Lecture Presentation Chapter 16 Superposition and Standing Waves

Lecture Presentation Chapter 16 Superposition and Standing Waves 2015 Pearson Education, Inc. Section 16.1 The Principle of Superposition 2015 Pearson Education, Inc.

The Principle of Superposition If two baseballs are thrown across the same point at the same time, the balls will hit one another and be deflected. 2015 Pearson Education, Inc. Slide 16-22 The Principle of Superposition Waves, however, can pass through one another. Both observers would hear undistorted sound, despite

the sound waves crossing. 2015 Pearson Education, Inc. Slide 16-23 The Principle of Superposition 2015 Pearson Education, Inc. Slide 16-24 The Principle of Superposition

To use the principle of superposition, you must know the displacement that each wave would cause if it were alone in the medium. Then you must go through the medium point by point and add the displacements due to each wave at that point. 2015 Pearson Education, Inc. Slide 16-25 Constructive and Destructive Interference The superposition of two waves is called interference. Constructive interference occurs when both waves are

positive and the total displacement of the medium is larger than it would be for either wave separately. 2015 Pearson Education, Inc. Slide 16-26 Constructive and Destructive Interference The superposition of two waves is called interference. Constructive interference occurs when both waves are positive and the total displacement of the medium is larger than it would be for either wave separately.

2015 Pearson Education, Inc. Slide 16-27 Constructive and Destructive Interference Destructive interference is when the displacement of the medium where the waves overlap is less than it would be due to either of the waves separately. During destructive interference, the energy of the wave is in the form of kinetic energy of the medium. 2015 Pearson Education, Inc.

Slide 16-28 Constructive and Destructive Interference Destructive interference is when the displacement of the medium where the waves overlap is less than it would be due to either of the waves separately. During destructive interference, the energy of the wave is in the form of kinetic energy of the medium. 2015 Pearson Education, Inc. Slide 16-29

Constructive and Destructive Interference Destructive interference is when the displacement of the medium where the waves overlap is less than it would be due to either of the waves separately. During destructive interference, the energy of the wave is in the form of kinetic energy of the medium. 2015 Pearson Education, Inc. Slide 16-30 QuickCheck 16.1 Two wave pulses on a string

approach each other at speeds of 1 m/s. How does the string look at t = 3 s? 2015 Pearson Education, Inc. Slide 16-31 QuickCheck 16.1 Two wave pulses on a string approach each other at speeds of 1 m/s. How does the string look at t = 3 s?

C. 2015 Pearson Education, Inc. Slide 16-32 QuickCheck 16.2 Two wave pulses on a string approach each other at speeds of 1 m/s. How does the string look at t = 3 s? 2015 Pearson Education, Inc.

Slide 16-33 QuickCheck 16.2 Two wave pulses on a string approach each other at speeds of 1 m/s. How does the string look at t = 3 s? B. 2015 Pearson Education, Inc.

Slide 16-34 QuickCheck 16.3 Two waves on a string are moving toward each other. A picture at t = 0 s appears as follows: How does the string appear at t = 2 s? 2015 Pearson Education, Inc. Slide 16-35 QuickCheck 16.3

Two waves on a string are moving toward each other. A picture at t = 0 s appears as follows: How does the string appear at t = 2 s? A. 2015 Pearson Education, Inc. Slide 16-36 Section 16.2 Standing Waves 2015 Pearson Education, Inc.

Standing Waves Waves that are trapped and cannot travel in either direction are called standing waves. Individual points on a string oscillate up and down, but the wave itself does not travel. It is called a standing wave because the crests and troughs stand in place as it oscillates. 2015 Pearson Education, Inc. Slide 16-38

Superposition Creates a Standing Wave As two sinusoidal waves of equal wavelength and amplitude travel in opposite directions along a string, superposition will occur when the waves interact. 2015 Pearson Education, Inc. Slide 16-39 Superposition Creates a Standing Wave 2015 Pearson Education, Inc.

