TOPIC 8 Wave Mechanics Updated 2.pptx PDF

TOPIC 8 Wave Mechanics MECHANICAL WAVES  WAVE Disturbance of a system from equilibrium that propagates from one region of the system to another  Impulsive waves  Periodic waves PROPERTIE S OF MECHANICA L WAVES  PROPERTIES Mechanical Waves –  Disturbance trav...

TOPIC 8 Wave Mechanics MECHANICAL WAVES  WAVE Disturbance of a system from equilibrium that propagates from one region of the system to another  Impulsive waves  Periodic waves PROPERTIE S OF MECHANICA L WAVES  PROPERTIES Mechanical Waves –  Disturbance travels or require a medium in propagates with definite speed order to propagate through the medium  The medium does not travel through space – each particle of the medium undergo oscillation along their equilibrium point  Waves transport Kinds of Mechanica l Waves  Transverse Wave  Particlesin the medium vibrate perpendicular to the direction of the propagation of the wave  Longitudinal Wave  Particles in the medium vibrate parallel to the direction of the wave Crest, trough, amplitude, wavelength Frequency, Period Period – time elapsed between two successive crests passing the same point in space Frequency – number of cycles that pass a given point per unit time Wavelength λ– distance between two crests or two troughs or between two points on back-to-back cycles of a wave. Wave number k – number of waves in a unit distance Angular frequency - angular displacement per unit time  Wave Speed  The distance that disturbance travels per unit time. 𝜆 𝑣 =𝜆 𝑓 = 𝑇  Dependent on the properties of the medium in which it travels. Wave Speed 𝑣= √ 𝑟𝑒𝑠𝑡𝑜𝑟𝑖𝑛𝑔 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑣= √ 𝐹 𝜇 Linear mass density:  Type of wave Wave speed Transverse wave on a string Longitudinal waves in solid Longitudinal waves in fluids Sound in an ideal gas MATHEMATICAL REPRESENTATIO WAVE FUNCTION N OF A FOR A TRAVELING SINUSOIDAL WAVE WAVE  Sinusoidal wave Periodic wave with simple harmonic motion 𝑦 = 𝐴 sin 𝜔 𝑡 𝑡= 𝑥 𝑣 𝑦 = 𝐴 cos 𝜔 𝑡 Sine Wave 𝑥 (𝑡 − ) 𝑣 +x direction Sine Wave -x direction Cosine Wave +x direction Cosine Wave -x direction  Sample Problem 1 A transverse wave on the string has a wave speed of 12.0 m/s, amplitude 0.0500 m, and wavelength 0.400 m. The waves travel in the +x-direction, and at t=0, the x=0 end of the string has zero displacement and is moving upward.  (a) Find: f, T, k, ω A transverse wave 𝑣 12.00 𝑚 / 𝑠 𝑓= = =30. 0 𝐻𝑧 𝜆 0. 400 𝑚 on the string has a wave speed of 1 𝑇= = 1 =0.033 𝑠 𝑓 30.0 𝐻𝑧 12.0 m/s, 2𝜋 2 𝜋 𝑟𝑎𝑑 amplitude 0.0500 𝑘= 𝜆 = 0. 400𝑚 =15. 71 𝑟𝑎𝑑 /𝑚 m, and wavelength 𝜔=2 𝜋 𝑓 =2 𝜋 ( 30.0 𝐻𝑧 )=188.50 𝑟𝑎𝑑 / 𝑠 0.400 m. (b) Write a wave function describing the wave The waves travel in the +x-direction, and at t=0, the x=0 end of the string has zero displacement and is moving upward. 𝑦 ( 𝑥 , 𝑡 ) = 𝐴sin ( 𝜔 𝑡 −𝑘𝑥 ) 𝑦 ( 𝑥 , 𝑡 ) =0.0500 𝑚 sin ( (60. 0 𝜋 𝑟𝑎𝑑/ 𝑠)𝑡 −(5.0 𝜋 𝑟𝑎𝑑/𝑚) 𝑥 )  (c) Find the transverse displacement of a point at x=0.250 m at time t=0.150 s. 𝑦 ( 𝑥 , 𝑡 ) =0.0500 𝑚 sin ( (60. 