PHYS-110 Waves Chapter 15 PDF

Chapter 15 Waves Any Picture you think it can represent this chapter 1  Outline 15.1 WAVE DERIVATION OF MOTION WAVE EQUATION Chapter...

Chapter 15 Waves Any Picture you think it can represent this chapter 1  Outline 15.1 WAVE DERIVATION OF MOTION WAVE EQUATION Chapter 15.4 15.2 (15) COUPLED OSCILLATORS 15.3 MATHEMATICAL DESCRIPTION OF WAVES 2  Learning Objectives After studying this chapter, you will be able to: 1- Define waves. 2- Classify waves according to medium and propagation. 3- Differentiate between different wave types. 4- Apply the wave equations with different examples. 3  Waves A wave is an excitation that propagates through space or some medium as a function of time but does not generally transport matter with it. According to 1- Electromagnetic waves Waves medium 2- Mechanical waves classification According to 1- Transverse waves propagation 2- Longitudinal waves 4  Waves Types of waves according to medium: (1) Electromagnetic waves (light, radio waves, microwaves, X-ray…..) do not need a medium in which to propagate. They can move through empty space (vacuum). (2) Mechanical waves (sound wave, water wave….) need a medium in which to propagate. Many examples of waves surround us in everyday life. Light energy moving from the computer screen to your eye moves as light waves. Sound energy moving from a radio to your ear moves as sound waves. In this chapter, we will study the mechanical waves. 5  Coupled Oscillators Types of waves according to propagation: (1) Transverse Waves: Move in a direction perpendicular to the direction in which the oscillators move. Travel through a medium in the form of crests and troughs. Crest Crest Crest The up-and- down motion of Trough Trough Trough the rope is perpendicular to the direction of the wave. Direction of wave 6  Coupled Oscillators Type of waves according to propagation: (2) Longitudinal waves: ▪ Propagate along the direction in which the oscillators move. ▪ Travel through a medium in the form of compression and rarefaction. 7  Mechanical Description of Waves A wave can be represented as a function of position (x) and time (t). The figure represents a wave function. We can write the mathematical description of a sinusoidal wave as 𝑦 𝑥, 𝑡 = 𝐴𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡 + ∅0 ) Vertical position Phase difference Amplitude Angular frequency Wave number 8  Mechanical Description of Waves A number of parameters can be defined to describe a periodic wave: The amplitude A : ✓ Is the maximum displacement of points on a wave from the equilibrium position in units of 𝑐𝑚 or 𝑚. ▪ The wavelength λ : ✓ Is the length of one oscillation. ✓ It is the distance between two identical adjacent points in a wave, such as two adjacent crests or troughs in a waveform (in units of 𝑐𝑚 or 𝑚). The wave number k : ✓ Is the number of wavelengths that fit in a distance of 2π (in units of rad/cm or 2𝜋 rad/m). 𝑘= λ x 9  Mechanical Description of Waves The period T : ✓ Is the time of one oscillations (in units of s). ✓ It is the time between two identical adjacent points in a wave, such as two adjacent crests or troughs in a waveform (in units of 𝑠). The frequency f : ✓ Is the number of oscillations per second in the wave in units of 𝑠 −1 or 𝐻𝑧. 1 𝑓= 𝑇 The angular frequency ω : ✓ counts the number of oscillations in each time interval of 2π (in units of rad/s). 2𝜋 𝜔 = 2𝜋𝑓 = 𝑇 t 10  Mechanical Description of Waves The velocity of a wave 𝒗 : ✓ it is measured by (m/s). ∆𝒙 λ 𝒗= = = λf ∆𝒕 𝑻 𝟐𝝅 𝝎 𝝎 𝒗 = λf = = 𝑲 𝟐𝝅 𝑲 Phase difference ∅𝟎 : ✓ Is used to describe the difference in degrees or radians when two or more alternating quantities reach their maximum or zero values. Phase ∅ : ✓ it is measured by (rad). ✓ Is given by 11  Mechanical Description of Waves Notice that : A sin(+ kx – ωt + ϕ0) represents a wave that travels in the positive x- direction, as the signs of the (kx) and (ωt) terms are opposite. A sin(+ kx + ωt + ϕ0) represents a wave that travels in the negative x- direction, as the signs of the (kx) and (ωt) terms are the same. 12 Sample problem 15.1 A wave on a string is given by: 𝑟𝑎𝑑 𝑟𝑎𝑑 𝑦 𝑥, 𝑡 = 0.002𝑚 𝑠𝑖𝑛 78.8 𝑥 + 346 𝑡 𝑚 𝑠 where y and x are in meters and t is in seconds. a) What are the angular frequency and the wave number of this wave? b) What is the amplitude? c) What is the velocity of this wave? d) What is the wavelength of this wave? e) What is its period? f) In which direction does the wave travel? g) What is the phase difference? h) What is the phase at (x = 2.3 × 10-3 m and t = 2.1 s)? 