PH 221-3A Fall Waves - I. Lectures Chapter 16 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition)

Transcription

1 PH 1-3A Fall 010 Waves - I Lectures 5-6 Chapter 16 (Halliday/Resnick/Walker, Fundaentals of Physics 8 th edition) 1

2 Chapter 16 Waves I In this chapter we will start the discussion on wave phenoena. We will study the following topics: Types of waves Aplitude, phase, frequency, period, propagation speed of a wave Mechanical waves propagating along a stretched string Wave equation Principle of superposition of waves Wave interference Standing waves, resonance

3 Waves A wave is a vibrational, trebling otion in an elastic, deforable body. The wave is initiated by soe external force that acts on soe parts of the body and defors it. The elastic restoring force counicates this initial disturbance fro one part of the body to the next, adjacent part. The disturbance gradually spreads outward through the entire elastic body. The elastic body in which the wave propagate is called the ediu. When a wave propagates through a ediu, the particles in the ediu vibrate back and forth, but the ediu as a whole does not perfor translational otion. 3

4 Transverse Wave Motion 1 up If we shake one end of the tightly stretched elastic string up and down with a flick of the wrist a disturbance travels along the string. String is regarded as a row of particles connected by sall, assless springs. nd up 1 down down 1 particle up => it will pull the nd particle up. Transverse wave otion the direction of the disturbance propagation is perpendicular to the back and forth vibration of the particles. 4

5 Longitudinal Wave Motion Disturbance generated by pushing the 1 st particle towards the nd Longitudinal wave - direction of disturbance otion is parallel to the vibration of the particles.

6 The wave is a propagation of the disturbance through the ediu without any net displaceent of the ediu. Consider a horizontal string connected to the ass executing siple haronic otion in the vertical direction. Each particle transits the otion to the next particle along the entire length of the string. The resulting wave propagates in the horizontal direction with a velocity v, while any one particle of the string executes siple haronic otion in the vertical direction. The particle of the string is oving perpendicular to the direction of wave propagation and is not oving in the direction of the wave. A transverse wave is a wave in which the particles of the ediu execute siple haronic otion in a direction perpendicular to its direction of propagation. 6

7 A wave is defined as a disturbance that is self-sustained and propagates in space with a constant speed Waves can be classified in the following three categories: 1. Mechanical waves. These involve otions that are governed by Newton s laws and can exist only within a aterial ediu such as air, water, rock, etc. Coon exaples are: sound waves, seisic waves, etc.. Electroagnetic waves. These waves involve propagating disturbances in the electric and agnetic field governed by Maxwell s equations. They do not require a aterial ediu in which h to propagate but they travel through vacuu. Coon exaples are: radio waves of all types, visible, infra-red, and ultra-violet light, x- rays, gaa rays. All electroagnetic waves propagate in vacuu with the sae speed c = 300,000 k/s 3. Matter waves. All icroscopic particles such as electrons, protons, neutrons, atos etc have a wave associated with the 7 governed by Schroedinger s equation.

8 Transverse and Longitudinal waves Waves can be divided into the following two categories depending on the orientation of the disturbance with respect to the wave propagation velocity v. If the disturbance associated with a particular wave is perpendicular to the wave propagation velocity, this wave is called " transverse". An exaple is given in the upper figure which depicts a echanical wave that propagates p along a string. The oveent of each particle on the string is along the y-axis; the wave itself propagates along the x-axis. A wave in which the associated disturbance is parallel to the wave propagation velocity is known as a " longitudinal wave". An exaple of such a wave is given in the lower figure. It is produced by a piston oscillating in a tube filled with air. The resulting wave involves oveent of the air olecules along the axis of the tube which is also the direction of the wave propagation velocity v 8.

