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Waves. Harmonic Motion. Periodic Motion. Repeat in a regular cycle Examples Period T Time required to complete one cycle Frequency f Cycles that occur in one second Frequency is in hertz, Hz, which is 1/sec. Periodic Motion. Springs as periodic motion
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Waves Harmonic Motion
Periodic Motion • Repeat in a regular cycle • Examples • Period T • Time required to complete one cycle • Frequency f • Cycles that occur in one second • Frequency is in hertz, Hz, which is 1/sec
Periodic Motion • Springs as periodic motion • Always tries to return to equilibrium • Hooke’s Law • F=-kx • Potential Energy in a Spring • PEsp=1/2 kx^2
Periodic Motion • Pendulum as simple harmonic motion • Always tries to return to equilibrium Distance from equilibrium as a function of time http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=11
Wave Properties Types of Waves and Measuring Waves
Wave Properties • Amplitude Maximum distance from equilibrium
Wave Properties • Peak-to-Peak Amplitude Distance from Crest to Trough (or from low peak to high peak)
Wave Properties • Wavelength λ
Wave Properties • Period/frequency • Essentially the same variable, just inverted • T = 1/f Seconds per cycle • f = 1/T Cycles per second • (just remember frequency always has 1 sec as the denominator) Cycles per second and seconds per cycle Crash
Wave Properties • Phase • A way to describe the relationship between two points on waves ¼ of the distance from the other wave ¼ x 360 = 90 degrees out of phase
Wave Properties • Equations • Frequency: f = 1/T • Wavelength: λ = v/f • flambda: v = f λ (speed or velocity of a wave) ¼ wavelength meters x(meters) t(seconds)
Problems • In Class • Pg 397 • 64, 65, 66, 67, 68, 69, 71, 73, 75, 76, 77, 79, 80 • Homework • Pg 398 • 81, 82, 88, 89, 90, 97, 102
Wave Behavior Multiple waves, multiple boundaries
Wave Behavior • Multiple waves can exist in the same place at the same time. • Multiple particles cannot exist in the same place at the same time • Note: This is where the dilemma occurs when waves and particles are not clearly separated.
Boundaries • Incident waves • Waves that strike the boundary • Reflected waves • Waves that return from the boundary Boundary
Reflected Waves • Fixed/Rigid • Inverted reflection • Loose • Non-inverted reflection • No end • No reflection • Reflected waves do not change speed (ever). Speed is a function of the medium. http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=11
Principle of superposition • The displacement of the medium caused by two or more waves is the algebraic sum of the displacements caused by the individual waves. • Interference: The result of the superposition of two or more waves.
Interference • Constructive • Result is bigger • Destructive • Result is smaller
Interference • Node: Has zero displacement, does not move at all. • Antinode: Has maximum displacement. What is the pattern? (# of nodes vs. antinodes)
Standing Wave • Standing wave: Appears to be standing still. Is the interference of two or more waves.
Waves in two dimensions • Wave front: A line that represents the crest or peak of a wave in 2D. Can be any shape (e.g. straight, circular, etc.).
Law of reflection • Law of reflection • The angle of incidence is equal to the angle of reflection
Refraction • Refraction: The change in the direction of waves at the boundary between different media. • Examples are echoes and rainbows
Rules to Remember • Speed of a wave only changes when the medium changes (i.e. water depth changes or wave crosses a boundary into a different medium). • Speed will not change because of wavelength or frequency • Frequency only changes when the source changes • Frequency will never change when the wave crosses a boundary
Read the Graph • Maximum Speed, Zero Force, No Acceleration Equilibrium Point • Maximum Displacement, Maximum Amplitude, Maximum Force, Zero Velocity, No Motion
Things you need to know E=PE+KE Work is Energy - work done is equivalent to the change in potential energy All lengths need to be in meters
Problems • Consider a 10.0-kg pendulum clock that has a period of 1s on Earth. If the clock is moved to a location where it weighs 98 N, how many minutes will the minute hand move in 1 h? • Write the equation for Earth • Write the equation for the new location • Set up the ratios
Problems • Spring A with a spring constant of 253 N/m is stretched by a distance of 18.0 cm when a block is suspended from its end. An object is suspended from another spring B with a spring constant of 169 N/m. If the elastic potential energy in both the springs is the same, how far does spring B stretch? • Two parts • Find k with F=-kx • Compare with PE= ½ kx^2
Problems • A 150-g object subject to a restoring force F = –kx is undergoing simple harmonic motion. Shown below is a plot of the potential energy, PE, of the object as a function of distance, x, from its equilibrium position. The object has a total mechanical energy of 0.3 J. a. What is the farthest the object moves along the x-axis in the positive direction? Explain your reasoning. b. What is the object’s potential energy when its displacement is +4.0 cm from its quilibrium position? c. Determine the object’s kinetic energy when its position is x = –8.0 cm. d. What is the object’s speed at x = 0.0 cm?
