Traveling Waves. Wave variables are λ v 1) Wavelength, λ y 2) Period, T 3) Frequency, f=1/t 4) Amplitude, A x 5) Velocity, v T

Transcription

1 1 Traveling Waves Having discussed simple harmonic motion, we have learned many of the concepts associated with waves. Particles which participate in waves often undergo simple harmonic motion. Section 20.1, 2 Wave motion Examples of waves: Ocean waves, waves in crops, earthquakes, speech is a wave in the air, waves in ropes.. Electromagnetic waves travel in a vacuum and involve electric and magnetic fields. Covered in P102. We will largely be talking about mechanical waves in Physics 101 but there will be some discussion of light waves later. Classic example of a wave is the wake following a boat Waves have oscillatory behaviour and operate in both space and time. i.e. at any instant in time, the wave has a characteristic shape and at any point in space, the wave has a time dependence. e.g. the bobbing we experience when we are moved up and down by the wake of as boat. A wave is a disturbance which propagates through a system. Although individual masses oscillate about their equilibrium positions, they do not move with the wave. A wave transports energy not matter. Wave variables are λ v 1) Wavelength, λ y 2) Period, T 3) Frequency, f=1/t 4) Amplitude, A x 5) Velocity, v T y Arrangement in space at a fixed time. Time course at a fixed point in space. y t A single wave pulse. Boat wake. x or t Wave velocity: The wave travels a distance λ in time T; then v = λ/t = λf These are called traveling waves.

2 2 Transverse and Longitudinal waves A travelling wave that causes the particles of the disturbed medium to move perpendicular to the wave motion is called a transverse wave. e.g. waves on a rope A travelling wave that causes the particles of the medium to move parallel to the direction of wave motion is called a longitudinal wave. e.g waves in a spring, sound Sound waves involve longitudinal movement of air particles which results in compression and expansion and density changes. Ocean waves are a combination of transverse and longitudinal, each particle undergoes circular motion. There is no net displacement of molecules in an ocean wave or a sound wave. Note: Ocean waves are surface waves. Earthquakes have two types of waves: 1) P (primary or pressure) longitudinal waves which travel through the earth at about 7-8 km/s and 2) S (secondary or shear) transverse waves which travel along the earth's surface at 4-5 km/s. Transverse waves are not conducted by a fluid since a fluid cannot transmit shear forces. This is why secondary earthquake waves are not transmitted through the earth's core. It is possible to estimate the distance to the earthquake by recording the times of arrival of the two types of wave. Note that waves can be one dimensional, two dimensional or three dimensional. Examples of 1D waves: waves on a rope, air moving along a tube 2D waves: surface waves like ocean waves, crop waves 3D waves: earthquakes, sound in the lecture theatre, light

3 3 Now consider a general wave or wave pulse. e. g. a one dimensional travelling wave (on a rope), We assume that the shape of the wave doesn't change as it moves down the rope. y A v t=0 P x vt P y v t=t The -vt part of the function ensures that the wave moves to the right with velocity v. If you consider a point P on the rope, it is distance x from the wave crest at time 0 and a distance x-vt at time t. x The wave function, D, represents the y coordinate (x coordinate if we were talking about a longitudinal wave) of any point of the medium at time t. The wave has a velocity v. At t= 0, D= D(x), At t= t, D = D(x-vt) when the wave is travelling to the right, assuming that v is positive. Example wave function: y = 2 Show plots for t=0,2,5. (x-5t) What do we write if the wave pulse were traveling to the left? What determines the velocity of a wave? For example, suppose I send a wave pulse down a rope. What determines how fast the wave travels to the end of the rope? What determines how long it takes my voice to reach the back of the lecture theatre? Is it affected by how fast I jerk the rope? Or how loud I shout? Will a wave pulse move faster along a thin rope than along a thick rope?

