A wave is a disturbance that propagates energy through a medium without net mass transport.

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1 Waes A wae is a disturbance that propagates energy through a medium without net mass transport. Ocean waes proide example of transerse waes in which if we focus on a small olume of water, at a particular location, we would see that it moes (mostly) up and down as the wae moes by.

2 Sound waes propagating through the atmosphere proide example of a longitudinal wae. Consider a speaker oscillating at 440 Hz (middle C). For one half cycle it pushes the air in front of it, compressing the air, thereby increasing the pressure just in front of it. This pressurized olume of air pushes on the adjacent olume of air pressurizing it, and so on. The half cycle has launched a disturbance of pressurized air propagating out from the speaker. Before this disturbance has traeled ery far, the diaphragm quickly draws back, in the second half of the cycle now rarifying the air in front of it.

3 This train of pressurized/rarified regions propagate through the air. Here the air molecules slosh back and forth along the direction of the wae motion but not ery far as the waes moe to the right. pressure We can plot these pressure ariations ersus position and immediately see the similarity with other waes to recognize that these hae a common description.

4 A conenient way to represent waes is to draw wae fronts. Wae fronts are points of constant phase on a wae, often chosen as the crests of the propagating wae (moing with it). A point source radiates its wae energy radially outwards in all directions. wae fronts moing radially outwards λ

5 Since waes transport energy from one place to another and energy is consered the radiation pattern (shape of the waefronts) is important. To be concrete we consider a spherical speaker that launches sound waes radially outwards in all directions. The speaker radiates sound energy with a power (energy radiated/time) P. If we draw a spherical shell of radius r around the speaker then (ignoring energy losses to friction in the air) since the shell intercepts all the radiated power, the total power intercepted is the emitted power P. r Something that will be true for any radius shell that we draw (independent of r).

6 But a sound detector (like an ear) doesn t typically surround a source. Rather it has a typically (much) smaller area that intercepts only a portion of any passing waes. This makes it sensible to define for waes the intensity which is the (power crossing)/(unit area) (with the area perpendicular to the waes direction of propagation): I = P A For a spherically radiating source the intensity at distance r from the source is, P source power I = 4πr area of shell of radius r r The intensity falls off with the distance from the source squared.

7 Suppose that the source radiates with P = 0 mw. Then at meter the intensity is while at 0 meters the intensity is If the detector is an ear with an area of cm = 0-4 m then the power it intercepts, P, is P = I A ear 3 P 0 0 W 4 W I = = = πr 4 π( m) m 3 P 0 0 W 6 W I = = = πr 4 π(0 m) m So at m, 4 W 4 8 P = (0 m ) = W m and at 0 m, 6 W 4 0 P = (0 m ) = W m R

8 Note that this fall off of the intensity with the inerse square distance from a spherically radiating source is really a consequence of the conseration of energy. I P = 4 π r As the waes spread out spherically they distribute the same power (the emitted source power) oer an eer increasing area (the surface area of increasing radius shells). If instead of spherical waes we had plane waes the sound could go much further without reduction of the intensity. Plane wae fronts

9 The demo in the lobby with the two large parabolic reflectors does just that. If you whisper in your softest oice into one reflector it conerts the hemi-spherical waes emanating from your mouth into plane waes. Despite the long distance across the lobby to the other dish the plane waes get there, little affected by the distance. At the second dish the plane waes are reconerted to spherical waes that focus down to an ear, where what you said can be heard.

10 To deelop some of the characteristic features of waes we consider waes induced on a string by a ibrating rod. ibrating rod The rod here ibrates with a constant frequency (cycles/s) inducing transerse waes in the string that propagate to the right. Any particular point (P) on the string executes simple harmonic oscillations, up and down with an amplitude A (the maximum displacement), a period T (time per cycle) and a frequency f (cycles/s).

11 The period of this oscillation is set by the period with which the rod ibrates. Since frequency and period are inersely related (f = /T) the frequency with which point P oscillates is also set by the frequency with which the rod oscillates. So f (and thus T) are established by the source (generally true). Now consider two adjacent crests of the resulting wae. The distance between them is defined as the waelength λ.

12 The time taken for the wae to moe by one waelength is clearly one cycle which gies the wae elocity as: = λ T Since f = T This can also be expressed as, = λf The wae elocity is gien by the waelength times the frequency.

13 Now in contrast to the frequency or period which is set by the source. The wae elocity is set by properties of the medium. If the tension in the string is F and its linear density (mass/length) is µ then the wae speed down the string is gien by (GRR p. 397) m F kg N m = s = = µ kg kg s m m

14 Example A 00 cm long taught string has a mass of 6.00 g. A wae is found to trael at 0.0 m/s on this string. A second string has the same length and tension but half the mass of the first. What will the speed of the wae be on the second string? F = µ F = µ Diide F µ µ = = F µ µ = µ µ

15 But told that m = m Then Also, gien that m 0 s µ = µ = m m = L L = So, µ µ µ = = µ µ = µ m m = (0.0 ) = 4. s s

16 Example A transerse wae on a taught string has amplitude A, waelength λ and speed. A point of the string only moes in the transerse direction. Call its maximum speed Tmax. In terms of A and λ what is? T max T

17 T Since each point of the string undergoes simple harmonic oscillations the transerse displacement of a point is, y = Acosωt And its transerse speed is, T = ωasin ωt The sine function repeatedly oscillates between ± so T max = ωa

18 T But ω= πf so, =ω A = πfa T max The wae speed is = λf So T max πfa A = = π λf λ

19 HITT F = -kx m x(t) For the simple harmonic oscillator the acceleration of the mass has maximum magnitude at point: x max A D x(t) t x max B C

20 HITT The piston in an internal combustion engine executes approximately simple harmonic motion. When the engine is running at 3000 rpm the period for the piston s motion is? A) s B) 0.00 s C) 50 s D) 3000 s

21 Reflection of waes at boundaries Two cases: boundary fixed or free Fixed end case Reflection inerted Reflection not inerted

22 Wae interference Consider two single pulse waes on the same rope propagating in opposite directions. They pass through each other and continue on completely unaffected by the collision. During the time when they occupy the same segment of rope they obey the principle of superposition. That is the net displacement at each point becomes simply the sum of the indiidual displacements at each point.

23 The superposition principle is a signed summation so that if one of the pulses has a negatie displacement then the resultant wae (the sum) can be smaller than one or both waes as they cross. If the two waes are mirror images of each other then as they cross the net amplitude is zero.

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