Section 16.6 The Interference of Waves from Two Sources (cont.)

Section 16.6 The Interference of Waves from Two Sources (cont.)
© 2015 Pearson Education, Inc.

Interference Along a Line
If d1 and d2 are the distances from the loudspeakers to the observer, their difference is called the path-length difference. Two waves will be in phase and will produce constructive interference any time their path-length difference is a whole number of wavelengths. © 2015 Pearson Education, Inc.

When the speakers are separated by half a wavelength, the waves are out of phase by 180º. The sum of the two waves is zero at every point; this is destructive interference. © 2015 Pearson Education, Inc.

Interference For two sources of (in-phase) waves
constructive interference occurs when the path-length difference is Destructive interference occurs when the path-length difference is For interference to be totally destructive, the waves must have the same amplitude. © 2015 Pearson Education, Inc.

Interference Example Two speakers are emitting identical sound waves with a wavelength of 4.0 m. The speakers are 8.0 m apart and directed toward each other, as in the following diagram. At each of the noted points in the previous diagram, the interference is Constructive. Destructive. Something in between. Answer: A. (a, c, e) B. (b, d) © 2015 Pearson Education, Inc.

Interference Example Two speakers are emitting identical sound waves with a wavelength of 4.0 m. The speakers are 8.0 m apart and directed toward each other, as in the following diagram. At each of the noted points in the previous diagram, the interference is Constructive. Destructive. Something in between. (a, c, e) (b, d) © 2015 Pearson Education, Inc.

QuickCheck 16.14 Two loudspeakers emit in-phase sound waves with the same wavelength and the same amplitude. The waves are shown displaced, for clarity, but assume that both are traveling along the same axis. At the point where the dot is, The interference is constructive. The interference is destructive. The interference is somewhere between constructive and destructive. There’s not enough information to tell about the interference. Answer: C © 2015 Pearson Education, Inc. 8

QuickCheck 16.14 Two loudspeakers emit in-phase sound waves with the same wavelength and the same amplitude. The waves are shown displaced, for clarity, but assume that both are traveling along the same axis. At the point where the dot is, The interference is constructive. The interference is destructive. The interference is somewhere between constructive and destructive. There’s not enough information to tell about the interference. © 2015 Pearson Education, Inc. 9

QuickCheck 16.15 Two loudspeakers emit in-phase sound waves with the same wavelength and amplitude. Which of the following would cause there to be constructive interference at the position of the dot? (Again, assume the speakers are on the same axis.) Move speaker 2 forward (right) 1.0 m Move speaker 2 forward (right) 0.5 m Move speaker 2 backward (left) 0.5 m Nothing – the interference is already constructive. Trick question. Constructive interference is not possible without changing the position of the dot. Answer: C © 2015 Pearson Education, Inc. 10

QuickCheck 16.15 Two loudspeakers emit in-phase sound waves with the same wavelength and amplitude. Which of the following would cause there to be constructive interference at the position of the dot? (Again, assume the speakers are on the same axis.) Move speaker 2 forward (right) 1.0 m Move speaker 2 forward (right) 0.5 m Move speaker 2 backward (left) 0.5 m Nothing – the interference is already constructive. Trick question. Constructive interference is not possible without changing the position of the dot. Answer: C © 2015 Pearson Education, Inc. 11

Example 16.10 Interference of sound from two speakers
Susan stands directly in front of two speakers that are in line with each other. The farther speaker is m from her; the closer speaker is 5.0 m away. The speakers are connected to the same Hz sound source, and Susan hears the sound loud and clear. The frequency of the source is slowly increased until, at some point, Susan can no longer hear it. What is the frequency when this cancellation occurs? Assume that the speed of sound in air is 340 m/s. © 2015 Pearson Education, Inc.

Example 16.10 Interference of sound from two speakers (cont.)
prepare We’ll start with a visual overview of the situation, as shown in FIGURE The sound waves from the two speakers overlap at Susan’s position. The path-length difference—the extra distance traveled by the wave from speaker 1—is just the difference in the distances from the speakers to Susan’s position. In this case, ∆d = d2  d1 = 6.0 m 5.0 m = 1.0 m © 2015 Pearson Education, Inc.

At 680 Hz, this path-length difference gives constructive interference. When the frequency is increased by some amount, destructive interference results and Susan can no longer hear the sound. © 2015 Pearson Education, Inc.

solve The path-length difference and the sound wavelength together determine whether the interference at Susan’s position is constructive or destructive. Initially, with a 680 Hz tone and a 340 m/s sound speed, the wavelength is © 2015 Pearson Education, Inc.

