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Chapter 14: Wave Motion Types of mechanical waves Mechanical waves • are disturbances that travel through some material or substance called medium for the waves. • travel through the medium by displacing particles in the medium • travel in the perpendicular to or along the movement of the particles or in a combination of both transverse waves: longitudinal waves: waves in a string etc. sound waves etc. waves in water etc. Types of mechanical waves (cont’d) Longitudinal and transverse waves sound wave = longitudinal wave C = compression air compressed R = rarefaction air rarefied Types of mechanical waves (cont’d) Longitudinal-transverse waves Types of mechanical waves (cont’d) Periodic waves • When particles of the medium in a wave undergo periodic motion as the wave propagates, the wave is called periodic. wavelength l A amplitude t=0 x=0 t=T/4 t=T period x Mathematical description of a wave Wave function • The wave function describes the displacement of particles in a wave as a function of time and their positions: y y ( x, t ) ; y is displaceme nt at x, t • A sinusoidal wave is described by the wave function: sinusoidal wave moving in y ( x, t ) A cos[ (t x / v )] +x direction A cos[ ( x / v t )] angular frequency velocity of wave, NOT of A cos 2 f ( x / v t ) particles of the medium 2 f fl v A cos 2 ( x / l t / T ) wavelength y ( x, t ) A cos[ (t v / x )] period f 1/ T sinusoidal wave moving in -x direction v->-v phase velocity Mathematical description of a wave (cont’d) Wave function (cont’d) y ( x, t ) A cos 2 ( x / l t / T ) l wavelength y( x l , t) y ( x, t T ) t=0 x=0 t=T/4 t=T period x Mathematical description of a wave (cont’d) Wave number and phase velocity wave number: k 2 / l y ( x, t ) A cos( kx t ) phase The speed of wave is the speed with which we have to move along a point of a given phase. So for a fixed phase, kx t const. dx / dt / k v phase velocity y ( x, t ) A cos( kx t ) A cos[ k ( x vt)] Mathematical description of a wave (cont’d) Particle velocity and acceleration in a sinusoidal wave y ( x, t ) A cos( kx t ) u in textbook v y ( x, t ) y ( x, t ) / t A sin( kx t ) velocity a y ( x, t ) 2 y ( x, t ) / t 2 2 A cos( kx t ) 2 y ( x, t ) Also acceleration y ( x, t ) / x k A cos( kx t ) k y ( x, t ) 2 2 2 2 2 y ( x, t ) / x 2 (k 2 / 2 ) 2 y ( x, t ) / t 2 y ( x, t ) / v t 2 2 2 wave equation Mathematical description of a wave (cont’d) General solution to the wave equation 2 y ( x, t ) k 2 2 y ( x, t ) 2 y ( x, t ) 2 2 2 2 2 x t v t Solutions: y ( x, t ) f ( x vt) such as wave equation cos( kx t ) The most general form of the solution: y ( x, t ) f ( x vt) g ( x vt) Speed of a transverse wave Wave speed on a string F2 y •The mass of the segment is m x. F1x F F1 F2 •Consider a small segment of string whose length in the equilibrium position is x. F1 y F2 x F • The x component of the force (tension) at both x x x x Newton’s 2nd law ends have equal in magnitude and opposite in direction because this is a transverse wave. • F1 y / F (y / x) x , F2 y / F (y / x) x x • The total y component of the forces is: Fy F1 y F2 y F [( y / x ) x x (y / x) x ] x( 2 y / t 2 ) mass acceleration Speed of a transverse wave (cont’d) Wave speed on a string (cont’d) F2 y F F1 • The total y component of the forces is: F2 F F1 y x x Fy F1 y F2 y F [( y / x ) x x (y / x) x ] x( 2 y / t 2 ) [( y / x ) x x (y / x ) x ] / x ( / F )( 2 y / t 2 ) x 0 2 y / x 2 ( / F )( 2 y / t 2 ) v F / ( restoring force) /(inertia ) wave eq. Energy in wave motion Total energy of a short string segment of mass dm dx F2 y F F1 F2 F a • At point a, the force F1 y does work on the work done string segment right of point a. • Power is the rate of work done : P( x, t ) F1 y ( x, t )( y ( x, t ) / t ) t 0 F (y ( x, t ) / x)(y ( x, t ) / t ) F1 y y ( x, t ) A cos( kx t ) (y / x ) kAsin( kx t ) x x (y / t ) A sin( kx t ) P( x, t ) dE / dt FkA2 sin 2 (kx t ) v 2 A2 sin 2 (kx t ) vk, v F / 2 Pmax F 2 A2 sin 2 (kx t ) Energy in wave motion (cont’d) Maximum power of a sinusoidal wave on a string: Pmax F 2 A2 Average power of a sinusoidal wave on a string • The average of sin 2 ( kx t ) over a period: 1 2 2 0 sin 2 d • The average power: 1 2 Pave (1/ 2) F 2 A2 Wave intensity Wave intensity for a three dimensional wave from a point source: P I 2 4r r1 in units of W/m 4r I 4r I 2 1 1 r2 I1 r22 2 I 2 r1 2 2 2 2 power/unit area Wave interference, boundary condition, and superposition The principle of superposition • When two waves overlap, the actual displacement of any point at any time is obtained by adding the displacement the point would have if only the first wave were present and the displacement it would have if only the second wave were present: y( x, t ) y1( x, t ) y2 ( x, t ) Wave interference, boundary condition, and superposition (cont’d) Interference • Constructive interference (positive-positive or negative-negative) • Destructive interference (positive-negative) Wave interference, boundary condition, and superposition (cont’d) Reflection • Free end incident wave reflected wave y ( x, t ) A cos( k x t ) B cos( k x t ) For x<xB xB At x=xB (y ( x, t ) / x ) x xB 0 B A Vertical component of the force at the boundary is zero. Wave interference, boundary condition, and superposition (cont’d) Reflection (cont’d) • Fixed end y ( x, t ) A cos( k x t ) B cos( k x t ) For x<xB At x=xB y ( x, t ) x x B 0 B A Displacement at the boundary is zero. Wave interference, boundary condition, and superposition (cont’d) Reflection (cont’d) • At high/low density Wave interference, boundary condition, and superposition (cont’d) Reflection (cont’d) • At low/high density Standing waves on a string Superposition Superposition of two waves moving in the same direction of two waves moving in the opposite direction Standing waves on a string (cont’d) Superposition of two waves moving in the opposite direction creates a standing wave when two waves have the same speed and wavelength. incident reflected y( x, t ) y1 ( x, t ) y2 ( x, t ) A cos(kx t ) A cos(kx t ) 2 A(sin k x)(sin t ) sin kx 0 when kx n or x n / k nl / 2 N=node, AN=antinode (n 0,1,2,..) Normal modes of a string There are infinite numbers of modes of standing waves l1 / 2 Ln fundamental l 2 (n 1,2,3,...) l2 first ln 2 L / n overtone 3l3 / 2 v 1 F fn n f1 2L 2L second overtone 2l4 third overtone fixed end L fixed end Sound waves Sound • Sound is a longitudinal wave in a medium • The simplest sound waves are sinusoidal waves which have definite frequency, amplitude and wavelength. • The audible range of frequency is between 20 and 20,000 Hz. Sound waves (cont’d) Sound wave (sinusoidal wave) Sinusoidal sound wave function: y ( x, t ) A cos( kx t ) Change of volume: V S ( y2 y1 ) S[ y ( x x, t ) y ( x, t )] y1 y( x, t ) y2 y( x x, t ) undisturbed cyl. of air disturbed cyl. of air Pressure: bulk modulus S pressure B p( x, t ) /( dV / V ) x x dV / V y ( x, t ) / x (V Sdx) x+x p( x, t ) B(y ( x, t ) / x ) BkA sin( kx t ) Pressure amplitude and ear Pressure amplitude for a sinusoidal sound wave p( x, t ) BkA sin( kx t ) • Pressure: • Pressure amplitude: Ear pmax BkA Perception of sound waves Fourier’s theorem and frequency spectrum • Fourier’s theorem: Any periodic function of period T can be written as y (t ) n [ An sin( 2f nt ) Bn cos(2f nt )] fundamental freq. where f1 1 / T , f n nf1 (n 1,2,3,...) • Implication of Fourier’s theorem: Perception of sound waves Timbre or tone color or tone quality Frequency spectrum noise music piano piano Speed of sound waves (ref. only) velocity of wave The speed of sound waves in a fluid in a pipe movable piston pA longitudinal momentum carried by the fluid in motion pA fluid in original volume of the fluid in equilibrium motion change in volume of the fluid in motion vt v yt ( p p ) A v y vy vy vy fluid in motion ( vtA)v y Avt velocity of fluid Av y t p bulk modulus B: -pressure change/frac. vol. change ( Av t ) /( Avt) y pA change in pressure in the fluid in motion fluid at rest boundary moves at speed of wave p B vy v Speed of sound waves (ref. only) (cont’d) The speed of sound waves in a fluid in a pipe (cont’d) longitudinal impulse = change in momentum pAt B vy v speed of a longitudinal wave in a fluid The At vtAvy v B speed of sound waves in a solid bar/rod v Y ,Y Young’s modulus Speed of sound waves (cont’d) The speed of sound waves in gases B p0 bulk modulus of a gas p0 In textbook speed of a longitudinal wave in a fluid v ratio of heat capacities equilibrium pressure of gas - P in textbook (background p0 pressure). - density M RT R gas constant 8.314472 J/(mol K) T temperature in Kelvin M molar mass Sound level (Decibel scale) Decibel scale As the sensitivity of the ear covers a broad range of intensities, it is best to use logarithmic scale: I , I 0 10 12 W/m 2 Definition of sound intensity: (10 dB) log I0 ( unit decibel or dB) Sound intensity in dB Intensity (W/m2) Military jet plane at 30 m 140 102 Threshold of pain 120 1 Whisper 20 10-10 Hearing thres. (100Hz) 0 10-12 Standing sound waves Sound wave in a pipe with two open ends Standing sound waves Standing sound wave in a pipe with two open ends Standing sound waves Sound wave in a pipe with one closed and one open end Standing sound waves Standing wave in a pipe with two closed ends Displacement Normal modes Normal modes in a pipe with two open ends 2nd normal mode ln 2L L n or ln (n 1,2,3,...) 2 n v fn n (n 1,2,3,...) 2L Normal modes Normal modes in a pipe with an open and a closed end (stopped pipe) ln 4L L n or ln (n 1,3,5,...) 4 n v fn n ( n 1,3,5,...) 4L Resonance Resonance • When we apply a periodically varying force to a system that can oscillate, the system is forced to oscillate with a frequency equal to the frequency of the applied force (driving frequency): forced oscillation. When the applied frequency is close to a characteristic frequency of the system, a phenomenon called resonance occurs. • Resonance also occurs when a periodically varying force is applied to a system with normal modes. When the frequency of the applied force is close to one of normal modes of the system, resonance occurs. Interference of waves Two sound waves interfere each other destructive constructive d1 d2 d 1 d 2 nl ( n 1 / 2 )l n 0,1,2,.... (constructive) (destructive) Beats Two interfering sound waves can make beat Two waves with different frequency create a beat because of interference between them. The beat frequency is the difference of the two frequencies. Beats (cont’d) Two interfering sound waves can make beat (cont’d) Suppose the two waves have frequencies f a and f b . For simplicity, consider two sinusoidal waves of equal intensity: ya (t ) A sin 2f a t ; yb (t ) A sin 2f bt Then the resulting combined wave will be: 1 1 ya (t ) yb (t ) 2 A sin[ ( 2 )( f a f b )t ] cos[ (2 )( f a f b )t ] 2 2 1 1 ( sin a sin b 2 sin (a b) cos (a b)) 2 2 As human ears does not distinguish negative and positive amplitude, they hear two max. or min. intensity per cycle, so 2 x (1/2)|fa-fb|= |fa-fb| is the beat frequency fbeat. Doppler effect Moving listener Source at rest Listener moving right Source at rest Listener moving left Doppler effect (cont’d) Moving listener (cont’d) •The wavelength of the sound wave does not change whether the listener is moving or not. • The time that two subsequent wave crests pass the listener changes when the listener is moving, which effectively changes the velocity of sound. freq. listener hears freq. source generates velocity of sound at source velocity of listener fL fs v vL fL v vL l v vL v / fs - for a listener moving away from + for a listener moving towards the source. Doppler effect (cont’d) Moving source When the source moves Doppler effect (cont’d) Moving source (cont’d) • The wave velocity relative to the wave medium does not change even when the source is moving. • The wavelength, however, changes when the source is moving. This