4 4 Velocity of transverse waves Consider a transverse travelling wave moving with velocity v in a string under tension T s and with mass per unit length of µ (kg/m). Consider a small arc of the traveling wave from a reference frame which moves with the wave at velocity v to the right. The acceleration of the string in this moving frame is v 2 /R (centripetal acceleration). T s θ R θ θ Note that for a travelling wave the medium oscillates about a fixed point while the wave moves at velocity v to the right. T r θ T s In this frame, the segment of string is moving to the left with velocity v and following a curved arc. The arc represents an angle change of 2θ relative to the centre of curvature of the path. The horizontal components of the tensions cancel. The net force is down in the radial direction, equivalent to circular motion about the centre of the arc. ΣT r = 2T s sinθ, use sinθ = θ for a small section of arc. Then set T r = ma: 2T s θ = mv 2 /R Mass of rope segment m = sµ = 2Rθµ, Then 2T s θ = 2Rθµv 2 /R Make a right angle triangle with the hypotenuse from the centre of the axis. Then it is straightforward to work out that the lower angle is θ. Large µ thick rope; small µ thin rope v = T s /µ The velocity of the wave increases with increasing string tension and decreases with increasing string mass per unit length. Velocity of longitudinal waves in a fluid (e.g sound in a liquid or gas) Consider a wave pulse traveling in a fluid in a long tube of cross-sectional area A, with a piston in one end. At time t=0, the piston starts to move to the right at a speed v'. In time t, it has moved a distance v't and the longitudinal wave it produces will move at the speed of sound, v, a distance vt. vt v't A P 0 + P v' Sound wave front P o, ρ v Tube area A Note that the speed of the wave, as before, is a property of the medium and not determined by the mechanism which is producing the wave. We assume v < v;e.g. v < the speed of sound

5 5 Net force on the compressed region of the fluid: F net = (P o + P)A - P o A = A P The impulse given to the compressed fluid is F net t = mv' (recall that impulse = change in momentum) Here, (ρavt) is the mass of fluid that was given the velocity v' when the piston moved. A Pt = (ρavt)v' => P = ρvv' The Bulk modulus of the fluid is defined as B = - P/( V/V o ) = -ρvv'/( V/V o ) The bulk modulus is the ratio of the change in pressure to the fractional volume change. Note that an increase in pressure leads to a decrease in volume. See Chapter 15 (15-6) for more details. Where V o is the original volume, Avt, and V = -Av't The assignment of V o and V is obvious from the figure. So, B = -ρvv'/(-av t/avt) = ρv 2 And v = B/ρ Problem 2.00 m θ L/4 L/4 T 1 A B L/2 T s At point A: Mg Light string of mass 10.0g and length L = 3.00m. Walls separated by 2.00 m Each object has mass 2.00 Kg How long does it take a wave to travel from A to B? µ = m/l =.01Kg/3.00m = Kg/m Sin θ =[(2.00m L/2)/2]/L/4 = [(2m-3/2m)/2]/[3m/4] = 0.25m/0.75m θ = 19.4 o Cosθ = Fy = T 1 cosθ-mg =0 T 1 = mg/cosθ = (2.00kg)(9.81kg)/(0.943) = N F x = T 1 sinθ -T s =0 then T s = T 1 sinθ = 20.81N/3 =6.935N

6 6 Then the speed of travel of a travelling wave in this rope is v = T s /µ = 6.935N/ Kg/m = 45.64m/s. Then the time required for a pulse to travel from A to B is t = (3m/2)/45.64m/s = 32.9ms Section 20.3 Sinusoidal waves Consider these two representations of the same travelling wave. We want an equation which will take into account both the spatial and temporal aspects of the wave motion. D D λ T v x We use the letter D for displacement in order to have the treatment general for both transverse and longitudinal waves. t Note that the sinusoidal wave repeats itself (i.e. covers 2π radians) every wavelength spatially and every period temporally. Then D(x,t) = D M sin [(2π/λ) x - (2π/T) t + φ ] where φ is the phase at t=0 Recall that 2π/T = ω and we now define 2π/λ = k the wave number. Then D(x,t) = D M sin(kx - ωt + φ) Note that v = λ/t = λf = (2π/k)(ω/2π) = ω/k Note that here, k = the wave number, not the spring constant, k, from Hooke s Law. The transverse speed of a particle in this traveling wave is given by: v y = dd(x,t) = -ωd M cos(kx - ωt + φ) dt Note: v y is not the propagation velocity, v.

