3.  We can pull the mass to the right and then release it to begin its motion:  Or we could push it to the left and release it:  Both motions would have the same values for T and f.  However, the resulting motion will have a phase difference of half a cycle. Start stretched Start compressed The two motions are half a cycle out of phase. x x Phase difference

4. Start stretched and then release Start unstretched with a push left x x Phase difference

5. Start stretched and then release Start unstretched with a push right x x Phase difference

6. x Conditions for simple harmonic motion EXAMPLE: A spring having a spring constant of 125 N m -1 is attached to a 5.0-kg mass, stretched +4.0 m as shown, and then released from rest. (a) Using Hooke’s law, show that the acceleration a of a mass-spring system is related to the spring’s displacement x by the proportion a  -x. (b) Taylor your equation to this example, and find the acceleration of the mass when x = -2.0 m.  a = -(k / m) x  a = -25x  a = -25(-2.0) = +50. ms -2. (c) What is the displacement of the mass when the acceleration is -42 ms -2 ?  a = -25x  x = -a/25  x = +1.7 m

7. Sketching and interpreting graphs of SHM examples EXAMPLE: The displacement x vs. t for a 2.5-kg mass on a spring having spring constant k = 4.0 Nm -1 is shown in the sinusoidal graph. SOLUTION: The period is the time for a complete cycle. (a)Find the period and frequency of the motion. of the motion. (b) Find the amplitude of the motion.  From the graph it is x MAX = 2.0 mm = 2.0  10 -3 m. SOLUTION: The amplitude is the maximum displacement.  From the graph it is T = 6.0 ms = 6.0  10 -3 s.  Then f = 1 / T = 1 / 0.006 = 170 Hz. (c) Sketch the graph of x vs. t for the situation where the amplitude is cut in half. SOLUTION: For SHM, the period is independent of the amplitude.

8. (c) The blue graph shows an equivalent system in SHM. What is the phase difference between the red and blue? SOLUTION:  We see that it is T / 6 (= 360  / 6 = 60  = 2  / 6 rad).

9. Sketching and interpreting graphs of SHM examples EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown. a system undergoing SHM is shown. The system consists of a 0.125-kg mass on a spring. The system consists of a 0.125-kg mass on a spring. (a) Determine the maximum velocity of the mass. SOLUTION: When the E k is max, the velocity is also max. (b) Sketch E P and determine the total energy of the system. Thus the E P graph will be the “inverted” E K graph ETETETET EPEPEPEP EKEKEKEK

10. (c) Determine the spring constant k of the spring. SOLUTION: E P = (1/2)kx 2.  E K = 0 at x = x MAX = 2.0 cm. E K + E P = E T = CONST E T = 0 + (1/ 2)kx MAX 2  4.0 = (1/ 2)k 0.020 2 E T = 0 + (1/ 2)kx MAX 2  4.0 = (1/ 2)k 0.020 2 k = 20000 Nm -1. k = 20000 Nm -1. (d) Determine the acceleration of the mass at x = 1.0 cm. SOLUTION:  From Hooke’s law, F = -kx F = -20000(0.01) = -200 N. F = -20000(0.01) = -200 N.  From F = ma -200 = 0.125a -200 = 0.125a a = -1600 ms -2. a = -1600 ms -2.

11. Sketching and interpreting graphs of SHM examples EXAMPLE: A 4.0-kg mass is placed on a spring’s end and displaced 2.0 m to the right. and displaced 2.0 m to the right. The spring force F vs. its displacement x from The spring force F vs. its displacement x from equilibrium is shown in the graph. equilibrium is shown in the graph. (a) How do you know that the mass is undergoing SHM? SOLUTION: In SHM, a  -x. Since F = ma, then F  -x also.  The graph shows that F  -x. Thus we have SHM. (b) Find the spring constant of the spring. SOLUTION: Hooke’s law: F = -kx.  Pick any F and any x. Use k = -F / x. F = -5.0 N x = 1.0 m  k = -(-5.0 N) / 1.0 m = 5.0 Nm -1. (c) Find the total energy of the system. SOLUTION: E T = (1/2)kx MAX 2. Then E T = (1/2)kx MAX 2 = (1/2)  5.0  2.0 2 = 10. J. (d) Find the maximum speed of the mass. SOLUTION: E T = (1/2)mv MAX 2. 10. = (1/2)  4.0  v MAX 2 10. = (1/2)  4.0  v MAX 2 v MAX = 2.2 ms -1 v MAX = 2.2 ms -1 (e) Find the speed of the mass when its displacement is 1.0 m. when its displacement is 1.0 m. SOLUTION: E T = (1/2)mv 2 + (1/2)kx 2. Then 10. = (1/2)(4)v 2 + (1/2)(5)1 2 Then 10. = (1/2)(4)v 2 + (1/2)(5)1 2 v = 1.9 ms -1. v = 1.9 ms -1.