is because, when the source generates the next crest, the the distance between the previous and next crest i.e. the wavelength changed by the speed of the source. The source at rest v l s fs When the source is moving l v vs v vs fs fs fs + for a receding source - for a approaching source Doppler effect (cont’d) Moving source and listener fL v vL l v vL fs v vs - for a listener moving away from + for a listener moving towards the source. + for a receding source - for a approaching source The signs of vL and vS are measured in the direction from the listener L to the source S. Effect of change of source v vs speed v vs Doppler effect (cont’d) Example 1 • A police siren emits a sinusoidal wave with frequency fs=300 Hz. The speed of sound is 340 m/s. a) Find the wavelength of the waves if the siren is at rest in the air, b) if the siren is moving at 30 m/s, find the wavelengths of the waves ahead of and behind the source. a) l v / f s 340 m/s /300 Hz 1.13 m. b) In front of the siren: l (v vs ) / f s (340 m/s - 30 m/s)/300 Hz 1.03 m Behind the siren: l (v vs ) / f s (340 m/s 30 m/s)/300 Hz 1.23 m Doppler effect (cont’d) Example 2 • If a listener l is at rest and the siren in Example 1 is moving away from L at 30 m/s, what frequency does the listener hear? v 340 m/s fL fs (300 Hz) 276 Hz. v vs 340 m/s 30 m/s Example 3 • If the siren is at rest and the listener is moving toward the left at 30 m/s, what frequency does the listener hear? v vL 340 m/s - 30 m/s) fL fs (300 Hz) 274 Hz. v 340 m/s Doppler effect (cont’d) Example 4 • If the siren is moving away from the listener with a speed of 45 m/s relative to the air and the listener is moving toward the siren with a speed of 15 m/s relative to the air, what frequency does the listener hear? v vL 340 m/s 15 m/s fL v vs fs 340 m/s 45 m/s (300 Hz) 277 Hz. Example 5 • The police car with its 300-MHz siren is moving toward a warehouse at 30 m/s, intending to crash through the door. What frequency does the driver of the police car hear reflected from the warehouse? 340 m/s Freq. reaching fW v f s (300 Hz) 329 Hz. v vs 340 m/s 30 m/s the warehouse Freq. heard by the driver fL v vL 340 m/s 30 m/s fW (329 Hz) 358 Hz. v 340 m/s Exercises Problem 1 A transverse wave on a rope is given by: y ( x, t ) (0.750 cm) cos [( 0.400 cm 1 ) x (250 s 1 )t ] (a) Find the amplitude, period, frequency, wavelength, and speed of propagation. (b) Sketch the shape of the rope at the following values of t : 0.0005 s, and 0.0010 s. (c) Is the wave traveling in the +x or –x direction? (d) The mass per unit length of the rope is 0.0500 kg/m. Find the tension. (e) Find the average power of this wave. Solution (a) y ( x, t ) A cos 2 ( x / l t / T ) A=0.75 cm, l=2/0.400 = 5.00 cm, f=125 Hz, T=1/f=0.00800 s and v=lf=6.25 m/s. (b) Homework (c) To stay with a wave front as t increases, x decreases. Therefore the wave is moving in –x direction. (d) v ( F / ) , the tension is F v 2 (0.050 kg / m)(6.25 m / s)2 19.6 N . (e) Pav (1 / 2) F 2 A2 54.2 W . Exercises Problem 2 A triangular wave pulse on a taut string travels in the positive +x direction with speed v. The tension in the string is F and the linear mass density of the string is . At t=0 the shape of the pulse I given by 0 h( L x ) / L h( L x ) / L 0 y ( x ,0) for x L for L x 0 for 0 x L for x L (a) Draw the pulse at t=0. (b) Determine the wave function y(x,t) at all times t. (c) Find the instantaneous power in the wave. Show that the power is zero except for –L < (x-vt) < L and that in this interval the power is constant. Find the value of this constant. Solution y (a) h -L L x Exercises Problem 2 (cont’d) Solution (b) The wave moves in the +x direction with speed v, so in the experession for y(x,0) replace x with –vt: y ( x, t ) 0 h( L x vt) / L h( L x vt) / L 0 for x L for L x 0 for 0 x L for x L (c) y y P ( x, t ) F x t F (0)0 0 for x L F (h / L)( hv / L) Fv(h / L)2 for L x 0 F ( h / L)( hv / L) Fv(h / L)2 F (0)( 0) 0 for 0 x L for x L