7 7 Problem from December 2004 Final 8 B D(cm) -4-8 A x (m) v Zero crossings at 4, 9, 14, 19, 24, 29 The plot above represents the displacement of a wave at t = 0.10 s. This wave has a frequency of 5.0 Hz and is traveling in the negative x direction along a string (a) What is the speed of the wave? (b) Write an equation describing the wave including all constants (c) What is the transverse speed at point A? at point B? (d) Sketch the displacement at t=0 sin(π-φ) = sinφ, hence we consider π/6 and 5π/6 (a) From inspection of the plot. λ = 0.10m, then v = fλ = 5.0 Hz 0.10m = 0.50 m/s (b) Amplitude = 8.0 cm, k = 2π/λ = 20π m, ω = 2πf = 10π rad/s -1 At x=0, and t=0.1s, we have -4.0cm =8.0cm sin(ωt + φ); ωt =π. Note sin (π + φ) = -sinφ, φ = sin -1 (½); wave moving to the left, so φ = π/6 or 5π/6; Consider how the plot evolves from x=0 at t=0.1s, it is at phase π+π/6 and going down, If we used 5π/6, the curve would go up, then φ= π/6. Then x = 8cm sin(20πx + 10πt + π/6) (c) at A, the transverse speed is ωa = (10π s -1 )(8cm) = 80π cm/s; at B v = 0; (d) Move the plot to the right a distance v(0.1s) = 0.05m. Now the plot starts at displacement 4 cm and increases. Consider a wave traveling on a rope which is made up of two sections with different diameters. Then the speed of the wave is v 1 in the left side and v 2 on the right side. v 1 /v 2 = (µ 2 /µ 1 ) ½ using v i = T s /µ i. T s T s t=0 t=t v 1 v 2 Actually this is a µ 1 µ 2 bit more complicated because some of the wave will be reflected back at λ 1 = v 1 T λ 2 = v 2 T the boundary. On the right hand side, the wave crests move at velocity v 1 and wave crests travel one wavelength in time v 1 T = λ 1.

8 8 Imagine that a wave crest reaches the junction at t=0. The crest will undergo a discontinuous change in velocity at the junction from v 1 to v 2. A time T later, the junction will undergo the next wave crest. In this time, the previous wave crest will have moved a distance v 2 T = λ 2. Then λ 1 /λ 2 = v 1 /v 2 and the period (and frequency) is unchanged. The period cannot change at the junction, since if it did, adjacent particles in the rope would get out of phase with each other and the rope would break! Problem A cord with two sections T s µ 1 = 0.10 kg/m µ 2 = 0.20 kg/m λ 1 λ 2 T s Incident wave function: D = (0.050m) sin(6.0x t) Find (a) λ 1, (b) T s and (c) λ 2 The wave function on the other side will be different From the wave function, D = D M sin(kx - ωt) where k=2π/λ and ω = 2π/T (a) Then λ 1 = 2π/k 1 = 2π/6.0 m -1 = 1.047m, k 1 = 6.00 m -1 (b) v 1 = ω 1 /k 1 = T s /µ 1 => T s = µ 1 [ω 1 /k 1 ] 2 = (0.10 kg/m)[12.0 s -1 /6.0 m -1 ] 2 = 0.40N (c ) Since the tensions are equal, and the period (or ω) is equal on each side, what changes are v and λ. v 2 = T s /µ 2 = ω 2 /k 2 = ω 2 λ 2 /2π => λ 2 = 2π (T s /µ 2 )/ω 2 λ 2 =2π (0.40N/0.20kg/m)/12.0 s -1 = m