12. PRACTICE: A spring is moved in SHM by the hand as shown. The hand moves through 1.0 complete cycle in 0.25 s. A metric ruler is placed beside the waveform. (a) What is the wavelength? = 4.7 cm = 0.047 m. (b) What is the period? T = 0.25 s. (c) What is the wave speed? v = / T = 0.047 / 0.25 = 0.19 m s -1. 1 2 34 5 6 7 8 910 11 12 13 14 CM Solving wave speed and wavelength problems

13. Either graph gives the correct amplitude. Sketching and interpreting displacement vs time graphs EXAMPLE: Graph 1 shows the variation with time t of the displacement d of a traveling wave. Graph 2 shows the variation with distance x along the same wave of its displacement d. (a) Use the graphs to determine the amplitude of the wave motion.  Amplitude (maximum displacement) is 0.0040 m. (b) Use the graphs to determine the wavelength.  Wavelength is measured in meters and is the length of a complete wave. = 2.40 cm = 0.0240 m. Graph 2 must be used since its horizontal axis is in cm (not seconds as in Graph 1). (c) Use the graphs to determine the period.  Period is measured in seconds and is the time for one complete wave. T = 0.30 s. (d) Use the graphs to find the frequency.  f = 1 / T = 1 / 0.30 = 3.3 Hz. [3.333 Hz] Graph 1 must be used since its horizontal axis is in s (not cm as in Graph 2). (e) Use the graphs to find the wave speed.  v = / T = 0.024 / 0.30 = 0.080 m s -1.

14. Graph 2 must be used for since its horizontal axis is in cm. Sketching and interpreting displacement vs time graphs PRACTICE: Graph 1 shows the variation with time t of the displacement y of a traveling wave. Graph 2 shows the variation with distance x along the same wave of its displacement. (a)Use the graphs to determine the amplitude and wavelength of the wave motion.  Amplitude (maximum displacement) is y = 0.0020 m.  Wavelength is y = 0.30 cm =.0030 m. (b) Use the graphs to determine the period and the frequency. (c) Use the graphs to determine the wave speed.  Period (cycle time) is 0.25 ms = 0.00025 s.  Frequency is f = 1 / T = 1 / 0.00025 = 4000 Hz.  Wave speed is a calculation.  v = / T = 0.0030 / 0.00025 = 12 m s -1. Graph 1 must be used for T since its horizontal axis is in ms.

15. Sketching and interpreting displacement vs time graphs EXAMPLE: longitudinal Graph 1 shows the variation with time t of the displacement x of a single particle in the medium carrying a longitudinal wave in the +x direction. (a) Use the graph to determine the period and the frequency of the particle’s SHM.  The period is the time for one cycle. T = 0.20 s.  f = 1 / T = 1 / 0.20 = 5.0 Hz. Graph 2 shows the variation of the displacement x with distance d from the beginning of the wave at a particular instant in time. (b) Use the graph to determine the wavelength and wave velocity of the longitudinal wave motion.  = 16.0 cm = 0.160 m.  v = / T = 0.160 / 0.20 = 0.80 m s -1.

16. Sketching and interpreting displacement vs time graphs (c) The equilibrium positions of 6 particles in the medium are shown below. Using  ’s, indicate the actual position of each particle at the instant shown above. (d) In the diagram label the center of a compression with a C and the center of a rarefaction with an R. C R

17. A sound wave produced by a clock chime is heard 515 m away 1.5 s later. (a)What is the speed of sound in the air there? (b) The sound wave has a frequency of 436 Hz. What is the period of the wave? (c) What is the wave's wavelength? (a) v = d/t = 515/1.5 = 343 m/s (b) f = 1/T = 1/436 = 2.29x10 -3 s (c) v = f → = v / f = 0.87 m

18. A hiker shouts toward a vertical cliff 465 m away. The echo is heard 2.75 s later. (a) What is the speed of sound in air there? (b) The wavelength of the sound is 0.75 m. What is the frequency of the wave? (c) What is its period? (b) v = f → f = v/ = 338/0.75 = 451 Hz (a) v = distance/time = 2d/t = 2·465/2.75 = 338 m/s (c) T = 1/f = 2.22x10 -3 s If you wanted to increase the wavelength of waves in a rope should you shake it at a higher or lower frequency? v = f v depends only on the medium. Therefore for given medium it is constant. So if wavelength increases the frequency decreases. You should shake it at lower frequencies. CHECK IT, PLEASE

19. A stone is thrown onto a still water surface and creates a wave. A small floating cork 1.0 m away from the impact point has the following displacement—time graph (time is measured from the instant the stone hits the water): Find (a) amplitude (b) the speed of the wave (c) the freq. (d) wavelength (a)A = 2 cm (b)v = d/t = 1/1.5 = 0.67 m/s (c) f = 1/T = 1/0.3 = 3.33 Hz (d)λ = v/f = 0.666/3.333 = 0.2 m