9 Consider the transverse wave propagating along the string as shown in the figure. The position of any point on the string can be described by a function y h( x, t). Further along the chapter we shall see that function h has to have a specific for to describe a wave. Once such suitable function is: yxt (, ) y sin kx- t Such a wave which h is described d by a sine (or a cosine) function is known as " haronic wave". The various ters that appear in the expression for a haronic wave are identified in the lower figure Function y (xt, ) depends on x and t. There are two ways to visualize it. The first is to "freeze" tie (i.e. set t to). This is like taking a snapshot of the wave at t t. y y x, t The second is to set x o o x. In this case y yx, t. o o 9

10 yxt (, ) y sin kxt The aplitude y is the absolute value of the axiu displaceent fro the equilibriu position. The phase is defined as the arguent kx t of the sine function. The wavelength is the shortest distance between two repetitions of the wave at a fixed tie. We fix t at t 0. We have the condition: y( x1,0) y ( x1,0) ysin kx1 ysin kx1 ysin kx1k Since the sine function is periodic with period k k A period T is the tie it takes (with fixed x ) the sine function to coplete one oscillation. We take x 0 y(0, t) y(0, t T) ysin t ysin tt ysin ttt T

11 v k The speed of a traveling wave In the figure we show two snapshots of a haronic wave taken at ties t and tt. During the tie interval t the wave has traveled a distance x. The wave speed x v. One ethod of finding v is to iagine that t we ove with the sae speed along the x-axis. In this case the wave will see to us that it does not change. Since yxt (, ) y sin kxt this eans that the arguent of the sine function is constant. kx t constant. We take the derivative with respect to t. dx dx dx k 0 The speed v dt dt k dt k A haronic wave that propagates along the negative x-axis is described by the equation: yxt (, ) y sin kx t. The function yxt (, ) hkx t describes a general wave that propagates along the positive x-axis. A general wave that propagates along the negative x-axis is described by the equatio n: y( xt, ) hkx t 11

12 A particular wave is given by y = (0.00)sin[( )x (8.0rad/s)t] Find: a) A e) f h) y at x = and t = 0.500s b) K f) t c) λ g) v 1

13 Proble 5. If yxt (, ) (6.0 )sin( kx(600 rad/ st ) ) describes a wave traveling along a string, how uch tie does any given point on the string take to ove between displaceents y=+.0 and y=-.00? Let y 1 =.0 (corresponding to tie t 1 ) and y =.0 (corresponding to tie t ). Then we find and kx + 600t 1 + = sin 1 (.0/6.0) kx +600t 1 + = sin (.0/6.0) 0). Subtracting equations gives Thus we find t 1 t = s (or 1.1 s). 600(t 0) 1 t ) = sin 1 (.0/6.0) sin 1 (.0/6.0). 0) 13

14 Proble 9. A transverse sinusoidal wave is oving along a string in the positive direction of an x axis with a speed of 80 /s. At t=0, the string particle at x=0 has a tranverse displaceent of 4.0 c fro its equilibriu position and is not oving. The axiu transverse speed of the string particle at x=0 is 16 /s. (a) What is the frequency of the wave? (b) What is the wavelength of the wave? If the wave equation is of the for yxt (, ) y sin( kxt) what are (c) y,( d) k, ( e), (f), and (g) the correct choice of sign in front of? (a) Recalling fro Ch. 1 the siple haronic otion relation u = y,wehave rad/s. Since = f, we obtain f = 64 Hz. (b) Using v = f, we find = 80/64 = (c) The aplitude of the transverse displaceent is (d) The wave nuber is k = / = 5.0 rad/. y 4.0 c (e) The angular frequency, as obtained in part (a), is (f) The function describing the wave can be written as y s i n 5x 400t 16 / rad/s. where distances are in eters and tie is in seconds. We adjust the phase constant to satisfy the condition y = at x = t = 0. Therefore, sin = 1, for which the siplest root is = /. Consequently, the answer is y sin 5x 400 t. (g) The sign in front of is inus. 14

15 Proble 9. Use the wave equation to find the speed of a wave given by yxt x s t (, ) (00 (.00 )[(0 ) (4.0 (40 )]. The wave yxt (, ) (.00 )[(0 ) x (4.0 s ) t] 1 1 1/ is of the for hkx ( t) 1 angular wave nuber k 0 and angular frequency 4.0 rad/s. Thus, the speed of the wave is 1 v/ k(4.0 rad/s)/(0 ) 0.0 /s. with 15

16 v Wave speed on a stretched string Below we will deterine the speed of a wave that propagates along a string whose linear ass density is. The tension on the string is equal to. Consider a sall section of the string of length. The shape of the eleent can be approxiated to be an arc of a circle of radius R whose center is at O. The net force in the direction of O is F sin. Here we assue that 1 sin F (eqs.1) R R v v The force is also given by Newton's second law: F = (eqs.) R R v If we copare equations 1 and we get: v R R Note : The speed v depends on the tension and the ass density but not on the wave frequency f. 16