9 9 Section 20.5 Sound and light (Read Giancoli Chapter 16 sections 1,2,3 on sound) Sound (light later) Creation of a sound pulse by moving a piston in a gas chamber High density region v We already derived a relation for the velocity of sound. Regardless of how we set up a sound wave, its propagation velocity depends only upon the medium. For a liquid or gas or a solid volume, use v = B/ρ where B is the bulk modulus of the medium and ρ is the density of the medium. See table 20-1, page 629, for velocities of sound for various materials. v s varies from 343m/s on air to 1480m/s in fresh water to > 6000 m/s in rock and aluminum. Sounds have pitch (frequency) and loudness (intensity). Three categories of sound waves (based on pitch): 1) Infrasonic - low frequency (< 20Hz) 2) Audible - 20 to 20,000 Hz 3) Ultrasonic - high frequency (> 20,000 Hz) Infrasonic waves are created by earthquakes, thunder, and vibrating heavy machinery. Ultrasonic waves have several important applications. E.g ultrasonic imaging and auto focus for cameras.

10 10 Mathematical representation of longitudinal waves Note that the displacements change direction at the points of maximum and minimum displacement The higher pressure regions are called compressions and the lower pressure regions are called expansions. Both regions move with the speed of sound. The distance between two successive compressions is called the wavelength λ. This is just like a slinky; air pressure is scaled by the brightness of blue. Displacement D along the wave: x Note that this displacement is along the direction of motion of the wave. There are displacement minima (i.e. zero) when the pressure is minimum or maximum. Pressure difference P along the wave Note that the pressure change is 90 o out of phase with the displacement. D = D M sin(kx - ωt + φ) P = - P M cos(kx - ωt + φ) As always, sin (θ + ½π) = cos(θ). Derivation of the equation for P: Consider a volume A x of gas of thickness x and cross-sectional area A A For each position along the tube, there is a gas displacement D from the equilibrium position. D is the change in this displacement, across the distance x. x D x The change in volume V (=A x) accompanying the pressure change is V (= A D) where D is the difference between D at x and at x + x. Then P = -B V/V = -BA D/A x = -B D x This is the definition for bulk modulus: See Section 15-6 Partial derivatives are used here we are considering the system at a fixed time.

11 11 using D = D M sin(kx - ωt), we get P = -BkD M cos(kx - ωt) = - P M cos(kx-ωt) where P M is the pressure amplitude. Using B = ρv 2 and k = v/ω, P M = ρv 2 (ω/v)d M = ρvωd M Section 20.6 Power and Intensity Waves and energy are two of the key concepts in Physics. v Area A, width l = vt We draw a small volume of the medium of area A and width vt, All the molecules in this volume participate in the wave motion. Consider a sinusoidal transverse (or longitudinal) wave moving at velocity v through an elastic medium. Each particle in the medium moves with simple harmonic motion and each particle has an energy E = ½kD m 2 where D m is the maximum amplitude of the motion for that particle. Recall that ω 2 = k/m or k =ω 2 m. Then E = ½ω 2 md m 2 For 3D waves, m = ρv where V = Al =Avt, so we can write E = ½ω 2 ρavtd m 2 This simple equation shows us that the energy transported by a wave is proportional to the square of the amplitude, the wave velocity and to the square of the frequency. If we divide the Energy by time, we get the average power (or rate of energy transfer) P = E/t = ½ω 2 ρavd m 2 Energy is like money, power is like a salary and intensity across unit area perpendicular to the flow I = P/A = ½ω 2 ρvd m 2 For a 3D wave in an isotropic medium (e.g. sound), known as a spherical wave, I = P/A =P/4πr 2 So I α 1/r 2 the 'so called' inverse square law. For two points as distances r 1 and r 2 from the source, we have I 1 /I 2 = r 2 2 /r 1 2

