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Waves and Harmonic Motion AP Physics M. Blachly. Review: SHO Equation Consider a SHO with a mass of 14 grams: Positions are given in mm.

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Presentation on theme: "Waves and Harmonic Motion AP Physics M. Blachly. Review: SHO Equation Consider a SHO with a mass of 14 grams: Positions are given in mm."— Presentation transcript:

1 Waves and Harmonic Motion AP Physics M. Blachly

2 Review: SHO Equation Consider a SHO with a mass of 14 grams: Positions are given in mm

3 SHO Equation Find the following: T k At t = 3, find y E total a

4 Period of a Harmonic Oscillator For a system with a mass m, and a linear restoring force characterized by k, the period is:

5 Pendulum For small angles, gravity acts as a linear restoring force.

6 Pendulum

7 A 500 g. mass is suspended from a 2 m long string. It is then pulled to a position of x=4 cm and released. Write the equation for the position of this pendulum at any time.

8 Damping The damped Harmonic Oscillator is a system where energy is not conserved. Applet: http://www.lon-capa.org/~mmp/applist/damped/d.htm http://www.lon-capa.org/~mmp/applist/damped/d.htm In the real world, there is always some damping, from friction, air resistance, internal deformations of the medium, etc.

9 Damped SHO Examples Car Spring and shocks Lots of damping is good Radio receivers Very little damping is good

10 Wave Motion Types of Waves: Transverse Longitudinal Torsional Characteristics of Waves Frequency Period Amplitude Wavelength velocity

11 Wave Speed

12 Wave Phenomena Superposition Interference Beats

13 Types of Waves Longitudinal: The medium oscillates in the same direction as the wave is moving Sound Slinky demo Transverse: The medium oscillates perpendicular to the direction the wave is moving. Water (more or less) Slinky demo 8

14 Slinky TnR Suppose that a longitudinal wave moves along a Slinky at a speed of 5 m/s. Does one coil of the slinky move through a distance of five meters in one second? 1. Yes 2. No 5m 12

15 Harmonic Waves Wavelength Wavelength: The distance between identical points on the wave. Amplitude: The maximum displacement A of a point on the wave. Amplitude A A 20 y(x,t) = A cos(  t –kx) Angular Frequency  :  = 2  f x y Wave Number k: k = 2  / Recall: f = v /

16 Period and Velocity l Period: The time T for a point on the wave to undergo one complete oscillation. Speed: The wave moves one wavelength in one period T so its speed is v =  / T. 22

17 TnR Suppose a periodic wave moves through some medium. If the period of the wave is increased, what happens to the wavelength of the wave assuming the speed of the wave remains the same? 1. The wavelength increases 2. The wavelength remains the same 3. The wavelength decreases correct = v T 26

18 l The wavelength of microwaves generated by a microwave oven is about 3 cm. At what frequency do these waves cause the water molecules in my burrito to vibrate ? (a) 1 GHz (b) 10 GHz (c) 100 GHz 1 GHz = 10 9 cycles/sec The speed of light is c = 3x10 8 m/s ACT 29

19 Recall that v = f. 1 GHz = 10 9 cycles/sec The speed of light is c = 3x10 8 m/s HH O Makes water molecules rotate Solution 30

20 Absorption coefficient of water as a function of frequency. f = 10 GHz Visible “water hole” 31 Water and Waves

21 Interference and Superposition When too waves overlap, the amplitudes add. Constructive: increases amplitude Destructive: decreases amplitude 34

22 Reflection A slinky is connected to a wall at one end. A pulse travels to the right, hits the wall and is reflected back to the left. The reflected wave is A) InvertedB) Upright Fixed boundary reflected wave is inverted Free boundary reflected wave upright 37

23 Standing Waves Fixed Endpoints Fundamental n=1 n = 2L/n f n = n v / (2L)

24 Wave Speed

25 Standing Waves: f 1 = fundamental frequency (lowest possible) L  / 2 A guitar’s E-string has a length of 65 cm and is stretched to a tension of 82N. If it vibrates with a fundamental frequency of 329.63 Hz, what is the mass of the string? v = f = 2 (0.65 m) (329.63 s -1 ) = 428.5 m/s v 2 = F /   = F / v 2 m= F L / v 2 = 82 (0.65) / (428.5) 2 = 2.9 x 10 -4 kg


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