Answer • a. Answer: 10.0 cm or 0.100 m. The total mechanical energy is E = 0.3 J, and total mechanical energy is given by E = KE + PE. Since the maximum potential energy cannot be greater than the total mechanical energy, the maximum potential energy is also 0.3 J. From the graph, PE = 0.3 J at x = 10.0 cm. Hence the maximum possible position in the +x-direction is 10.0 cm or 0.100 m. The particle stops at this point (i.e. the kinetic energy KE = 0 at this point) because all of the energy is in the form of potential energy. • b. PE ≈ 0.05 J • Method 1: • Read off the value from the graph: PE ≈ 0.05 J • Method 2: • Since the restoring force is given by Hooke’s law F = –kx, we can use the following equation for the potential energy of simple harmonic motion: • PEsp = ½ kx^2 • The maximum value of potential energy is PEmax = 0.3 J. This occurs at a distance of xmax = 10.0 cm = 0.100 m. Use these values to determine the value of the spring constant k: • k = 2PEmax/(x max^2) = 2(0.3J)/(0.100m)2 = 60M/n • Then, to determine the value of PE at x = 14.0 cm = 0.040 m: • PEsp = ½ kx^2 = ½ (60N/m)(0.040m)^2 = 0.05J
Answer • c. 0.1 J • Method 1: • Read off the value from the graph: PE ≈ 0.2 J The total mechanical energy is the sum of kinetic energy and potential energy: • E = KE + PE • Find the kinetic energy: • KE = E – PE = 0.3 J – 0.2 J = 0.1 J • Method 2: • Use the value of the spring constant k found in part b: • k = 60 N/m • Determine the value of PE at x = 28.0 cm = 20.080 m: • PEsp = ½ kx^2 = (60N/m)(20.080m)^2 = 0.2J • Then, find the kinetic energy: • E = KE + PE • KE = E – PE • = 0.3 J – 0.2 J = 0.1 J • d. 2 m/s • At x = 0.0cm,PEsp = ½ kx^2 = 0 • E = KE + PE = KE + 0 • so • KE = E; ½ mv2 = E • Solving for the speed, • v = sqrt(2E/m) • v =sqrt[2(0.3J)/(0.15 kg)] = 2m/s
Problem • Read the graph • Given: the incident wave travels at 1.0m/s • How much time has passed in graph b? • How far has the transmitted wave traveled in that time? • What is the velocity of the transmitted wave in graph b?
Problem • Given: 3 nodes and 2 antinodes in a 2-meter span • What happens when the frequency is doubled? • What is the wavelength? • Does the velocity change if the frequency changes? • Write an equation relating the first frequency and wavelength to the second frequency and wavelength.
Problems • In Class • Pg 391 • 27, 28, 29, 30, • Homework • Pg 398 • 84, 85, 86, 87, 91, 93
Boundaries • Fixed • Loose • No end • Incident waves • Reflected waves: Do waves slow down after reflection? • Principle of superposition • Wave interference • Node • Antinode • Standing waves • 2-D reflection • 2-D refraction • Normal • Law of reflection: angle of incidence is equal to angle of reflection http://phet.colorado.edu/simulations/sims.php?sim=Wave_on_a_String
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