12 12 Note that if the Intensity decreases as 1/r 2 then the amplitude, D m must decrease as 1/r for a spherical wave. Intensity of Sound: Decibels (Not in Knight) The expressions we derived above for transverse waves hold equally well for longitudinal waves. The only difference is the direction of the displacement. From before: I = P/A = ½ρv(ωD M ) 2 or P M 2 /2ρv The faintest sounds the ear can hear have I = 1.00x watts/m 2 Using the above equations and ρ air = 1.29 kg/m 3, v = 343 m/s and ω = 1.00 khz, we find P M = (2ρvI) ½ = [2(1.29 kg/m 3 )(343m/s)(1.00x10-12 watts/m 2 )] ½ =2.97x10-5 N/m 2 compared to P o of 1.013x10 5 N/m 2 D M = P M /(ρvω) = 2.97x10-5 N/m 2 )/[(1.29kg/m 3 )(343m/s)(2π(1.00x10 3 )] =1.07 x10-11 m = about 1/10 the size of a hydrogen atom. The loudest sounds the ear can process have an intensity of 1.0 watts/m 2. This corresponds to a pressure P M of 29.7 N.m 2 and D M of 1.07x10-5 m or 11 µ. The ear is a sensitive device. The same is true of our noses and our eyes! We hear a range of displacements spanning 10 6 and intensities spanning 10 12! Sound level in decibels The ear can hear sounds with intensities as low as 1x10-12 watts/m 2 and as high as about 1.0 watts/m 2. Also, the ear hears logarithmically, i.e. when the sound intensity is increased by a factor of 10, we perceive a factor of two in loudness. Because this range is so large, a logarithmic scale has been devised. The sound level, β, is defined by: β = 10 log(i/i o ) then I = I o 10 β/10 where I o is the reference intensity, taken to be 1.00x10-12 watts/m 2, i.e. the threshold of hearing. The units are decibels or db The threshold of hearing I o corresponds to 0 db and the threshold of pain, 1.00 watts/m 2 corresponds to 120 db. Look at Table 16.2 to compare sound levels from different sources. Rock concerts are too loud (120 db) and cause ear damage. Whisper 20 db, Conversation 65dB, car interior at 90km/hr 75dB,

13 13 Problem: A loudspeaker is placed between two observers who are 110m apart along the line connecting them. If one observer records a sound level of 60.0 db and the other a sound level of 80.0 db, how far is the speaker from each observer? I 1 I 2 r 1 r 2 Using β = 10 log(i/io), or β/10 = log(i/i o ) and I/I o = 10 β/10 I 1 = Io10 (60.0/10) = (1.00x10-12 watts/m 2 )x 10 6 = 1.00x10-6 watts/m 2 I 2 = (1.00x10-12 watts/m 2 )x 10 8 = 1.00 x 10-4 watts/m 2 I 1 /I 2 = r /r 1 then r 1 /r 2 = (I 2 /I 1 ) ½ = [(1.00x10-4 )/(1.00x10-6 )] ½ = 10 Then r 1 + r 2 = 110m and r 1 /r 2 = 10 so r 1 = 100m and r 2 = 10m. Section 20.7 The Doppler Effect Consider an observer O and a sound source S. The air is stationary and the observer moves with velocity, v o. The source velocity is v S. The sound has frequency f o, wavelength λ and velocity v. The observer registers sound frequency by the number of wave fronts which they detect per unit time. O S v O λ v S Wave fronts move from source to observer with velocity v, the speed of sound relative to the air which is stationary. When v O and v S are both zero, the observer counts f o wave fronts per second. This corresponds to the frequency of sound that the observer hears. If v O = 0 and v S 0, the wavefronts are no longer separated by λ. When v S > 0, i.e. the source moves towards the observer, λ' = λ - λ = λ - v S T = λ - v S /f o The sound velocity is unchanged Then the frequency heard by the stationary observer is f' = v/λ' = v / (λ - v S /f o ) = v/ [ v/f o - v S /f o ]. Then f = f o = f o v (1 - v S /v) (v v S ) The perceived frequency increases if the source is moving toward the observer