17 Proble : The drawing shows a 15.0 kg ball being whirled in a circular path on the end of a string. The otion occurs on a frictionless horizontal table. The angular speed of the ball is ω = 1.0 rad/s. The string has a ass of kg. How uch tie does it take for a wave on the string to travel fro the center of the circle to the ball? 17

18 The Speed of a Wave on a String Proble To easure the acceleration due to gravity on a distant planet, an astronaut hangs a kg ball fro the end of a wire. The wire has length of 1.5 and a linear density of 3.1 x 10-4 kg/. Using electronic equipent, the astronaut easures the tie for a transverse pulse to travel the length of the wire and obtains a value of s. The ass of the wire is negligible copared to the ass of the ball. Deterine the acceleration due to gravity. 18

19 The Speed of a Wave on a String Proble A horizontal wire is under a tension of 315 N and has a ass per unit length of 6.50 x 10-3 kg/. A transverse wave with an aplitude of.50 and a frequency of 585 Hz is traveling on this wire. As the wave passes, a particle of the wire oves up and down in siple haronic otion. Obtain: a) The speed of the wave b) The axiu speed with which the particle oves up and down 19

20 Proble 3. A 100 g wire is held under a tension of 50 N with one end at x=0 and the other at x=10.0. At tie t=0, pulse 1 is sent along the wire fro the end at x=10.0. At tie t=30.0 s, pulse is sent along the wire fro the end at x=0. At what position x do the pulses begin to eet? The pulses have the sae speed v. Suppose one pulse starts fro the left end of the wire at tie t = 0. Its coordinate at tie t is x 1 = vt. The other pulse starts fro the right end, at x = L, where L is the length of the wire, at tie t = 30 s. If this tie is denoted by t 0 then the coordinate of this wave at tie t is x = L v(t t 0 ). They eet when x 1 = x, or, what is the sae, when vt = L v(t t 0 ). We solve for the tie they eet: t = (L + vt 0 )/v and the coordinate of the eeting point is x = vt = (L + vt 0 )/. Now, we calculate the wave speed: L (50 N)(10.0) v 158/s kg Here is the tension in the wire and /L is the linear ass density of the wire. The coordinate of the eeting point is x (158/s)( s) This is the distance fro the left end of the wire. The distance fro the right end is L x = ( ) =.63. 0

21 Rate of energy transission Consider a transverse wave propagating along a string which is described by the equation: yxt (, ) y sin kx- t. The transverse velocity y u - y cos kx- t At point a both t u and K are equal to zero. At point b both u and K have axia. 1 In general the kinetic energy of an eleent of ass d is given by: dk dv 1 dk dx - cos - The rate at which kinetic energy propagates y kx t dk 1 along the string is equal to - dt v y cos kx t The average rate dk 1 1 v y cos - kx t v y As in the case of the oscillating dt avg 4 avg spring-ass syste du dk du dk P dt dt dt dt avg avg avg avg avg 1 v y

22 Proble 6. A string along which waves can travel is.70 long and has a ass of 60 g. The tension in the string is 36.0 N. What ust be the frequency of traveling waves of aplitude 770fortheaveragepowertobe850W? 7.70 to 85.0 P avg 1 v y Using Eq for the average power and Eq for the speed of the wave, we solve for f = /: v f 1 P 1 (85.0 W) 198 Hz. / (36.0 N)(0.60 kg /.70 ) avg 3 y ( )

23 y 1 y x v t θ The wave equation Consider a string of ass density and tension A transverse wave propagates along the string. The transverse otion is described by yxt (, ) Consider an eleent of length dx and ass d dx The forces F F 1 the net force along the y-axis is given by the equation: F F sin F sin sin sin Here we ynet y assue that 1 1 and 1 sin1 tan1 x and sin y y y tan Fynet x x x 1 1 y y y 1 Fro Newton's second law we have: Fynet day dx t x x y y y y y x x 1 y y 1 y dx x x 3 1 t dx x t v t

24 The Superposition of Waves A superposition principle: when two or ore waves arrive at any given point siultaneously, the resultant instantaneous deforation is the su of the individual instantaneous deforations. The waves do not interact, they have no effect on one another Each wave propagates as though hthe others were not present The net displaceent of the ediu is the vector su of the individual displaceents The superposition principle is very well satisfied for waves of low aplitude. For waves of very large aplitude the superposition principle fails, because the first wave alters the properties of the ediu and therefore affects the behavior of a second wave propagating on the sae string, air, etc. (ediu)