The ratio of the path-length difference to the wavelength is The path-length difference matches the constructive- interference condition ∆d = mλ with m = 2. We expect constructive interference, which is what we get—the sound is loud. © 2015 Pearson Education, Inc.

As the frequency is increased, the wavelength decreases and the ratio ∆d/λ increases. The ratio starts at 2.0. The first time destructive interference occurs is when the ratio reaches 2½, which matches the destructive-interference condition ∆d = (m + )λ with m = 2. So destructive interference first occurs when the wavelength is decreased to This corresponds to a frequency of © 2015 Pearson Education, Inc.

assess 850 Hz is an increase of 170 Hz from the original Hz, an increase of one-fourth of the original frequency. This makes sense: Originally, 2 cycles of the wave “fit” in the 1.0 m path-length difference; now, 2.5 cycles “fit,” an increase of one-fourth of the original. © 2015 Pearson Education, Inc.

If two loudspeakers are side by side, and one emits the exact inverse of the other speaker’s wave, then there will be destructive interference and the sound will completely cancel. Headphones with active noise reduction measure the ambient sound and produce an inverted version to add to it, lowering the overall intensity of the sound. © 2015 Pearson Education, Inc.

Interference of Spherical Waves
In practice, sound waves from a speaker or light waves emitted from a lightbulb spread out as spherical waves. © 2015 Pearson Education, Inc.

Interference occurs where the waves overlap. The red dot represents a point where two wave crests overlap, so the interference is constructive. The black dot is at a point where a crest overlaps a trough, so the wave interference is destructive. © 2015 Pearson Education, Inc.

Counting the wave fronts, we see that the red dot is three wavelengths from speaker 2 and two wavelengths from speaker 1. The path-length difference is Δr = r2 – r1 = λ The path-length of the black dot is Δr = ½ λ. © 2015 Pearson Education, Inc.

The general rule for identifying whether constructive or destructive interference occurs is the same for spherical waves as it is for waves traveling along a line. Constructive interference occurs when Destructive interference occurs when © 2015 Pearson Education, Inc.

QuickCheck 16.16 Two in-phase sources emit sound waves of equal wavelength and intensity. At the position of the dot, The interference is constructive. The interference is destructive. The interference is somewhere between constructive and destructive. There’s not enough information to tell about the interference. Answer: A © 2015 Pearson Education, Inc. 24

QuickCheck 16.16 Two in-phase sources emit sound waves of equal wavelength and intensity. At the position of the dot, The interference is constructive. The interference is destructive. The interference is somewhere between constructive and destructive. There’s not enough information to tell about the interference. © 2015 Pearson Education, Inc. 25

Example 16.11 Is the sound loud or quiet?
Two speakers are 3.0 m apart and play identical tones of frequency 170 Hz in phase. Sam stands directly in front of one speaker at a distance of 4.0 m. Is this a loud spot or a quiet spot? Assume that the speed of sound in air is 340 m/s. © 2015 Pearson Education, Inc.

Example 16.11 Is the sound loud or quiet? (cont.)
prepare We draw a figure. solve We first compute the path- length difference. r1, r2, and the distance between the speakers form a right triangle, so we can use the Pythagorean theorem to find Thus the path-length difference is ∆r = r2  r1 = 1.0 m © 2015 Pearson Education, Inc.

Example 16.11 Is the sound loud or quiet? (cont.)
Next, we compute the wavelength: The path-length difference is λ, so this is a point of destructive interference. Sam is at a quiet spot. © 2015 Pearson Education, Inc.

You are regularly exposed to sound from two separate sources: stereo speakers. You don’t hear a pattern of loud and soft sounds because the music is playing at a number of frequencies and the sound waves are reflected off the walls in the room. © 2015 Pearson Education, Inc.

Example Problem Two speakers emit identical (in-phase) sinusoidal waves. The speakers are placed 4.0 m apart. A listener moving along a line in front of the two speakers finds loud and quiet spots as shown in the following figure. The grid lines are spaced at 1.0 m. Is this possible? If so, compute the wavelength of the sound wave. If not, why? Answer: Not possible – the quiet spot is equal path length from both speakers which means that the path length difference is 0, and hence the point marked quiet should be a loud spot. © 2015 Pearson Education, Inc.

Beats The air oscillates against your eardrum at frequency
The beat frequency is the difference between two frequencies that differ slightly: fosc determines the pitch, fbeat determines the frequency of the modulations. © 2015 Pearson Education, Inc.