14 14 If the source moves away from the observer, the wavelength is increased to λ + v S T giving f = f o = f o v (1 + v S /v) (v + v S ) The perceived frequency decreases if the source is moving away from the observer. When v S = 0 and v O is not zero, the relative velocity of wave fronts to observer is v' = v O + v but the wavelength λ is unchanged so the frequency, f', increases: f' = v'/λ = (v + v O )/λ since λ = v/f o, we get f' = (1 + v O /v)f o = f o (v + v O )/v The perceived frequency increases if the observer is moving toward the source If the observer is moving away from the source, the relative speed of the sound waves and the observer is v' = v - v O and f' = (1 - v O /v)f o = f o (v v O )/v The perceived frequency decreases if the observer is moving away from the source. To summarise all the above derivations, we can write f' = (v ± v O ) f o (v Ŧ v S ) Toward the sound means an increase in frequency and away from the sound means a decrease in frequency. where the upper signs correspond to the source and observer moving towards each other and the lower signs correspond to motion away from each other. Problem: A bat moving at 5.0 m/s is chasing a flying insect. If the bat emits a 40.0 khz chirp and receives back an echo at 40.4 khz, at what relative speed is the bat moving toward or away from the insect? v B v I We shall assume that the insect moves away from the bat with speed v I. Then frequency heard by the insect is f I = (v - v I )f o (v - v B ) The insect is listening. The insect is moving away from the sound, the bat is moving towards the sound. The frequency heard back by the bat again after the sound reflects from the insect is f B = (v + v B )f I (v + v I ) Now the bat is listening. Then f B = (v - v I )(v + v B )f o (v + v I )(v - v B ) Solve to get 40.4 = 40.0 ( )(343 - v I )

15 15 ( ) (343 + v I ) (343 + v I ) = (40.0/40.4)(348/338)(343- v I ) = (343 - v I ) v I ( ) = 343( ); v I = 6.651/ = 3.29 m/s Then the bat is closing in on the insect at 5.00m/s m/s = 1.71 m/s Echolocation in Bats Bats make use of sound waves to locate insects which they eat. Bat species make use of two approaches. 1) They emit short sound pulses of duration a one to two ms. These pulses are both amplitude and frequency modulated. The frequency changes from 70kHz to 30 khz during the pulse. When searching for insects, bats emit these pulses about 4 times per second. Once they have found a reply, the rate of sending pulses can increase to as high as 200 per second. They maximise the reply until they can scoop up the insect. Considerations: The frequencies of 70kHz to 30kHz correspond to λ = v s /f = 5mm to 11mm; the approximate size of the insect. Given that the pulses are sent out 4 times /s, at what range must an insect be located in order that its reflection of the sound gets back to the bat before the next pulse? There are 0.250s between pulses. A sound wave will travel x = vt = (343m/s)(0.250s) = 86 m. This wave travels to the insect and back so the distance to the insect is 86m/2 = 43m. Another factor is the amount of power reflected back from the insect compared to the threshold of hearing on the bat. If the bat produces P = 10-5 Watts, and its hearing threshold is 1x10-12 watts/m 2 (human); what is the maximum distance from which it can detect an echo. Assume the insect has cross sectional area 1x10-4 m 2 (1cmx1cm). The P of 10-5 watts is spread over all directions in a spherical wave. The fraction which strikes the insect is P i = P(a insect /4πr 2 ) where r is the distance between the bat and the insect. The intensity of reflected sound from the insect follows I i = P i. /4πr 2. Then the intensity returning to the bat is I B = Pa insect /(4πr 2 ) 2. If the threshold intensity is 1x10-12 watts/m 2 ; Then r 4 = a insect P/(I B (4π) 2 ) = (1x10-4 m 2 )(1x10-5 watts)/[(1x10-12 watts/m 2 )(4π) 2 ] = 6.33m 4 And r = 1.6m a more realistic minimum distance from which a bat can detect an echo from its meal. (Some bats can direct their sound waves in a narrow beam so that more of the power gets to the insect.) 2) The second type of echolocation makes use of the Doppler effect. This species of bat uses a long pulse of about 0.1s long and frequency 60.0 khz and takes advantage of the Doppler effect to detect the reflection at 61.8 khz from the insect. By maximising the

16 intensity of the Doppler reflected sound which he can hear while sending out his pulse, the bat can move to the location of his prey. 16