2. Waves Harmonic Motion

3. 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

4. Periodic Motion Springs as periodic motion Always tries to return to equilibrium Hooke’s Law F=-kx Potential Energy in a Spring PE sp =1/2 kx^2

5. Periodic Motion Pendulum as simple harmonic motion Always tries to return to equilibrium http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=11 Distance from equilibrium as a function of time

6. Wave Properties Types of Waves and Measuring Waves

7. Wave Properties Amplitude Maximum distance from equilibrium

8. Wave Properties Peak-to-Peak Amplitude Distance from Crest to Trough (or from low peak to high peak)

9. Wave Properties Wavelength λ

10. 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

11. 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

12. Wave Properties Equations Frequency: f = 1/T Wavelength: λ = v/f flambda: v = f λ (speed or velocity of a wave) meters x(meters) t(seconds) ¼ wavelength

13. Wave Properties

15. 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

16. Wave Behavior Multiple waves, multiple boundaries

17. 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.

18. Boundaries Incident waves –Waves that strike the boundary Reflected waves –Waves that return from the boundary Boundary

19. 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

20. 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.

21. Interference Constructive –Result is bigger Destructive –Result is smaller

22. Interference Node: Has zero displacement, does not move at all. Antinode: Has maximum displacement. What is the pattern? (# of nodes vs. antinodes)

23. Standing Wave Standing wave: Appears to be standing still. Is the interference of two or more waves.

24. Waves in two dimensions

25. Wave front: A line that represents the crest or peak of a wave in 2D. Can be any shape (e.g. straight, circular, etc.).

26. Law of reflection –The angle of incidence is equal to the angle of reflection

27. Refraction Refraction: The change in the direction of waves at the boundary between different media. –Examples are echoes and rainbows

28. 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

29. Read the Graph Maximum Speed, Zero Force, No Acceleration Equilibrium Point Maximum Displacement, Maximum Amplitude, Maximum Force, Zero Velocity, No Motion

30. 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

32. 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

33. 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

34. 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?

35. 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

36. 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

37. 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?

38. 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.

39. Wave table or Ripple tank

40. Problems In Class –Pg 391 27, 28, 29, 30, Homework –Pg 398 84, 85, 86, 87, 91, 93

41. 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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