows:

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:

l

NodeAntinode Antinode

Waves - 4 By Aditya Abeysinghe13

For the first overtone, l = λ1 . So, f1 = v/ λ1

Thus, f1 = v/ lTherefore, f1 = 2 f0

1st Overtone

l

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


END – CORRECTION OF PIPES

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.

l

Node

Antinode

Antinode

lAntinode

Node

ee

e


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.

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