Sections Covered in the Text: Selections from Chapter 20, except Figure Examples of a transverse wave (pulse) on a string.

Transcription

1 Traveling Waves Sections Covered in the Text: Selections from Chapter 20, except 20.7 Whole courses in physics are devoted to the study of waves (PHYB20H3S, to name one). Waves occur throughout nature from mechanical waves on a string, to sound waves, to heat waves, to electromagnetic waves. Waves take many forms and move in various kinds of media (or none, as in the case of electromagnetic waves). It is remarkable that the same basic mathematical language can be used in the description of all of them. In this note we survey the properties of traveling waves. We shall continue in Note 20 with the study of the superposition of waves. A Few Words About Medium For the most part, we shall think of a wave as a disturbance that travels through a medium. For example, a mechanical wave can travel on a stretched string the continuum of particles making up the string is the medium. By resultant disturbance of the wave we mean the resultant displacement of the string. In this note we shall also apply these ideas, appropriately modified, to waves on water and to sound waves in air. Eventually, we shall want to apply these ideas to electromagnetic waves that travel through a vacuum. 1 A Wave A wave can be described as a disturbance or a displacement in a medium that changes in form or in position or in both as time goes by. Though the medium supports the propagation of the disturbance from place to place, the material of the medium itself does not move along with the wave. There are two basic types of waves: transverse and longitudinal. In a transverse wave the displacement of the medium is at right angles to the direction of wave propagation (Figure 19-1). In a longitudinal wave, the displacement of the medium is in the direction of the wave propagation (Figure 19-2). A wave may involve a combination of longitudinal and transverse motion. For example, a wave propagating on the surface of water moves the particles of the water in a circular motion (Figure 19-3). We shall not be concerned with this unusual motion here. An important characteristic of a wave is that it can convey energy from one point to another even though no net movement of matter occurs. Waves are often 1 In the case of electromagnetic waves we shall associate the resultant disturbance with the electromagnetic field and not with a medium. We shall cover this subject if time permits. employed to transmit information from one point to another, sound waves and radio waves being good examples. Figure Examples of a transverse wave (pulse) on a string. Figure Example of a longitudinal wave on a slinky. Figure When a wave moves along the surface of water the water particles move in circles. We begin our mathematical description of waves with the simplest kind of wave, a hypothetical transverse wave on a line. 19-1

2 A Transverse Wave on a Line We shall show in due course that the most general expression for a wave pulse moving with speed v along a string or a line, say the x axis, is D = D + + D = f(x vt) + g(x+vt). Here D represents the displacement of the line a transverse displacement if the line is a string and f and g are arbitrary functions of their arguments (x ± vt). What this expression means physically can be seen by studying the two terms separately. The first term is: D + = f(x vt). [19-1] Suppose for convenience that D represents the transverse displacement of a string. And suppose that at t = 0 the displacement is as shown at the left of Figure Thus the function D = f(x) has a peak when x = 0. You should be able to see that the transverse displacement at an arbitrary later time t is given by the function D + = f(x vt). For the moment we consider a wave consisting of only one component, and for convenience the component moving to the right. We shall consider a wave having both components in Note 20. Speed of a Transverse Wave: Calculated From Dimensional Analysis In a moment we shall calculate the speed of a wave on a string by applying Newton s laws. But we can also calculate the speed using the method of dimensional analysis (introduced in Note 02) by making certain assumptions. We assume the speed v depends only on the mass m of the string, its length l and tension T. Since we propose only three variables we can write the speed most generally as v = (const)m a l b T c, [19-2] where a, b and c are constants to be determined in the analysis. This equation is valid only if the dimensions of the left and right hand sides of eq[19-2] are equal. Hence we need to prove [v] = [m a l b T c ], or L.T 1 = M a L b (MLT 2 ) c, [19-3] Figure A general wave pulse represented by the function f(x) at two clocktimes t = 0 and t. At a later time t (at the right side of the figure), the string looks exactly the same as at time t = 0, but the peak has moved to the value of x for which (x vt) = 0. In other words, the peak is now at the position x = d = vt. 2 Since the peak moves a distance d = vt in an elapsed time t, the peak moves in the +x direction with speed v. Therefore, D + = f(x vt) describes a wave of arbitrary (pulse) form propagating with speed v in the positive x direction. By a similar argument the function D = g(x+vt) describes a wave of arbitrary (pulse) form moving with speed v in the negative x direction. The most general form for a wave includes both components D + and D, though we shall usually focus on a wave consisting of only one of the two components. 2 Assuming of course that no dissipative or frictional forces damp or otherwise remove energy from the disturbance where square brackets indicate the operation of taking dimensions. (Tension T is different from dimension T.) In order for eq[19-3] to hold it follows that 0 = a + c (from the powers of M on both sides) 1 = b + c (from the powers of L on both sides) 1 = 2c (from the powers of T on both sides) These equations have a consistent solution. It is a = 1/2, b = 1/2, c = 1/2. So eq[19-3] is true if the speed of the wave is v = (const) Tl m = (const) T µ, [19-4] where µ is the mass per unit length of the string (m/l). We shall see in a moment that a rigorous treatment of this subject will confirm eq[19-4] and yield a value of 1 for the constant. An especially important kind of wave is the sinusoidal wave, which we consider next.