20. SOLUTION:  = c / f = 3.00  10 8 / 1.67  10 15 = 1.80  10 -7 m. (a) What is its frequency, and what part of the spectrum is it from? SOLUTION: T = 6.00  10 -16 s.  f = 1 / T = 1 / 6.00  10 -16 s = 1.67  10 15 Hz.  This is from the ultraviolet part of the spectrum. (b) What is the wavelength of this light wave? PRACTICE: The graph shows one complete oscillation of a particular frequency of light. (c) Determine whether or not this light is in the visible spectrum. SOLUTION: The visible spectrum is from about 400 nm to 700 nm. = 1.80  10 -7 m = 180  10 -9 m = 180 nm. NO! UV. 700600500400 Wavelength / nm

21. Solving problems involving amplitude and intensity EXAMPLE: A 200. watt speaker projects sound in a spherical wave. Find the intensity of the sound at a distance of 1.0 m and 2.0 m from the speaker.  Note that whatever power is in the wavefront is spread out over a larger area as it expands.  The area of a sphere of radius x is A = 4  x 2.  For x = 1 m: I = P / (4  x 2 ) = 200 / (4  1.0 2 ) = 16 W m -2.  For x = 2 m: I = P / (4  x 2 ) = 200 / (4  2.0 2 ) = 4.0 W m -2.  Doubling your distance reduces the intensity by 75%!

22. Solving problems involving amplitude and intensity EX: At a distance of 18.5 m from a sound source the intensity is 2.00  10 -1 W m -2. (a) Find its intensity at a distance of 26.5 m.  Then I 1  x 1 -2 and I 2  x 2 -2 so that I 2 / I 1 = x 2 -2 / x 1 -2 = x 1 2 / x 2 2 = (x 1 / x 2 ) 2.  Thus I 2 = I 1 (x 1 / x 2 ) 2 = 2.00  10 -1 (18.5 / 26.5) 2 = 0.0975 W m -2. SOLUTION: We can just use I  x -2 and dispense with finding the actual power as an intermediate step. (b) Compare the amplitudes of the sound at 18.5 m and 26.5 m.

23. PRACTICE: Two pulses are shown in a string approaching each other at 1 m s -1. Sketch diagrams to show each of the following: (a) The shape of the string at exactly t = 0.5 s later. (b) The shape of the string at exactly t = 1.0 s later. 1 m Since the pulses are 1 m apart and approaching each other at 1 ms -1 they will overlap at this instant, temporarily canceling out. Since the pulses are 1 m apart and approaching each other at 1 m s -1, at this time they will each have moved ½ m and their leading edges will just meet. Superposition

24. EXAMPLE: Two waves P and Q reach the same point at the same time, as shown in the graph. The amplitude of the resulting wave is A. 0.0 mm. B. 1.0 mm. C. 1.4 mm. D. 2.0 mm.  In the orange regions P and Q do not cancel.  In the purple regions P and Q partially cancel.  Focus on orange. Try various combos…  Blue (0.7 mm) plus green (0.7 mm) = 1.4 mm.  Note that the crests to not coincide, so NOT 2.0. Superposition

25. Observe the light intensity through two polarizing filters as the top one is rotated through 180º. Which filter is the analyzer? Which filter is the polarizer? RED FILTER BLACK FILTER

26. Solving problems involving Malus’s law PRACTICE: The preferred directions of two sheets of Polaroid are initially parallel. (a) Calculate the angle through which one sheet needs to be turned in order to reduce the amplitude of the observed E-field to half its original value. (b) Calculate the effect this rotation has on the intensity. (c) Calculate the rotation angle needed to halve the intensity from its original value. SOLUTION: (a) Solve cos  = 1 / 2   = 60º. (b) I = I 0 cos 2  = I 0 (1/2) 2 = I 0 / 4. (c) I 0 / 2 = I 0 cos 2   cos 2  = 1 / 2. Thus cos  = (1 / 2) 1/2   = 45 0.

27. Solving problems involving Malus’s law  In general, light sources produce waves having their E-fields oriented in many random directions.  Polarized light is light whose waves have their E-fields oriented in one direction.  I = I 0 cos 2  ⇒ I = I 0 cos 2 60 0 ⇒ I = 0.25 I 0.

28. I 0 cos 2 0º = I 0 I 0 cos 2 60º = 0.25I 0 I 0 cos 2 90º = 0 I 0 cos 2 120º = 0.25I 0 I 0 cos 2 180º = I 0  In general, light sources produce waves having their E-fields oriented in many random directions.  Polarizing sunglasses only allow waves in one direction through, thereby reducing the intensity of the light entering the eye.  Reflecting surfaces also polarize light to a certain extent, thereby rendering polarizing sunglasses even more effective.