0 𝜋 𝑟𝑎𝑑/ 𝑠)𝑡 −(5.0 𝜋 𝑟𝑎𝑑/𝑚) 𝑥 ) 𝑦 ( 𝑥 , 𝑡 ) =0.0500 𝑚sin ( (60. 0 𝜋 𝑟𝑎𝑑/ 𝑠)(0.150 𝑠)−(5.0 𝜋 𝑟𝑎𝑑/𝑚)(0. 250𝑚)) 𝑦 ( 𝑥=0. 250𝑚 ,𝑡=0. 150 𝑠 ) =−0. 035m  Sample Problem 2 Transverse waves on a string have speed 8.00 m/s, amplitude 0.0700 m and wavelength 0.320 m. The waves travel in –x-direction and at t=0 the x=0 end of the string has its maximum upward displacement. Find the wave number, angular velocity, and write a wave function describing the wave. Wave number k = 2π/λ = 19.63 m-1 Angular velocity/angular frequency ω= 2πf = 157 rad/s Wave function  Seatwork (15 mins) The right end of a string is attached to a stationary support. The other end is wiggled up and down with frequency 2.0 Hz and amplitude 0.25 m. The wave speed is observed to be v = 10 m/s. At time t = 0, the displacement is zero and is moving in the +x-direction. Assume that no waves bounces back from the far end. Find the following: A. amplitude, wavelength, wave number, angular frequency. B. wave equation describing the wave  A. 2𝜋 𝐴=0. 25 𝑚 𝑘= =1.26 𝑟𝑎𝑑 / 𝑚 λ 𝑟𝑎𝑑 𝑣 10 𝑚 / 𝑠 𝜔=2 𝜋 𝑓 =2 𝜋 ( 2. 0 𝐻𝑧 ) =4 𝜋 𝑜𝑟 12. 6𝑟𝑎𝑑 / 𝑠 λ= = =5. 0 𝑚 𝑠 𝑓 2. 0 𝐻𝑧 B.𝑦 ( 𝑥 , 𝑡 ) = 𝐴𝑠𝑖𝑛( 𝜔 t − kx ) 𝑦 ( 𝑥 , 𝑡 ) =0.25𝑚 sin((4¿ 𝜋 𝑟𝑎𝑑/ 𝑠)𝑡 −(1.26 𝑟𝑎𝑑/𝑚) 𝑥)¿ REFLECTION AND TRANSMISSION The reflection of a traveling pulse at the fixed end of a stretched string. The reflected pulse is inverted, but its shape is otherwise unchanged The reflection of a traveling pulse at the free end of a stretched string. The reflected is not inverted.  WAVE INTERFERENCE The resulting effect when two or more waves overlap in the same region of space. The Principle of Superposition  When two or more waves overlap, the displacement at any point is the algebraic sum of the displacement of the corresponding points of the overlapping waves. Constructive Interference Resultswhen the waves are in phase  Constructive Interference Pathdifference between the two waves is 𝒓 𝟐 − 𝒓 𝟏 =𝒏 𝝀 𝑛=0 , ±1 , ±2 ,±3  Destructive Interference  Results when the waves are out phase Destructive Interference Path difference between the two waves is ( 𝑟 2 − 𝑟 1 = 𝑛+ 1 2 𝜆) 𝑛=0 , ±1 , ±2 ,±3  Standing wave  Occurswhen two waves of the same frequency and amplitude but traveling in opposite direction interfere  Does not appear to be moving in either direction  Standing wave  Node – points that never move at all; manifests destructive interference  Antinode – points that have greatest amplitude; manifests constructive interference  Does not transfer energy – there is local flow of energy from each node to the next; but average energy transfer  𝑦 1 ( 𝑥 , 𝑡 ) =− 𝐴 𝑠𝑖𝑛 ¿ 𝑦 2 ( 𝑥 , 𝑡 ) = 𝐴𝑠𝑖𝑛 ¿ + 𝑦 ( 𝑥 , 𝑡 ) =2 𝐴sin 𝑘𝑥 cos 𝜔 𝑡 𝐴 𝑆𝑊 =2 𝐴 Amplitude for standing wave Location of the displacement nodes 𝑛=0 , 1 , 2 ,3 ,… 2𝜋 since 𝑘= 𝜆 Location of the displacement antinodes Normal Modes Of A String Normal Modes Of A String  Length  Wavelength  Frequency Normal Modes Of A String 𝑣 𝑓 1= 1st harmonic 2 𝐿 2 𝑣 2 nd 𝑓 2= harmonic 1st overtone 2 𝐿 3 𝑣 𝑓 3= 3rd harmonic 2nd overtone 2 𝐿 HARMONIC SERIES – sequence of frequencies in which each frequency is an integer of a fundamental OVERTONE – any frequency greater than the fundamental frequency  Sample Problem 3 A steel wire 1.0 m long having a mass of 5.0 g is under tension of 500 N. What is the fundamental frequency of vibration? 