3 A Sinusoidal Wave A sinusoidal wave is the kind of wave that would be set up on a very long string if one end of it were given a periodic displacement (moved up and down continuously), or in air by a loudspeaker driven by a sinusoidal voltage source. Each peak of the wave moves to the right with a constant velocity, and the wave is periodic both in space (at a fixed time t) and in time (at a fixed position x). A snapshot graph at t = 0 of such a wave is shown in Figure Note 19 wave moving in the x-ve x direction we need to substitute the expression (x vt) for x. In doing so we obtain the most general form of the displacement D(x,t) = Asin 2π x vt λ + φ 0 = Asin 2π x λ t T + φ 0 [19-7] A history graph of the wave at x = x 1 and a snapshot graph of the wave at t = t 1 are shown in Figures Here A, λ, and φ 0 have their meanings as just described. T is the period of the wave, or the time (in s) for the wave to repeat itself. Figure Snapshot graph of a sinusoidal wave at a time t = 0. The transverse displacement of the wave at time t = 0 is given by the function D(x,t = 0) = Asin 2π x λ + φ 0. [19-5] You can see that the wave has a number of attributes. A or D max is the amplitude of the wave, or the maximum transverse displacement (in m). λ (in m) is the wavelength of the wave; it is the distance between equivalent points on the wave. In other words, it is the distance between adjacent peaks or the distance between adjacent troughs of the wave. It is also the distance between every other zero of the wave. φ 0 is called the phase constant. The phase constant sets the initial condition of the wave, or the transverse displacement at x = 0. From eq[19-5] we have Figure History and snapshot graphs of the wave. The angular frequency ω (in units of rad.s 1 ), linear frequency f (in units of cycles per second or Hz) and period T (in units of seconds) are related by 3 ω = 2πf = 2π T. [19-8] D(0 m,0 t) = Asinφ 0 so φ 0 = sin 1 D(0,0). A [19-6] If we define the wave number k as k = 2π λ, [19-9] The wave as shown in Figure 19-5 and described by eq[19-5] is a snapshot in time at t = 0. To get the 3 These terms have the same meanings as defined for oscillatory motion in Note

