Chapter 8:Waves Transferring Energy
8.3Interference of Waves
Superposition of Waves
We learned from the demonstration in class that waves traveling toward each other DO pass through each other.The two (or more) waves traveling toward each other are called component waves. During the brief period of time that they are superimposed, we observe a resultant wave. To determine the displacement at any point in resultant wave, we simply add the displacements of the component waves, taking into consideration whether or not their displacements are positive (+) or negative (-) relative to the resting position.
When the resultant wave has a greater displacement than one (or in some cases, all) of its component waves, we have constructive interference.
Example:
+3 +3
BEFORE
+6
DURING
(Constructive interference)
+3 +3
AFTER
When the resultant wave has a smaller displacement than one (or in some cases, all) of its component waves, we have destructive interference.
Example:
+3
BEFORE
-3
0
DURING
(Destructive Interference)
+3
AFTER
-3
Standing Waves
When periodic waves with the same shape, amplitude and wavelength travel in opposite directions in a linear medium such as a rope or spring, they produce a distinct pattern in the medium that appears to be standing still. For this reason, these are called standing waves.
Points of maximum destructive interference are termed nodes (N). Points of maximum constructive interference are termed antinodes (AN). Figure 8.17 p.357 shows the evolution of a standing wave.
Example:
AN AN AN AN AN AN AN AN
N N N N N N N N N
NOTES:1. Standing waves do not have to be produced by two separate vibrating sources. Because a reflected wave possesses essentially the same characteristics as the original wave, standing waves can be produced by only one vibrating source and its’ reflected wave.
2. The distance between two consecutive nodes (or antinodes) is ½ while the distance between a consecutive node – antinode is ¼
Application of Standing Waves: Natural Frequencies and Overtones in Instruments
For every medium of fixed length (L), there are many natural frequenciesof a vibrating wave that produce resonance. The lowest natural frequency (corresponding the longest wavelength) that will produce resonance is called the fundamental mode of vibration and corresponds to a N – AN – N arrangement in the standing wave. The frequency at which this occurs is called the fundamental frequency. The nodes represent two fixed ends, for example in a guitar.
Example: L
NAN N
NOTE:L = ½ in this case.
Sample problem#1 (not in text):
What is the fundamental frequency of a string having a length a 60.0 cm if the speed of the wave is measured to be 10.0 m/s?
Given:L = 60.0 cm; v = 10.0 m/s
Find: f
Step 1: Note that the fundamental frequency corresponds to the case where L = ½
Thus, 2 L = 2 (60.0 cm) = 120. cm = 1.20 m.
Step 2: Rearrange the universal wave equation to solve for f.
v = ff = v = 10.0 m/s = 8.33 Hz
1.20 m
Overtones
All natural frequencies higher than the fundamental frequency are called overtones. For the example earlier:
L
NAN NAN N1st Overtone
(L =
N AN N AN N AN N2nd Overtone
(L = 1.5
3rd Overtone
(L = 2
Sample Problem #2 (not in text):
What are the 1st three overtones of the fundamental frequency in Sample Problem #1:
Given:L = 60.0 cm; v = 10.0 m/s
Find: f
Step 1: Note the relationship between L and for the first 3 overtones.
1st Overtone: L = = 60.0 cm = 0.600 m
2nd Overtone: L = 1.5 = L / 1.5 = 60.0 cm / 1.5 = 40.0 cm = 0.400 m
3rd Overtone L = 2 = L / 2 = 60.0 cm / 2 = 30.0 cm = 0.300 m
Step 2: Rearrange the universal wave equation to solve for f.
1st Overtonef = v = 10.0 m/s = 16.7 Hz = 2 x fundamental frequency
0.600 m
2nd Overtonef = v = 10.0 m/s = 25.0 Hz = 3 x fundamental frequency
0.400 m
3rd Overtonef = v = 10.0 m/s = 33.3 Hz = 4 x fundamental frequency
0.300 m
Assignment:
Section Review p. 362 Q. 1-7 + Q. 8 below:
8. A 20.0 cm long spring is fixed at both ends. If the speed of a wave in the spring is 25.0 m/s, calculate (a) the fundamental frequency; (b) the next 3 overtones of the fundamental frequency.