QuickCheck 16.18 You hear 2 beats per second when two sound sources, both at rest, play simultaneously. The beats disappear if source 2 moves toward you while source 1 remains at rest. The frequency of source 1 is 500 Hz. The frequency of source 2 is 496 Hz 498 Hz 500 Hz 502 Hz 504 Hz Answer: B © 2015 Pearson Education, Inc. 35

QuickCheck 16.18 You hear 2 beats per second when two sound sources, both at rest, play simultaneously. The beats disappear if source 2 moves toward you while source 1 remains at rest. The frequency of source 1 is 500 Hz. The frequency of source 2 is 496 Hz 498 Hz 500 Hz 502 Hz 504 Hz © 2015 Pearson Education, Inc. 36

Example 16.12 Detecting bats using beats
The little brown bat lives in North America, and emits echolocation pulses at a frequency of 40 kHz (above the range of human hearing). To allow observers to “hear” these bats, the bat detector shown combines the bat’s sound wave at frequency f1 with a wave of frequency f2 from a tunable oscillator. The beat frequency is isolated with a filter, then Sent to a loudspeaker. To what frequency should the oscillator be set to produce an audible beat frequency of 3 kHz? © 2015 Pearson Education, Inc.

Example 16.12 Detecting bats using beats
solve The beat frequency is , so the oscillator frequency and the bat frequency need to differ by 3 kHz. An oscillator frequency of either 37 kHz or 43 kHz will work nicely. © 2015 Pearson Education, Inc.

QuickCheck 16.17 Two speakers emit sounds of nearly equal frequency, as shown. At a point between the two speakers, the sound varies from loud to soft. How much time elapses between two successive loud moments? 0.5 s 1.0 s 2.0 s 4.0 s Answer: B © 2015 Pearson Education, Inc.

QuickCheck 16.17 Two speakers emit sounds of nearly equal frequency, as shown. At a point between the two speakers, the sound varies from loud to soft. How much time elapses between two successive loud moments? 0.5 s 1.0 s 2.0 s 4.0 s © 2015 Pearson Education, Inc.

Example Problem A typical police radar sends out microwaves at 10.5 GHz. The unit combines the wave reflected from a car with the original signal and determines the beat frequency. This beat frequency is converted into a speed. If a car is moving at 20 m/s toward the detector, what will be the measured beat frequency? Answer: The beat frequency is the absolute value of the difference of the emitted frequency with the received frequency which is the same as the shift in frequency on reflection. The car is moving at 20 m/s so the Doppler-shifted frequency is given by equation = 2*10.5E9*20/3E8 where 3E8 is the speed of the radar waves. The beat frequency is thus 1400 Hz. © 2015 Pearson Education, Inc.

Chapter 13 Preview Looking Ahead: Pressure in Liquids
A liquid’s pressure increases with depth. The high pressure at the base of this water tower pushes water throughout the city. You’ll learn about hydrostatics—how liquids behave when they’re in equilibrium. © 2015 Pearson Education, Inc.

Chapter 13 Preview Looking Ahead: Buoyancy
These students are competing in a concrete canoe contest. How can such heavy, dense objects stay afloat? You’ll learn how to find the buoyant force on an object in a fluid using Archimedes’ principle. © 2015 Pearson Education, Inc.

Chapter 13 Preview Looking Ahead: Fluid Dynamics
Moving fluids can exert large forces. The air passing this massive airplane’s wings can lift it into the air. You’ll learn to use Bernoulli’s equation to predict the pressures and forces due to moving fluids. © 2015 Pearson Education, Inc.

Chapter 13 Preview Looking Back: Equilibrium
In Section 5.1, you learned that for an object to be at rest—in static equilibrium—the net force on it must be zero. We’ll use the principle of equilibrium in this chapter to understand how an object floats. This mountain goat is in equilibrium: Its weight is balanced by the normal force of the rock. © 2015 Pearson Education, Inc.

Chapter 13 Preview Stop to Think
Three identical books are stacked vertically. The normal force of book 1 on book 2 is Equal to the weight of one book. Less than the weight of one book. Greater than the weight of one book. Stop to Think Answer: C © 2015 Pearson Education, Inc.

Fluids and Density A fluid is a substance that flows.
Liquids and gases are fluids. Gases are compressible; the volume of a gas is easily increased or decreased. Liquids are nearly incompressible; the molecules are packed closely, yet they can move around. © 2015 Pearson Education, Inc.

Density The mass density is the ratio of mass to volume:
The SI units of mass density are kg/m3. Gasoline has a mass density of 680 kg/m3, meaning there are 680 kg of gasoline for each 1 cubic meter of the liquid. © 2015 Pearson Education, Inc.

