Chapter 14 Lecture Notes

Chapter 14 The Principle of Superposition Often more than one wave is present in any region in space. To analyze how multiple waves combine, we need to apply the principle of superposition.

If two or more traveling waves are moving through a medium and combine at a given point, the resultant position of the element of the medium at that point is the sum of the positions due to the original waves.

If we look at two interacting pulses, we have:

As the two pulses approach each other at the same speed they will eventually combine to form a pulse of amplitude y1 + y2. The pulses will then pass through each other and return, undiminished, to their original form. The combination of two or more waves is called interference. Interference will only occur when the interacting waves are in the same region of space. Wave Interference

We now need to obtain a mathematical model for two waves interfering with each other. Imagine that we have two identical transverse waves that are out of phase. The wave functions for these two waves are: Physics 210 Santiago Canyon College

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Chapter 14 Lecture Notes

= y1 Asin ( kx − ωt )

= and y 2 Asin ( kx − ωt + φ)

When these waves interfere, the principle of superposition tells us that the amplitude of the resultant wave is the sum of the original waves.

= y A sin ( kx − ωt ) + sin ( kx − ωt + φ)

To simplify this expression, we apply the following trigonometric identity: a−b a+b sina + sinb = 2cos   sin    2   2 

So our resultant wave is:

φ  φ  y 2Acos   sin  kx − ωt +  = 2 2 

(1)

If φ= 0, 2π , 4 π , 6π ,  then the resultant wave will have its maximum possible amplitude. This is because the two interfering waves are in phase and we will have constructive interference.

If φ = π , 3π , 5π ,  then the cosine term will be zero. This means that the two interfering waves have cancelled each other out in a process called destructive interference.

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Typically the phase difference is due to the fact that while the interfering waves were produced by identical sources, they may not travel identical distances to some observer. The difference in their path lengths can be related to the phase angle by:

∆r=

φ λ 2π

(2)

Example 1: Two waves traveling on a string in the same direction both have a frequency of 100 Hz, a wavelength of 2 cm, and amplitude of 0.02 m. What is the amplitude of the resultant wave if the original waves differ in phase by (a) π π and (b) ? 6 3

Solution:

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Example 2: Two loudspeakers are placed on a wall 3.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 350 Hz. What is the phase difference between the two waves when they reach the observer? Solution:

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Chapter 14 Lecture Notes

Standing Waves We have previously looked at two waves interfering with each other where the waves were both moving in the same direction. Now we want to consider what happens when two identical waves head towards each other. Consider the following two wave functions: = y1 Asin ( kx − ωt )

= and y 2 Asin ( kx + ωt )

When these two waves combine, we have:

= y 2Asin ( kx ) cos ( ωt )

The equation above is not an equation for a traveling wave (it actually is very similar to a harmonic oscillator). This type of equation is for a standing wave.

Every element of the medium vibrates with the same frequency, but the amplitude of oscillation depends on the position is space. The points of maximum amplitude are called antinodes and the points of minimum amplitude are called nodes. Looking at our wave function, the nodes will occur when: sin ( kx ) = 0 ⇒ kx = nπ

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Chapter 14 Lecture Notes

In terms of wavelength, we have:

x=

nλ 2

Similarly, the antinodes are located when:

sin ( kx ) = ±1 ⇒ kx =

In terms of wavelength we have: Standing Waves in Strings

= x

(3)

nπ n = odd 2

nλ = n odd 4

One way to establish two identical waves is to combine an incoming and a reflected wave.

It turns out that a string has several normal modes. When the wave reflects off the wall certain continuity conditions (called boundary conditions) must be satisfied. This means that only certain frequencies (quantized frequencies) are possible. These frequencies look like: Physics 210 Santiago Canyon College

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Chapter 14 Lecture Notes

Mathematically, these frequencies are given by: = fn

nv n T = 2L 2L µ

(4)

When n = 1, we have the fundamental frequency. For n > 1, we have the different harmonic frequencies possible in the string.

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Example 3: The G string on a violin has a fundamental frequency of 196 Hz. The length of the sting is 32 cm and it has a mass of 0.68 grams. Under what tension must the string be placed? Solution:

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Example 4: An unfingered guitar string is 0.73 m long and is tuned to play E above middle C (330 Hz). How far from the end of the string must the finger be placed to play A above middle C (440 Hz)? Solution:

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Standing Waves in Tubes We can produce standing waves in a tube similar to those produced in the string. If the pipe is open at both ends (typically called an open pipe), then the harmonic frequencies will be: fn =

nv 2L

(5)

The difference between pipes and strings is that we can also have a pipe which is closed at one end (usually called a closed pipe). With this type of pipe, there is a node located at the closed end. This means that the frequencies produced in this type of pipe will be different from those in an open pipe. The frequencies for a closed pipe are: fn =

nv 4L

n= 1, 3, 5, 

(6)

The waves produced in these pipes look like:

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Chapter 14 Lecture Notes

Example 5: What resonant frequency would you expect from blowing across the top of an empty soda bottle that is 15 cm deep? How would that change if it was one-third full of soda?

Solution:

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Chapter 14 Lecture Notes

Example 6: Determine the length of an open organ pipe that emits middle C (262 Hz) when the temperature is 21oC.

Solution:

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Example 7: The human ear canal is approximately 2.5 cm long. It is open to the outside and closed at the other end by the tympanic membrane. If the air is at 20oC, what are the first three harmonic frequencies for the ear canal? Solution:

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Chapter 14 Lecture Notes

When two sound waves with slightly different frequencies interfere, they produce beats. The frequency of this fluctuation is: fB= f1 − f2

This is the beat frequency. For example, if a 2523 Hz sound wave interacts with a 2528 Hz wave then the beat frequency is: fB = f1 − f2 = 2523 Hz − 2528 Hz = 5 Hz

Beats are commonly used to tune instruments (stringed) like pianos and violins. You create the note that is desired with a tuning fork and play the same note with the instrument. You then adjust the tension in the string until the beat frequency goes to zero. When this happens, the note from the tuning fork and that from the instrument are identical. Example 7: A piano tuner hears one beat every 2.0 s when trying to adjust two strings, one of which is sounding 440 Hz. How far off in frequency is the other string? Solution:

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Example 8: Two violin strings are tuned to the same frequency 294 Hz. The tension in one string is then decreased by 2.0 percent. What beat frequency will be heard when the two strings are played together? Solution:

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