25 y( x, t) y ( x, t) y ( x, t) 1 The principle of superposition for waves 1 The wave equation y y y even though x t v t it was derived for a transverse wave propagating along a string ti under tension, is true for all lltypes of waves. This equation is "linear" which eans that if y1 and y are solutions of the wave equation, the function cy 1 1 cy is also a solution. Here c1 and c are constants. The principle of superposition is a direct consequence of the linearity of the wave equation. This principle can be expressed as follows: Consider two waves of the sae type that overlap at soe point P in space. Assue that the functions y ( x, t) and y ( x, t) describe the displaceents 1 if the wave arrived at P alone. The displaceent at P when both waves are present is given by: y( x, t) y ( x, t) y ( x, t) 1 Note : Overlapping waves do not in any way alter the travel of each other 5

26 Interference of waves Consider two haronic waves of the sae aplitude and frequency which propagate along the x-axis. The two waves have a phase difference. We will cobine these waves using the principle of superposition. The phenoenon of cobing waves is knwon as interference and the two waves are said to interfere. The displaceent of the two waves 1 y sin y x t y kxt y y1 y, sin sinkx are given by the functions: y ( x, t) and (, ) sin. y x t y kx t y t y x, t y cos sin kxt The resulting wave has the sae frequency as the original waves, and its aplitude y ycos Its phase is equal to kx t 6

27 Constructive ti interference The aplitude of two interefering waves is given by: y y cos It has its axiu value if 0 In this case y y The displaceent of the resulting wave is: y x, t y sin kx t This phenoenon is known as fully constructive interference 7

28 Destructive interference The aplitude of two interefering waves is given by: y ycos It has its iniu value if In this case y 0 The displaceent of the resulting wave is: y x, t 0 This phenoenon is known as fully destructive interference 8

29 Interediate interference The aplitude of two interefering waves is given by: y ycos When interference is neither fully constructive nor fully destructive it is called interediate interference An exaple is given in the figure for 3 In this case y y The displaceent of the resulting wave is: y x, ty sinkxt 3 Nt Note: Soeties the phase difference is expressed as a difference in wavelength In this case reebre that: radians 1 9

30 Pr oble 3. What phase difference between two identical traveling waves oving in the sae direction along a stretched string, results in the cobined wave having an aplitude 1.50 ties that of the coon aplitude of the two cobining waves? Express your answer in (a) degrees, (b) radians, (c) wavelengths? y ycos (a) Let the phase difference be. Then fro Eq. 16 5, y cos(/) = 1.50y, which gives 1 150y 1.50y cos 8.8. y (b) () Converting to radians, we have = 1.45 rad. (c) In ters of wavelength (the length of each cycle, where each cycle corresponds to rad), this is equivalent to 1.45 rad/ = 0.30 wavelength. 30

31 Phasors A phasor is a ethod for representing a wave whose displaceent is: y x, t y sin kx t 1 1 The phasor is defined as a vector with the following properties: 1. Its agnitude is equal to the wave's aplitude y 1. The phasor has its tail at the origin O and rotates in the clockwise direction about an axis through O with angular speed. Thus defined, the projection of the phasor on the y-axis (i.e. its y-coponent) is equal to 1 y sin kxt A phasor diagra can be used to represent ore than one waves. (see fig.b). The displaceent of the second wave is: y x, t y sin kx t The phasor of the second wave fors an angle with the phasor of the first wave indicating that it lags behind wave 1 by a phase angle. 31

32 Wave addition using phasors Consider two waves that have the sae frequency but different aplitudes. They also have a phase difference. The displaceents of the two waves are: y xt, y sin kx t and 1 1 y x, t y sin kx t. The superposition of the two wavesyieldsawavethathasthesaeangular a has the sae angular frequency and is described by: y y sin kxt Here y is the wave aplitude and is the phase angle. To deterine y and we add the two phasors representing the waves as vectors (see fig.c). Note: The phasor ethod can be used to add vectors that have different aplitudes. 3