4 and use the relationship v = λf that applies for all waves we have v = λf = 2π k ω 2π = ω k, or ω = νk. [19-10] Substituting eqs[19-8] through [19-10] into eq[19-7] we have for the wave D(x,t) = Asin( kx ωt + φ 0 ). [19-11] Let us consider the analysis of an example sinusoidal wave. Example Problem 19-1 Analyzing a Sinusoidal Wave Thus φ 0 = π 2 rad (π rad.m 1 )(1.0 m) = π 2 rad. (b) With the information from part (a) the wave s displacement is D(x,t) = (1.0 cm) sin (3.14 rad.m 1 )x (628 rad.s 1 )t π 2 (rad) (c) We know that x = 1.0 cm is a wave crest at t = 0 s and that the wavelength is λ = 2.0 m. Because the origin is λ/2 away from the crest at x = 1.0 m, we expect to find a wave trough at x = 0. This is confirmed by calculating D(0 m, 0 t) = 1.0 cm. The snapshot graph of the wave is drawn in Figure A sinusoidal wave of amplitude 1.0 cm and frequency 100 Hz travels at 200 m.s 1 in the positive x direction. At t = 0 s, the point x = 1.0 m is on a crest of the wave. (a) Determine the values of A, v, λ, k, f, ω, T, and φ 0 for this wave. (b) Write the equation for the wave s displacement as it travels. (c) Draw a snapshot graph of the wave at t = 0 s. Solution: (a) From the statement of the problem we have A = 1.0 cm, v = 200 m.s 1, f = 100 Hz. With these values we can calculate λ = v/f = 200/100 = 2.00 m, k = 2π/λ = π rad.m 1 or 3.14 rad.m 1, ω = 2πf = 200π = 628 rad.s 1, T = 1/f = s = 10.0 ms. The phase constant φ 0 is determined by the initial conditions. We know that a wave crest, with displacement D = A, is passing x 0 = 1.0 m at t 0 = 0 s. Eq[19-6] at x 0 and t 0 is ( ) = A = Asin k( 1.0 m) + φ 0 D x 0,t 0 [ ]. This means that sin[k(1.0 m) + φ 0 ] = 1, which requires k(1.0 m) + φ 0 = π 2 rad. Figure A snapshot graph of the wave. Speed of a Transverse Wave: Calculated from Newton s Laws For a wave moving along a string the displacement D of the string, eq[19-11], is just the y-coordinate of a particle in the string at the position x, that is, for such a wave eq[19-11] reduces to y(x,t) = Asin( kx ωt + φ 0 ). [19-12] The transverse velocity of a particle in the string at position x is the time derivative of eq[19-12]: v y (t) = dy(t) dt = ωacos( kx ωt + φ 0 ) [19-13] The transverse velocity vectors of particles in the 19-4

5 string are drawn in Figure Each particle vibrates up and down with simple harmonic motion with a time dependence given by eq[19-13]. Be careful to distinguish the movement of the wave from the movement of the string particles. The string particles move up and down; the wave moves to the right. Note 19 The point of maximum displacement of the string at the peak of the crest is also the point of maximum acceleration of the string particles. This acceleration is given by the time derivative of eq[19-13]: a y (t) = dv y (t) dt = ω 2 Asin( kx ωt + φ 0 ). [19-15] The maximum value of this acceleration is a y = ω 2 A. From eq[19-10], ω = vk where k is the wavenumber. Thus the maximum acceleration is a y = ω 2 A = v 2 k 2 A. [19-16] Figure A snapshot graph of a wave on a string with vectors showing the instantaneous transverse velocity vectors of the string particles at various points. We are now ready to derive an expression for the wave speed. Consider a small segment of the string with length x << λ at a wave crest (Figure 19-9). The tension in the string at either end of this segment pulls downward on the segment tending to bring it back to the equilibrium position. Applying Newton s second law to the segment yields This shows that a large wave speed v causes the string particles to oscillate more quickly and thus to have a larger acceleration. The y-component of the net force acting on the segment of string is ( F net ) y = 2T S sinθ. [19-17] The angle θ is very small ( x << λ) so we can put sinθ tanθ: ( F net ) y 2T S tanθ. [19-18] At this instant of time t = 0, with the crest of the wave at x = 0, the equation of the string is given to a good approximation by y(x) = Acos(kx). Now tanθ = dy(x) dx at x= Δx / 2 = kasin(kx) at x= Δx / 2 = kasin kδx 2 ka kδx = k 2 AΔx, [19-19] 2 2 Figure A small segment of string at the crest of a wave. ( F net ) y = ma y = (µδx)a y [19-14] where m = µ x and µ is the mass per unit length of the string. using the small angle approximation. Substituting eq[19-19] into eq[19-18] we get ( F net ) y = k 2 AT S Δx. [19-20] Using eq[19-16] for a y we have 19-5

