2. Simple Harmonic Oscillator (SHO) Any situation where the force exerted on a mass is directly proportional to the negative of the object’s position from an equilibrium point is a simple harmonic oscillator. English: A simple harmonic oscillator is an object that gets pushed toward some point, and the farther the object is from that point, the stronger the force pushing it back.

3. An Example of a SHO: The spring X = 0 m FGFG FNFN Equilibrium Position

4. An Example of a SHO: The spring

9. For all SHO: Where: Sinusoidal Curve Generator

10. Let: * is measured in radians Since,

11. The Simple Pendulum FGFG FTFT Equilibrium Position m = mass of bob L = length of cord F G = weight, mg F T = tension

12. The Simple Pendulum FGFG FTFT m = mass of bob L = length of cord F G = weight, mg F T = tension is measured along the arc

13. The Simple Pendulum FGFG FTFT m = mass of bob L = length of cord F G = weight, mg F T = tension is measured along the arc The Assumption: For sufficiently small, F GY F GX F GY =mg cos(θ) F GX =-mg sin(θ) This is in the format of Hooke’s law, and thus we have a SHO

14. T of a Simple Pendulum (definition of a SHO) (restoring force of a pendulum) (eqn. 11-7) ( Period of a simple Pendulum )

15. Pendulum Lab Expected Results: ( Period of a simple Pendulum )

16. Pg. 343, #34: A fisherman notices that wave crests pass the bow of his anchored boat every 3.0 s. He measures the distance between two crests to be 8.5 m. How fast are the waves traveling? T=3.0 s λ=8.5 m v=? Eqn. (11-12) Oar...

17. Pg. 343, #35: A sound wave in air has a frequency of 262 Hz and travels with a speed of 330 m/s. How far apart are the wave crests (compressions)? f=262 Hz v= 330 m/s Eqn. (11-12)

18. Pg. 344, # 36: AM radio signals have frequencies between 550 kHz and 1600 kHz and travel with a speed of 3x10 8 m/s. On FM, the frequencies range from 88.0 MHz to 108 MHz and travel at the same speed. What are the wavelengths of these signals? f AM high =1600 kHz f AM low =550 kHz f FM low =88.0 MHz f FM high =108 MHz v= 3x10 8 m/s λ AM high f = 188 m λ AM low f = 545 m λ FM low f = 2.78 m λ FM high f = 3.41 m

19. Pg. 344, #39: A cord of mass 0.55 kg is stretched between two supports 30 m apart. If the tension in the cord is 150 N, how long will it take a pulse to travel from one support to the other? m = 0.55 kg L = 30 m = Δx F T = 150 N Δt=? Eqn. 11-13 v = 90.5 m/s

20. Pg. 344, #44: Compare the (a.) intensities and (b.) the amplitudes of an earthquake wave as it passes two points 10 km and 20 km from the source. Let the wave have power P. r 10 km = 10,000 m r 20 km =20,000 m = 2r 10 km I 10km is 4 times greater than I 20km

21. Pg. 344, #44: Compare the (a.) intensities and (b.) the amplitudes of an earthquake wave as it passes two points 10 km and 20 km from the source. Let the wave have power P. r 10 km = 10,000 m r 20 km = 20,000 m = 2r 10 km I 10km = 4 I 20km

22. Pg. 344, #45: The intensity of a particular earthquake wave is measured to be 2.0 x 10 6 J/m 2 s at a distance of 50 km from the source. a.) What was the intensity when it passed a point only 1 km from the source? I 50 km = 2.0 x 10 6 J/m 2 s r 50 km = 50,000 m r 1 km = 1000 m

23. Pg. 344, #45: The intensity of a particular earthquake wave is measured to be 2.0 x 10 6 J/m 2 s at a distance of 50 km from the source. b.) What was the rate energy passed through an area of 10.0 m 2 at 1.0 km? I 50 km = 2.0 x 10 6 J/m 2 s r 50 km = 50 km r 1 km = 1 km Area = 10.0 m 2

24. Pg. 344, #46: Show that the amplitude A of circular water waves decreases as the square root of the distance r from the source. Ignore damping. In other words, show that: 1. The same energy that passes through the small circle each second must pass through the big circle each second, so P for each whole circle is constant. 2. Since and, The only variable is r, so

25. Pg. 344, #50:

26. Standing Waves; Resonance Standing waves occur on a string of length L when the waves have a wavelength,λ, in which L is a multiple of.5λ

27. Standing Waves; Resonance Only certain wavelengths can create standing waves for a given length of cord, but there are an infinite number of wavelengths that can create a standing wave on any given cord.

28. Standing Waves; Resonance A little algebra to find resonant frequencies: Eqn. (11-12) Eqn. (11-19) For the first harmonic, n=1: so:

29. Pg. 344, #51: If a violin string vibrates at 440 Hz as its fundamental frequency, what are the frequencies of the first four harmonics? f n=1 = 440 Hz f n=2 = ? f n=3 = ? f n=4 = ? f n=2 =880 Hz f n=3 = 1320 Hz f n=4 = 1760 Hz (derived on pg. 336)

30. Sound Intensity The more powerful a wave is when it reaches the ear, the more energy per unit time it delivers to the ear, and as a consequence, a more powerful sound is perceived as being louder. The human ear can perceive a wide range of sound intensities: –10 -12 W/m 2 – threshold of hearing –1 W/m 2 – threshold of pain Notice that the quietest sound a human can hear is almost a trillion times less intense than a sound that is so intense that it causes physical pain.

31. The (deci) bel The human ear can hear a wide range of intensities, but can not distinguish between small changes in intensity. We connect the measurable quantity, Intensity, I, to the perceived quantity of loudness, called intensity level, β, using the equation: Where I 0 is the threshold of hearing, 1x10 -12 W/m 2

32. Logarithms Math Crash-course:

33. Arithmetic scale Logarithmic scale

34. Logarithms Math Crash-course: If then Logarithms are discussed in your text, Appendix A, pg. 1046 Logarithm identities:

35. The (deci) bel continued A change in the intensity level is associated with a change in a sound’s loudness. A human ear can perceive a change in intensity level of about 1 dB. –How great of an increase in intensity is an increase in intensity level of 1 dB? –How great of an increase in intensity is an increase in intensity level of 10 dB?