Example 13.1 Weighing the air in a living room
What is the mass of air in a living room with dimensions 4.0 m  6.0 m  2.5 m? © 2015 Pearson Education, Inc.

Example 13.1 Weighing the air in a living room
What is the mass of air in a living room with dimensions 4.0 m  6.0 m  2.5 m? prepare Table 13.1 gives air density at a temperature of 20°C, which is about room temperature. solve The room’s volume is V = (4.0 m)  (6.0 m)  (2.5 m) = 60 m3 The mass of the air is m = ρV = (1.20 kg/m3)(60 m3) = 72 kg assess This is perhaps more mass—about that of an adult person—than you might have expected from a substance that hardly seems to be there. For comparison, a swimming pool this size would contain 60,000 kg of water. © 2015 Pearson Education, Inc.

Pressure Liquids exert forces on the walls of their containers.
The pressure is the ratio of the force to the area on which the force is exerted: The fluid’s pressure pushes on all parts of the fluid itself, forcing the fluid out of a container with holes. © 2015 Pearson Education, Inc.

Pressure We can measure the pressure in a liquid with a simple device. We find that pressure is everywhere in the fluid; different parts of a fluid are pushing against each other. © 2015 Pearson Education, Inc.

Pressure in Liquids The force of gravity (the weight of the liquid) is responsible for the pressure in the liquid. The horizontal forces cancel each other out. The vertical forces balance: pA = p0A + mg © 2015 Pearson Education, Inc.

Pressure in Liquids The liquid is a cylinder of cross- section area A and height d. The mass is m = ρAd. The pressure at depth d is Because we assumed that the fluid is at rest, this pressure is the hydrostatic pressure. © 2015 Pearson Education, Inc.

Pressure in Liquids A connected liquid in hydrostatic equilibrium rises to the same height in all open regions of the container. In hydrostatic equilibrium, the pressure is the same at all points on a horizontal line through a connected liquid of a single kind. © 2015 Pearson Education, Inc.

QuickCheck 13.1 An iceberg floats in a shallow sea. What can you say about the pressures at points 1 and 2? p1 > p2 p1 = p2 p1 < p2 Hydrostatic pressure is the same at all points on a horizontal line through a connected fluid. © 2015 Pearson Education, Inc. 63

QuickCheck 13.2 What can you say about the pressures at points 1 and 2? p1 > p2 p1 < p2 p2 = p1 Hydrostatic pressure is the same at all points on a horizontal line through a connected fluid. © 2015 Pearson Education, Inc. 65

Example 13.3 Pressure in a closed tube
Water fills the tube shown in the figure. What is the pressure at the top of the closed tube? © 2015 Pearson Education, Inc.

Example 13.3 Pressure in a closed tube
Water fills the tube shown in the figure. What is the pressure at the top of the closed tube? prepare This is a liquid in hydrostatic equilibrium. The closed tube is not an open region of the container, so the water cannot rise to an equal height. Nevertheless, the pressure is still the same at all points on a horizontal line. In particular, the pressure at the top of the closed tube equals the pressure in the open tube at the height of the dashed line. Assume p0 = 1 atm. © 2015 Pearson Education, Inc.

Example 13.3 Pressure in a closed tube (cont.)
solve A point 40 cm above the bottom of the open tube is at a depth of 60 cm. The pressure at this depth is p = p0 + ρgd = (1.01  105 Pa) + (1000 kg/m3)(9.80 m/s2)(0.60 m) = 1.07  105 Pa = 1.06 atm assess The water column that creates this pressure is not very tall, so it makes sense that the pressure is only a little higher than atmospheric pressure. © 2015 Pearson Education, Inc.

QuickCheck 13.3 What can you say about the pressures at points 1, 2, and 3? p1 = p2 = p3 p1 = p2 > p3 p3 > p1 = p2 p3 > p1 > p2 p2 = p3 > p1 Hydrostatic pressure is the same at all points on a horizontal line through a connected fluid, and pressure increases with depth. © 2015 Pearson Education, Inc. 70 70

Atmospheric Pressure Gas is compressible, so the air in the atmosphere becomes less dense with increasing altitude. 99% of the air in our atmosphere is below 30 km. Atmospheric pressure varies with altitude and with changes in the weather. © 2015 Pearson Education, Inc.

Section 16.6 The Interference of Waves from Two Sources (cont.)

Similar presentations

Presentation on theme: "Section 16.6 The Interference of Waves from Two Sources (cont.)"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Section 16.6 The Interference of Waves from Two Sources (cont.)

Similar presentations

Presentation on theme: "Section 16.6 The Interference of Waves from Two Sources (cont.)"— Presentation transcript:

Similar presentations

About project

Feedback