29. Solving problems involving Malus’s law I = I 0 cos 2  ⇒ I = I 0 (1/2) 2 = I 0 / 4

30.  Diffraction allows waves to turn corners. Wave behavior  All waves diffract – not just sound.

31. Huygens says the wavelets will be spherical.

33. PRACTICE: What wave behavior is being demonstrated here?  Refraction. T or F: The period changes in the different media. T or F: The frequency changes in the different media. T or F: The wavelength changes in the different media. T or F: The wave speed changes in the different media. T or F: The sound wave is traveling fastest in the warm air. A sound pulse entering and leaving a pocket of cold air (blue). Wave behavior

34. PRACTICE: What wave behavior is being demonstrated here?  Reflection. This is an echo from a wall. T or F: The frequency of the reflected wave is different from the frequency of the incident wave? Fill in the blanks: The angle of ___________ equals the angle of ___________. Fill in the blank: In this example, both angles equal ___°.  Remember that angles are measured wrt the normal. incidence reflection 0 Wave behavior

35. PRACTICE: What wave behavior is being demonstrated here?  Total internal reflection.  This is the same principle behind fiber optics. Light is used instead of sound.  Light travels faster than electrical signals so fiber optics are used in high speed communications networks such as token rings whenever possible. Wave behavior

36. PRACTICE: Complete with sketch and formula.  Remember that there are always nodes on each end of a string.  Add a new well-spaced node each time.  Decide the relationship between and L.  We see that = (2/3)L.  Since v = f we see that f = v / (2/3)L = 3v / 2L. f 2 = v / L f 1 = v / 2L f 3 = 3v / 2L Sketching and interpreting standing wave patterns

37. Distinguishing between standing and traveling waves  A standing wave consists of two traveling waves carrying energy in opposite directions, so the net energy flow through the wave is zero. E E

38. Solving problems involving standing waves PRACTICE: A tube is filled with water and a vibrating tuning fork is held above the open end. As the water runs out of the tap at the bottom sound is loudest when the water level is a distance x from the top. The next loudest sound comes when the water level is at a distance y from the top. Which expression for is correct? A. = x B. = 2x C. = y-x D. = 2(y-x)  v = f and since v and f are constant, so is.  The first possible standing wave is sketched.  The second possible standing wave is sketched.  Notice that y – x is half a wavelength.  Thus the answer is = 2(y - x). y-x

39. PRACTICE: This drum head, set to vibrating at different resonant frequencies, has black sand on it, which reveals 2D standing waves.  Does the sand reveal nodes, or does it reveal antinodes?  Why does the edge have to be a node? Nodes, because there is no displacement to throw the sand off. The drumhead cannot vibrate at the edge. Solving problems involving standing waves

40.  Alternate lobes have a 180º phase difference. Solving problems involving standing waves

41.  Make a sketch. Then use v = f. antinode antinode L / 2 = L v = f = 2L f = v / f = v / (2L) Solving problems involving standing waves

42.  Reflection provides for two coherent waves traveling in opposite directions.  Superposition is just the adding of the two waves to produce the single stationary wave.

43.  The figure shows the points between successive nodes.  For every point between the two nodes f is the same.  But the amplitudes are all different.  Therefore the energies are also different. Solving problems involving standing waves

44.  Energy transfer via a vibrating medium without interruption.  The medium itself does not travel with the wave disturbance.  Speed at which the wave disturbance propagates.  Speed at which the wave front travels.  Speed at which the energy is transferred. Solving problems involving standing waves

45.  Frequency is number of vibrations per unit time.  Distance between successive crests (or troughs).  Distance traveled by the wave in one oscillation of the source. FYI: IB frowns on you using particular units as in “Frequency is number of vibrations per second.” FYI: There will be lost points, people! Solving problems involving standing waves

46.  The waves traveling in opposite directions carry energy at same rate both ways. NO energy transfer.  The amplitude is always the same for any point in a standing wave. Solving problems involving standing waves

47. L P / 4 = L = 4L L Q / 2 = L = 2L v = f f = v / f P = v / (4L) f Q = v / (2L) v = 4Lf P f Q = 4Lf P / (2L) f Q = 2f P Solving problems involving standing waves

48.  The tuning fork is the driving oscillator (and is at the top).  The top is thus an antinode.  The bottom “wall” of water allows NO oscillation.  The bottom is thus a node. Solving problems involving standing waves

49.  Sound is a longitudinal wave.  Displacement is small at P, big at Q. Solving problems involving standing waves

50.  If the lobe at T is going down, so is the lobe at U. Solving problems involving standing waves

51.  Pattern 1 is 1/2 wavelength.  Pattern 2 is 3/2 wavelength.  Thus f 2 = 3f 1 so that f 1 / f 2 = 1/3. Solving problems involving standing waves