Slide 16-40 Superposition Creates a Standing Wave 2015 Pearson Education, Inc. Slide 16-41 Superposition Creates a Standing Wave The two waves are represented by red and by orange in the previous figures. At each point, the net displacement of the medium is found by adding the red displacement and the orange displacement. The blue wave is the resulting

wave due to superposition. 2015 Pearson Education, Inc. Slide 16-42 Nodes and Antinodes In a standing wave pattern, there are some points that never move. These points are called nodes and are spaced /2 apart. Antinodes are halfway

between the nodes, where the particles in the medium oscillate with maximum displacement. 2015 Pearson Education, Inc. Slide 16-43 Nodes and Antinodes The wavelength of a standing wave is twice the distance between successive nodes or antinodes.

At the nodes, the displacement of the two waves cancel one another by destructive interference. The particles in the medium at a node have no motion. 2015 Pearson Education, Inc. Slide 16-44 Nodes and Antinodes At the antinodes, the two

waves have equal magnitude and the same sign, so constructive interference at these points give a displacement twice that of the individual waves. The intensity is maximum at points of constructive interference and zero at points of destructive interference. 2015 Pearson Education, Inc.

Slide 16-45 QuickCheck 16.4 What is the wavelength of this standing wave? A. B. C. D. E. 0.25 m 0.5 m 1.0 m

2.0 m Standing waves dont have a wavelength. 2015 Pearson Education, Inc. Slide 16-46 QuickCheck 16.4 What is the wavelength of this standing wave? A. B. C. D.

E. 0.25 m 0.5 m 1.0 m 2.0 m Standing waves dont have a wavelength. 2015 Pearson Education, Inc. Slide 16-47 Section 16.3 Standing Waves on a String

2015 Pearson Education, Inc. Reflections A wave pulse traveling along a string attached to a wall will be reflected when it reaches the wall, or the boundary. All of the waves energy is reflected; hence the amplitude of a wave reflected from a boundary

is unchanged. The amplitude does not change, but the pulse is inverted. 2015 Pearson Education, Inc. Slide 16-49 Reflections Waves also reflect from a discontinuity, a point where there is a change in the properties of the medium. At a discontinuity, some of the waves energy is transmitted forward and some is reflected.

2015 Pearson Education, Inc. Slide 16-50 Reflections When the string on the right is more massive, it acts like a boundary so the reflected pulse is inverted. 2015 Pearson Education, Inc. Slide 16-51 Try It Yourself: Through the Glass Darkly

A piece of window glass is a discontinuity to a light wave, so it both transmits and reflects light. To verify this, look at the windows in a brightly lit room at night. The small percentage of the interior light that reflects from windows is more intense than the light coming in from outside, so reflection dominates and the windows show a mirror-like reflection of the room. Now turn out the lights. With no more reflected interior light you will be able to see the transmitted light from outside. 2015 Pearson Education, Inc.

Slide 16-52 Creating a Standing Wave Standing waves can be created by a string with two boundaries where reflections occur. A disturbance in the middle of the string causes waves to travel outward in both directions.

[Insert Figure 16.11] The reflections at the ends of the string cause two waves of equal amplitude and wavelength to travel in opposite directions along the string. 2015 Pearson Education, Inc. Slide 16-53 Creating a Standing Wave Two conditions must be met in order to create

standing waves on the string: [Insert Figure 16.11 (repeated).] Because the string is fixed at the ends, the displacements at x = 0 and x = L must be zero at all times. Stated another way, we require nodes at both ends of the string.

We know that standing waves have a spacing of /2 between nodes. This means that the nodes must be equally spaced. 2015 Pearson Education, Inc. Slide 16-54 Creating a Standing Wave There are three possible standing-wave modes of a string. The mode number m helps quantify the number of possible waves in a standing

wave. A mode number m 1 indicates only one wave, m 2 indicates 2 waves, etc. 2015 Pearson Education, Inc. Slide 16-55 Creating a Standing Wave Different modes have different wavelengths. For any mode m the wavelength is given by the equation

A standing wave can exist on the string only if its wavelength is one of the values given by this equation. 2015 Pearson Education, Inc. Slide 16-56 Creating a Standing Wave The oscillation frequency corresponding to wavelength m is

The mode number m is equal to the number of antinodes of the standing wave. 2015 Pearson Education, Inc. Slide 16-57 Creating a Standing Wave The standing-wave modes are frequencies at which the wave wants to oscillate. They can be called resonant modes or resonances. 2015 Pearson Education, Inc.