Thus the instantaneous power is zero except for –L < (x-vt) < L where It has the constant value Fv(h/L)2. Exercises Problem 3 The sound from a trumpet radiates uniformly in all directions in air. At a distance of 5.00 m from the trumpet the sound intensity level is 52.0 dB. At what distance is the sound intensity level 30.0 dB? Solution The distance is proportional to the reciprocal of the square root of the intensity and hence to 10 raised to half of the sound intensity levels divided by 10: I P / 4d 2 , 10 log( I / I 0 ) I I 010 /10 , I d 2 I1 / I 2 101 /10 2 /10 (d 2 / d1 ) 2 d110( 1 /10 2 /10) / 2 d 2 (5.00 m)10( 5.203.00) / 2 62.9 m. Exercises Problem 4 An organ pipe has two successive harmonics with frequencies 1,372 and 1,764 Hz. (a) Is this an open or stopped pipe? (b) What two harmonics are these? (c) What is the length of the pipe? Solution (a) For an open pipe, the difference between successive frequencies is the fundamental, in this case 392 Hz, and all frequencies are integer multiples of this frequency. If this is not the case, the pipe cannot be an open pipe. For a stopped pipe, the difference between the successive frequencies is twice the fundamental, and each frequency is an odd integer multiple of the fundamental. In this case, f1 = 196 Hz, and 1372 Hz = 7f1 , 1764 Hz = 9f1 . So this is a stopped pipe. (b) n=7 for 1,372 Hz, n=9 for 1,764 Hz. (c) f1 v /(4L), so L v /(4 f1 ) (344 m / s) /(784 Hz ) 0.439 m. Exercises Problem 5 Two identical loudspeakers are located at points A and B, 2.00 m apart. The loudA speakers are driven by the same amplifier and produce sound waves with a frequency of 784 Hz. Take the speed of sound in air to be 344 m/s. A small microphone is moved out from Point B along a line perpendicular to the B line connecting A and B. (a) At what distances from B will there be destructive interference? (b) At what distances from B will there be constructive interference? (c) If the frequency is made low enough, there will be no positions along the line BC at which destructive interference occurs. How low must the frequency be for this to be the case? 2.00 m C x Exercises Problem 5 Solution (a) If the separation of the speakers is denoted by h, the condition for destructive interference is x 2 h 2 x l, where is an odd multiple of one-half. Adding x to both sides, squaring, canceling the x2 term from both sides and solving for x gives: x [h 2 /( 2l) l / 2]. Using l v / f and h from the given data yields: 9.01 m for 1 / 2, 2.71 m for 3 / 2, 1.27 m for 5 / 2, 0.53 m for 7 / 2, 0.026 m for 9 / 2. (b) Repeating the above argument for integral values for , constructive interference occurs at 4.34 m, 1.84 m, 0.86 m, 0.26 m. (c) If h l / 2 , there will be destructive interference at speaker B. If h l / 2, the path difference can never be as large as l / 2 . The minimum frequency is then v/(2h)=(344 m/s)/(4.0 m)=86 Hz. Exercises Problem 6 A 2.00 MHz sound wave travels through a pregnant woman’s abdomen and is reflected from fetal heart wall of her unborn baby. The heart wall is moving toward the sound receiver as the heart beats. The reflected sound is then mixed with the transmitted sound, and 5 beats per second are detected. The speed of sound in body tissue is 1,500 m/s. Calculate the speed of the fetal heart wall at the instance this measurement is made. Solution Let f0=2.00 MHz be the frequency of the generated wave. The frequency with which the heart wall receives this wave is fH=[(v+vH)/v]f0, and this is also the frequency with which the heart wall re-emits the wave. The detected frequency of this reflected wave is f’=[v/(v-vH )]fH, with the minus sign indicating that the heart wall, acting now as a source of waves, is moving toward the receiver. Now combining f’=[(v+vH)/(v-vH)]f0, and the beat frequency is: f beat f ' f 0 [( v vH ) /( v vH )] f 0 2vH f 0 /( v vH ). Solving for vH , vH v[ f beat /( 2 f 0 f beat )] (1500 m / s ){85 Hz /[ 2(2.00 106 Hz ) 85 Hz )} 3.19 102 m / s.