𝑓 𝑛= = = √ √ 𝑛𝑣 𝑛 𝐹 𝑛 𝐹 𝐿 2𝐿 2𝐿 𝜇 2𝐿 𝑚 𝑓 1= 1 √ (500 𝑁)(1.00 𝑚) 2(1.0𝑚) 5.00×10 −3 𝑘𝑔 =158.11 𝐻𝑧  Sample Problem 3 How many overtones can be heard by a person capable of hearing frequencies up to 9480 Hz? 𝑓𝑛 9480 𝐻𝑧 𝑛= = =60 𝑓1 158 ℎ𝑧 ¿ 𝑜𝑣𝑒𝑟𝑡𝑜𝑛𝑒𝑠=𝑛 − 1=60 − 1=59 Longitudinal Standing Waves Longitudinal Normal Modes: Open Pipe  Length  Wavelength  Frequency Longitudinal Normal Modes: Closed Pipe  Length  Wavelength  Frequency  Sample Problem 4 Find the fundamental frequency and the first two overtones of a pipe 40.0 cm long. (a) if the pipe is open at both ends; (b) if the pipe is closed at one end. Take the speed of sound in air to be 344 m/s.  Open pipe Closed pipe 𝑛𝑣 𝑛𝑣 𝑓= 𝑓= 2𝐿 4𝐿 (1)(344 𝑚 / 𝑠) (1)(344 𝑚 / 𝑠) 𝑓 1= =430 𝐻𝑧 𝑓 1= =215 𝐻𝑧 2(0. 40 𝑚) 4(0. 40 𝑚) 𝑓 𝑛 =𝑛 𝑓 1 𝑓 𝑛 =𝑛 𝑓 1 𝑓 2 =(2) 𝑓 1=860 𝐻𝑧 𝑓 3 =(3) 𝑓 1=645 𝐻𝑧 𝑓 3 =(3) 𝑓 1=1290 𝐻𝑧 𝑓 5 =(5) 𝑓 1=1075 𝐻𝑧 (c) For each of the cases above, what is the number of the highest harmonic that may be heard by a person who can hear frequencies from 20 Hz to 20,000 Hz? Open pipe Closed pipe 𝑓𝑛20 , 000 𝐻𝑧 𝑓𝑛 20 , 000 𝐻𝑧 𝑛= = ≅ 46 𝑛= = ≅93 𝑓1 430 𝐻𝑧 𝑓1 215 𝐻𝑧  Sample Problem 5 On the day when the speed of sound is 344 m/s, the fundamental frequency of the stopped pipe is 230 Hz. (a) How long is the stopped pipe? (b) What are the third, fourth, and fifth harmonics? (c) The third harmonics of this stopped pipe has the same wavelength as the fifth harmonic of another open pipe. How long is the open pipe? On the day when the speed of sound is 344 m/s, the fundamental frequency of the stopped pipe is 230 Hz. (a) How long is the stopped pipe? 𝑓 1 =23 0 𝐻𝑧 𝑛𝑣 𝑣 𝑓= 𝑓 1= 4𝐿 4𝐿 𝑣 344 𝑚/ 𝑠 𝐿= = =0. 37 𝑚 4 𝑓 1 4 (230 𝐻𝑧 ) On the day when the speed of sound is 344 m/s, the fundamental frequency of the stopped pipe is 230 Hz. (b) What are the third, fourth, (c) The third harmonic of this stopped pipe has the same and fifth harmonics? wavelength as the fifth harmonic of 𝑓 3 =( 3 ) 𝑓 1 =690 𝐻𝑧 another open pipe. How long is the open pipe? 𝑓 3 ( 𝑠𝑡𝑜𝑝𝑝𝑒𝑑)= 𝑓 5 (𝑜𝑝𝑒𝑛) 𝑛𝑜 𝑓 4 𝑣 𝑓 5 =( 5 ) 𝑓 1 =5 =690 𝐻𝑧 𝑓 5 =( 5 ) 𝑓 1 =1150 𝐻𝑧 2𝐿 𝑣 𝐿=5 =1.25 𝑚 / 𝑠 2𝑓 1  Seatwork (255 mins ) On a new world, a spaceship touches down. The team makes the decision to gauge the planet's atmosphere's sound speed. They discover that a 50 cm tube has two resonance frequencies: 2,520 Hz for the first and 2,940 Hz for the second. a. What is the speed of the sound? b.What is the number of each harmonic? c.What type of pipe is the crew using (open-open, open- closed, closed- closed)? Resonance A phenomena that occurs if the frequency is equal to the one of the normal modes of frequencies, the amplitude of the resulting force oscillation can become very large. Resonance