Slide 16-58 QuickCheck 16.5 What is the mode number of this standing wave? A. B. C. D. 4 5 6

Cant say without knowing what kind of wave it is 2015 Pearson Education, Inc. Slide 16-59 QuickCheck 16.5 What is the mode number of this standing wave? A. B. C. D.

4 5 6 Cant say without knowing what kind of wave it is 2015 Pearson Education, Inc. Slide 16-60 The Fundamental and Higher Harmonics The first mode of the standing-wave modes has the frequency

This frequency is the fundamental frequency of the string. 2015 Pearson Education, Inc. Slide 16-61 The Fundamental and Higher Harmonics The frequency in terms of the fundamental frequency is fm = mf1 m = 1, 2, 3, 4, . . .

The allowed standing-wave frequencies are all integer multiples of the fundamental frequency. The sequence of possible frequencies is called a set of harmonics. Frequencies above the fundamental frequency are referred to as higher harmonics. 2015 Pearson Education, Inc. Slide 16-62 Example 16.2 Identifying harmonics on a string A 2.50-m-long string vibrates

as a 100 Hz standing wave with nodes at 1.00 m and 1.50 m from one end of the string and at no points in between these two. Which harmonic is this? What is the strings fundamental frequency? And what is the speed of the traveling waves on the string? 2015 Pearson Education, Inc.

Slide 16-63 Example 16.2 Identifying harmonics on a string (cont.) PREPARE We begin with the visual overview in FIGURE 16.15, in which we sketch this particular standing wave and note the known and unknown quantities. We set up an x-axis with one end of the string at x 0 m and the other end at x 2.50 m. The ends of the string are nodes,

and there are nodes at 1.00 m and 1.50 m as well, with no nodes in between. 2015 Pearson Education, Inc. Slide 16-64 Example 16.2 Identifying harmonics on a string (cont.) We know that standing-wave nodes are equally spaced, so there must be other nodes on the string, as shown in Figure 16.15a. Figure 16.15b is a

sketch of the standing-wave mode with this node structure. 2015 Pearson Education, Inc. Slide 16-65 Example 16.2 Identifying harmonics on a string (cont.) SOLVE We count the number of antinodes of the standing wave to deduce the

mode number; this is mode m = 5. This is the fifth harmonic. The frequencies of the harmonics are given by fm = mf1, so the fundamental frequency is 2015 Pearson Education, Inc. Slide 16-66 Example 16.2 Identifying harmonics on a string (cont.)

The wavelength of the fundamental mode is 1 = 2L = 2(2.50 m) = 5.00 m, so we can find the wave speed using the fundamental relationship for sinusoidal waves: v = 1 f 1 = (20 Hz) (5.00 m) = 100 m/s 2015 Pearson Education, Inc. Slide 16-67

Example 16.2 Identifying harmonics on a string (cont.) ASSESS We can calculate the speed of the wave using any possible mode, which gives us a way to check our work. The distance between successive nodes is /2. Figure 16.15 shows that the nodes are spaced by 0.50 m, so the wavelength of the m = 5 mode is 1.00 m. The frequency of this mode is 100 Hz, so we calculate v = 5 f5 = (100 Hz) (1.00 m) = 100 m/s

This is the same speed that we calculated earlier, which gives us confidence in our results. 2015 Pearson Education, Inc. Slide 16-68 Example Problem A particular species of spider spins a web with silk threads of density 1300 kg/m3 and diameter 3.0 mm. A passing insect brushes a 12-cm-long strand of the web, which has a tension of 1.0 mN, and excites the lowest frequency standing wave. With what frequency will the strand vibrate?

