Simple Harmonic Oscillator (SHO)

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. FN FG An Example of a SHO: The spring X = 0 m Equilibrium Position

4. An Example of a SHO: The spring

5. An Example of a SHO: The spring

6. An Example of a SHO: The spring

7. An Example of a SHO: The spring

9. For all SHO: Where: Sinusoidal Curve Generator

10. Let: Since , * is measured in radians

11. FT FG The Simple Pendulum m = mass of bob L = length of cord FG = weight, mg FT= tension Equilibrium Position

12. is measured along the arc FT FG The Simple Pendulum m = mass of bob L = length of cord FG = weight, mg FT= tension

13. is measured along the arc FT FG The Assumption: For sufficiently small , The Simple Pendulum m = mass of bob FGY=mg cos(θ) L = length of cord FGX=-mg sin(θ) FG = weight, mg FT= tension FGX FGY This is in the format of Hooke’s law, and thus we have a SHO

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

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

16. Eqn. (11-12) 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=? Oar. . .

17. Eqn. (11-12) 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

18. Pg. 344, # 36: AM radio signals have frequencies between 550 kHz and 1600 kHz and travel with a speed of 3x108 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? λAM low f= 545 m fAM low=550 kHz fAM high=1600 kHz λAM high f= 188 m v= 3x108 m/s fFM low=88.0 MHz λFM low f= 2.78 m fFM high=108 MHz λFM high f= 3.41 m

19. Eqn. 11-13 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 FT = 150 N Δt=? 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. r10 km = 10,000 m r20 km=20,000 m = 2r10 km I10km is 4 times greater than I20km

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. r10 km = 10,000 m r20 km= 20,000 m = 2r10 km I10km = 4 I20km

22. Pg. 344, #45: The intensity of a particular earthquake wave is measured to be 2.0 x 106 J/m2s 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? I50 km= 2.0 x 106 J/m2s r50 km = 50,000 m r1 km= 1000 m

23. Pg. 344, #45: The intensity of a particular earthquake wave is measured to be 2.0 x 106 J/m2s at a distance of 50 km from the source. b.) What was the rate energy passed through an area of 10.0 m2 at 1.0 km? I50 km= 2.0 x 106 J/m2s r50 km = 50 km r1 km= 1 km Area = 10.0 m2

24. In other words, show that: 2. Since and , 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. 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. 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. Eqn. (11-19) Eqn. (11-12) Standing Waves; Resonance A little algebra to find resonant frequencies: 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? fn=1 = 440 Hz (derived on pg. 336) fn=2 =880 Hz fn=2 = ? fn=3 = 1320 Hz fn=3 = ? fn=4 = 1760 Hz fn=4 = ?

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/m2 – threshold of hearing • 1 W/m2 – 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 I0 is the threshold of hearing, 1x10-12 W/m2

32. Logarithms Math Crash-course:

33. Logarithms Math Crash-course: Arithmetic scale Logarithmic scale

34. Logarithms Math Crash-course: then If 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?