6 ( F net ) y = k 2 AT S Δx = (µδx)a y = v 2 k 2 AµΔx. [19-21] Waves in Two and Three Dimensions We have all had the experience of dropping a pebble through the still surface of a pond. Ripples move away from the point where the pebble enters the water surface (Figure 19-10a). The ripple waves are circular, moving in 2D space with peaks and troughs that are readily seen, peaks alternating with troughs. The waves are transverse waves (or at least appear so). Alternate wave peaks are spaced a wavelength apart and the waves travel with a definite speed that can be measured. The speed of such waves is NOT given by eq[19-22]. If we observe such waves a great distance from the source, say as waves on a beach, then we see them as parallel (Figure 19-10b). We see them as parallel wave fronts because the radius of the wave circles is very large. Waves on the ocean originate in storms (or dislocations in the Earth s crust) far out to sea and travel many miles before encountering the shore. Cancelling k 2 A x from both sides we are left with the expression for the wave speed: v = T µ. [19-22] Thus we confirm that the speed of the wave is as given in eq[19-4] with the constant equal to unity. Note that the speed depends only on the mass per unit length and the tension in the string. Specifically, it does not depend on the shape of the wave or pulse. Let us consider a typical calculation. Example Problem 19-2 The Speed of a Wave on a String Calculate the speed of a wave on a stretched string of mass per unit length kg.m 1 under a tension of 10.0 N. Solution: The speed of a wave on a stretched string is given by eq[19-22]: v = T µ = 10.0 N 10 3 kg.m 1 =100 m.s Figure The wave fronts of a circular or spherical wave. In 3D space, waves move away from the source in the form of spherical waves. An example is light. If we observe these waves a great distance from the source then we see them as plane parallel waves (Figure 19-11). Figure A plane wave.

7 Phase and Phase Difference The argument of the sine function, (kx ωt + φ 0 ), in eq[19-12] is called the phase of the wave. It is often denoted just φ. We shall see that the concept of a wave phase will be important in our discussion of interference in Note 20. In fact the wave fronts illustrated in Figures and are surfaces of constant phase. You can see this if you write the displacement as D(x,t) = Asinφ. Because each point on a wave front has the same displacement, it follows that φ must have the same value at every point on a wave front. Let us now examine the notion of phase difference φ between two different points on a sinusoidal wave. Two points are shown on the wave in Figure The phase difference between them is Δφ = φ 2 φ 1 = (kx 2 ωt + φ 0 ) (kx 1 ωt + φ 0 ) = k( x 2 x 1 ) = kδx = 2π Δx λ. [19-23] Note 19 Example Problem 19-3 The Phase Difference between Two Points on a Sound Wave A 100 Hz sound wave travels with a wave speed of 343 m.s 1 (20 C). (a) What is the phase difference between two points 60 cm apart along the direction the wave is traveling? (b) How far apart are two points whose phase differs by 90? Solution: (a) The phase difference between two points at positions x 1 and x 2 is given by eq[19-23]: Δφ = 2π Δx λ. Solving for the wavelength λ with x = 60 cm = 0.60 m we have λ = v f = 343 m.s Hz = 3.43 m. Thus Δφ = 2π 0.60 m = 0.350π rad = m (b) A phase difference of φ = 90 is π/2 radians. Thus solving for x: Δx = Δφ 2π λ = π /2 (3.43 m) = m. 2π Notice that the length of the sound wave (of a relatively low frequency) is of the order of a dimension of an average-sized living room. Figure The phase difference between two points on a sinusoidal wave. Thus the phase difference between any two points on a wave depends on the ratio of their separation x to the wavelength λ. For example, two points on a wave separated by x = λ/2 have a phase difference φ = π. The phase difference between any two adjacent wave fronts ( x = λ) is φ = 2π. Let us consider a numerical example of phase difference. Sound Waves If you examine the cone of a loudspeaker closely enough while the speaker is reproducing a low frequency sound you can actually see the speaker cone move in and out continuously. As the cone moves out it compresses the air molecules in front of it. As the cone moves in it causes the air molecules in front of it to become less dense. These regions of condensation and rarefaction of the air molecules move away from the speaker as a longitudinal wave (Figure 19-13). The distance between alternate condensations or between alternate rarefactions is the wavelength λ. Sound can travel in gases, liquids and in solids. The 19-7