2015 Pearson Education, Inc. Slide 16-69 Stringed Musical Instruments The fundamental frequency can be written in terms of the tension in the string and the linear density: When you pluck a bow or string of an instrument, initially you excite a wide range of frequencies; however the resonance sees to it that the only frequencies to persist are those of the possible standing waves.

On many instruments, the length and tension of the strings are nearly the same; the strings have different frequencies because they differ in linear density. 2015 Pearson Education, Inc. Slide 16-70 QuickCheck 16.6 A standing wave on a string vibrates as shown. Suppose the string tension is reduced to 1/4 its original value while the frequency and length are kept unchanged. Which standing wave pattern is produced?

2015 Pearson Education, Inc. Slide 16-71 QuickCheck 16.6 A standing wave on a string vibrates as shown. Suppose the string tension is reduced to 1/4 its original value while the frequency and length are kept unchanged. Which standing wave pattern is produced? C. The frequency is f m m v .

2L Quartering the tension reduces v by one half. Thus m must double to keep the frequency constant. 2015 Pearson Education, Inc. Slide 16-72 QuickCheck 16.7 Which of the following changes will increase the frequency of the lowest-frequency standing sound wave on a stretched string? Choose all that apply. A. B.

C. D. Replacing the string with a thicker string Increasing the tension in the string Plucking the string harder Doubling the length of the string 2015 Pearson Education, Inc. Slide 16-73 QuickCheck 16.7

Which of the following changes will increase the frequency of the lowest-frequency standing sound wave on a stretched string? Choose all that apply. A. B. C. D. Replacing the string with a thicker string Increasing the tension in the string Plucking the string harder Doubling the length of the string

2015 Pearson Education, Inc. Slide 16-74 Example 16.4 Setting the tension in a guitar string The fifth string on a guitar plays the musical note A, at a frequency of 110 Hz. On a typical guitar, this string is stretched between two fixed points 0.640 m apart, and this length of string has a mass of 2.86 g. What is the tension in the string? PREPARE Strings sound at their fundamental frequency, so 110 Hz is f1.

2015 Pearson Education, Inc. Slide 16-75 Example 16.4 Setting the tension in a guitar string (cont.) SOLVE The linear density of the string is We can rearrange Equation 16.5 for the fundamental frequency to solve for the tension in terms of the other variables:

2015 Pearson Education, Inc. Slide 16-76 Example 16.4 Setting the tension in a guitar string (cont.) ASSESS If you have ever strummed a guitar, you know that the tension is quite large, so this result seems reasonable. If each of the guitars six strings has approximately the same tension, the total force on the neck of the guitar is a bit more than 500 N. 2015 Pearson Education, Inc.

Slide 16-77 Example Problem Two strings with linear densities of 5.0 g/m are stretched over pulleys, adjusted to have vibrating lengths of 50 cm, and attached to hanging blocks. The block attached to String 1 has a mass of 20 kg and the block attached to String 2 has mass M. When driven at the same frequency, the two strings support the standing waves shown. A. What is the driving frequency? B. What is the mass of the block suspended from String 2?

2015 Pearson Education, Inc. Slide 16-78 Standing Electromagnetic Waves A laser establishes standing light waves between two parallel mirrors that reflect light back and forth. The mirrors are the boundaries and therefore the light wave must have a node at the surface of each mirror. 2015 Pearson Education, Inc.

Slide 16-79 Example 16.5 Finding the mode number for a laser A helium-neon laser emits light of wavelength = 633 nm. A typical cavity for such a laser is 15.0 cm long. What is the mode number of the standing wave in this cavity? PREPARE Because a light wave is a transverse wave, Equation 16.1 for m applies to a laser as well as a vibrating string. 2015 Pearson Education, Inc.

Slide 16-80 Example 16.5 Finding the mode number for a laser (cont.) SOLVE The standing light wave in a laser cavity has a mode number m that is roughly ASSESS The wavelength of light is very short, so wed expect the nodes to be closely spaced. A high mode number seems reasonable. 2015 Pearson Education, Inc.

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