33 Proble 35. Two sinusoidal waves of the sae frequency travel in the sae direction along a string. If y 3.0 c, y 4.0 c, 0, and / rad, what is the alitude of the resultant wave? 1 1 The phasor diagra is shown below: y 1 and y represent the original waves and y represents the resultant wave. The phasors corresponding to the two constituent waves ake an angle of 90 with each other, so the triangle is a right triangle. The Pythagorean theore gives y y1 y (3.0 c) (4.0 c) (5c). Thus y = 5.0 c. 33

34 y xt, y sinkx cost Standing Waves : Consider the superposition of two waves that have the sae frequency and aplitude but travel in opposite directions. The displaceents y1x t y kxt yx t y sin kx t y x t y1 x t y x t, sin sin sin cos of two waves are:, sin,, The displaceent of the resulting wave,,, y x t y kx t y kx t y kx t This is not a traveling wave but an oscillation that has a position dependent aplitude. It is known as a standing wave. 34

35 The displaceent of a standing wave is given by the equation: y x, t y sinkx cost The position dependant aplitude is equal to y sin kx a a Nodes : These are defined as positions where the standing wave aplitude vanishes. They occur when kx n n 0,1,, x n xn n n 0,1,,... n n n n a Antinodes : These are defined as positions where the standing wave aplitude is axiu. 1 They occur when kx n n 0,1,, xn x n n n 0,1,,... Note 1 : The distance between ajacent nodes and antinodes is / Note : The distance between a node and an ajacent antinode is /4

36 Standing waves and resonance Consider a string under tension on which is claped A B at points A and B separated by a distance L. We send a haronic wave traveling to the right. The wave is reflected at poi nt B and the reflected wave travels to A B A B the left. The left going wave reflects back at point A and creates a thrird wave traveling to the right. Thus we have a large nuber of overlapping waves hlf half of which travel to the right and the rest to the left. For certain frequencies the interference produces a standing wave. Such a standing wave is said to be at resonance. The frequencis at which the standing wave occurs are known as the resonant frequencie s of the syste. 36

37 A B Resonances occur when the resulting standing wave satisfies the boundary condition of the proble. These are that the Aplitude ust be zero at point A and point B and arise fro the fact that the string is claped at both points and therefore cannot ove. A B The first resonance is shown in fig.a. The standing 1 wave has two nodes at points A and B. Thus L A B 1 L. The second standing wave is shown in fig.b. It has three nodes (two of the at A and B) In this case L The third standing wave is shown in fig.c. It has four nodes (two of the at A and B) In this case L 3 3 The general expression for the resonant 3 L L v v wavelengths is: n n 1,,3,... the resonant frequencies fn n n L L n

38 Proble 43. A string fixed at both ends is 8.40 long and has a ass of 0.10 kg. It is subjected to a tension of 96.0 N and set oscillating. (a) What is the speed of the waves on the string? (b) What is the longest possible wavelength for a standing wave? (c) Find the frequency of the wave. (a) The wave speed is given by v, where is the tension in the string and is the linear ass density of the string. Since the ass density is the ass per unit length, = M/L, where M is the ass of the string and L is its length. Thus v L (96.0 N) (8.40 ) 8.0 /s. M 0.10 kg (b) The longest possible wavelength for a standing wave is related to the length of the string by L = /, so = L = (8.40 ) = (c) The frequency is f = v/ = (8.0 /s)/(16.8 ) = 4.88 Hz. 38

39 Proble 50. A rope, under a tension of 00N and fixed at both ends, oscillates in a second haronic standing wave pattern. The displaceent of the rope is given by y (0.10 )(sin x/ )sin1 t, where x=0 at one end of the rope, x is in eters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the wave on the rope, and (c) () the ass of the rope? (d) () If the rope oscillates in a third haronic standing wave pattern, what will be the period of oscillation? Since the rope is fixed at both ends, then the phrase second-haronic standing wave pattern describes the oscillation shown in Figure 16 3(b), where (see Eq ) L and f = v. L (a) Coparing the given function with Eq , we obtain k = / and = 1 rad/s. Since k = / then y x, t y sin kx cos t 4.0 L 4.0. (b) Since = f then f 1 rad/s, which yields f 6.0 H z v f 4 /s. (c) Using Eq. 16 6, we have v 4 /s which leads to = 1.4 kg. 00 N /(4.0 ) (d) () With The period is T = 1/f = 0.11 s. f 3v 3(4 /s) L (4.0 ) 9.0 Hz 39