8 speed of sound depends on the properties of the medium. For air at room temperature (20 C) the speed of sound is about 343 m.s 1. In liquids and solids the speed of sound is generally greater than this. The human ear is capable of detecting sound with frequencies between about 20 Hz and 20 khz. Ultrasonic devices operate at frequencies of 40 khz and above. 4 In most calculations we shall use c = 3.00 x 10 8 m.s 1. Notice from Figure that visible light has wavelengths in the range 700 nm to 400 nm (1 nm = 10 9 m). The colors of the corresponding light range from red to blue, respectively. What is referred to as white light, like the light from the sun, was first shown by Isaac Newton to be a superposition of light of all of these colors. Example Problem 19-4 An Example of an Electromagnetic Wave A satellite orbiting Jupiter transmits data to the Earth as a radio wave with a frequency of 200 MHz. (a) What is the wavelength of the electromagnetic wave? (b) How long does it take the signal to travel the 800 million kilometers from Jupiter to the Earth? Figure A representation of a sound wave. Electromagnetic Waves What is loosely referred to as light are waves that can be seen by the human eye. Light is an electromagnetic wave, an oscillation of the electromagnetic field. Radio waves and microwaves are examples of electromagnetic waves that happen, because of their frequencies, to be invisible to us. A section of the electromagnetic spectrum is shown in Figure Visible light occupies a thin slice of spectrum. Solution: Solving for the wavelength we have λ = c f = 3.00x108 m.s 1 = x10 8 m. Hz Solving for the elapsed time we have Δt = Δx c = 8.0 x 1011 m = 2700 s = x 10 8 min. 1 m.s This wavelength, 1.5 m, is about the same as the wavelength used in taxicabs and police radios. Figure The electromagnetic spectrum from 10 6 Hz to Hz. All electromagnetic waves, of whatever type, travel through a vacuum with the same speed. This speed, denoted c, has been measured to have the value 19-8 c = 299,792,458 m.s 1. 4 As does the ultrasonic motion detector used in the first year physics lab in the experiment Linear Motion. Power and Intensity All forms of traveling waves, be they mechanical, sound or electromagnetic, transfer energy from one point to another. The power of a wave is the rate, in joules per second, at which the wave transfers energy. The intensity I of a wave of power P is defined I = P a, [19-24] where a is the area on which the wave impinges (Figure 19-15). The unit of intensity is W.m 2. Obviously, a wave focused into a small area will have a greater intensity than a wave of equal power that is spread out over a large area. For example, sunlight on a summer day can be focussed by a lens onto a

9 pinpoint on a piece of paper to burn a hole in it. According to eq[19-24] if a source of spherical waves (Figure 19-16) of power P radiates uniformly in all directions then the power at distance r is spread uniformly over the surface of a sphere of radius r. Thus for a spherical wave Note 19 Eq[19-25] means that for a source of a certain power the intensity I 1 at a distance r 1 and intensity I 2 at a distance r 2 are related by I 1 2 = r 2 2 I 2 r 1 [19-26] For any wave type of amplitude A, the intensity can be written I = CA 2, [19-27] where C is a constant of the particular wave type. Figure Plane waves of power P impinge on area a with intensity I = P/a. I(r) = P source 4πr 2. [19-25] Thus the intensity decreases with the square of the distance. This is referred to as rectilinear propagation. An early researcher into the physics of sound was none other than the inventor of the telephone, Alexander Graham Bell ( ). In researching the physiology of hearing Bell found that the human ear responds to sound intensity in a logarithmic, rather than linear, fashion. Bell concluded that sound intensity I is most logically expressed as the logarithm of an intensity ratio, that is, as I # in db = 10log 10, I 0 [19-28] where db stands for decibel (named after Bell, the basic unit being the bel) and I o = W.m 2 is the approximate intensity of the weakest sound that can be heard by the human ear. The normal ear can distinguish sounds that differ in intensity by 1 db. An intensity of 120 db causes severe discomfort. This concludes our survey of wave attributes. We shall continue in Note 20 with the effects when waves coexist at the same point in space. Figure A source emitting uniform spherical waves. Shown here is a 2D slice. 19-9

10 To Be Mastered Definitions: medium, wave, transverse wave, longitudinal wave General expression for: speed of a wave on a string General expression for: sinusoidal wave Definitions: period T, amplitude A, wavelength λ, frequency f, angular frequency ω, wave number k, phase constant φ 0, phase φ, phase difference φ Calculation of: speed of a wave on a stretched string: v = T µ Calculation of: phase φ, phase difference φ Description of: sound waves, electromagnetic waves Definitions: power P, intensity I, rectilinear propagation Typical Quiz/Test/Exam Questions 1. Explain the difference between a longitudinal and a transverse wave. 2. Give one example of each of the above, and also mention one kind of wave that is neither longitudinal nor transverse