13 Sample problem 15.1 SKETCH The sketch shows the wave function t and x. Research Look at the given wave function: 𝑟𝑎𝑑 𝑟𝑎𝑑 𝑦 𝑥, 𝑡 = 0.002 𝑚 𝑠𝑖𝑛 78.8 𝑥 + 346 𝑡 𝑚 𝑠 Comparing with a wave traveling in the negative x-direction 𝑦 𝑥, 𝑡 = 𝐴 sin(𝑘𝑥 + 𝜔𝑡 + ∅0 ) 14 Sample problem 15.1 SOLUTION: By comparison we can find: a) The angular frequency and the wave number: ω = 346 rad/s k = 78.7 rad/m b) The amplitude A= 0.002 m We can then find: c) the velocity of the wavefrom: 𝜔 346 𝑣= = = 4.4 𝑚/𝑠 𝑘 78.7 d) the wavelength from the wave number: 2𝜋 2𝜋 λ= 𝑘 = 78.7 = 0.08 𝑚 e) the period from the angular frequency: 2𝜋 2𝜋 𝑇= = = 0.012 s 𝜔 346 15 Sample problem 15.1 SOLUTION: f) The wave traveling in the negative x-direction ➔ as the signs of the (kx) and (ωt) terms are the same. g) the phase difference of this wavefrom is: ∅0 = 0 h) the phase at (x = 2.3 × 10-3 m and t = 2.1 s) is found from: ∅ = 𝑘𝑥 + 𝜔𝑡 + ∅0 ∅ = (78.8 x 0.0023) + (346 x 2.1) +0=726.8 rad 16  Sample problem 15.1 Double-check To double-check our work, we look at our sketch: T We see that the wavelength is approximately 0.08 m, in agreement with our result. We also see that the period is approximately 0.018 s, also in agreement with our result. 17 Concept check Which of the following waves is traveling in the positive x-direction? A. y(x,t) = (−0.002 m) sin[(78.8 m-1)x +(346 s-1)t] B. y(x,t) = (0.002 m) sin[(78.8 m-1)x + (346 s-1)t − 1] C. y(x,t) = (−0.002 m) sin[(−78.8 m-1)x + (346 s-1)t] D. y(x,t) = (0.002 m) sin[(−78.8 m-1) + (346 s-1)] 18 Concept check Which of the following waves is traveling in the positive x-direction? A. y(x,t) = (−0.002 m) sin[(78.8 m-1)x + (346 s-1)t] B. y(x,t) = (0.002 m) sin[(78.8 m-1)x + (346 s-1)t − 1] C. y(x,t) = (−0.002 m) sin[(−78.8 m-1)x + (346 s-1)t] D. y(x,t) = (0.002 m) sin[(−78.8 m-1) + (346 s-1)] 19  Derivation of the wave equation Waves on a String String instruments form a large class of musical instruments. Suppose a string has a mass M, a length L and a radius r. 𝑀 𝜇= = 𝜌𝐴 = 𝜌(𝜋𝑟 2 ) 𝐿 Where: 𝜇 is the linear mass density (kg/m) 𝜌 is the density (kg/m3) A is the cross section area, (𝐴 = 𝜋𝑟 2 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔) (m2) We can get the wave velocity on a string in terms of the tension: 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑇 𝑣= = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑚𝑎𝑠𝑠 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝜇 Some implications of this equation are: – Increasing μ reduces v. – Increasing F increases v. 20  Example 15.1 Elevator Cable 38 Kg 61 m An elevator repairman (mass 73 kg) sits on top of an elevator cabin of mass 655 kg inside a shaft of a skyscraper. The cabin is suspended by a 61 m long steel cable of mass 38 kg. 655 Kg He sends a signal to his colleague at the top of the elevator shaft by tapping the cable with his hammer. Find a) The linear mass density. b) The velocity of the wave on the string. c) How long will it take for the wave pulse generated by the hammer tap to travel up the cable? 21  Example 15.1 SOLUTION: T The tension in the cable from the weight of the elevator + man is: 38 Kg 61 m T = Fg The linear mass density of the steel cable is: 73 655 The wave speed is: Fg= mg T The pulse travels up the 61 m long cable in: 22 Extra Exercise A copper wire has a density of ρ = 8920 kg/m3, a diameter of 2.40 mm, and a length L. The wire is held under a tension of 10.00 N. Transverse waves are sent down the wire. (a) What is the linear mass density of the wire? (b) What is the speed of the waves through the wire? 23 Extra Exercise A copper wire has a density of ρ = 8920 kg/m3, a diameter of 2.40 mm, and a length L. The wire is held under a tension of 10.00 N. Transverse waves are sent down the wire. (a) What is the linear mass density of the wire? (b) What is the speed of the waves through the wire? SOLUTION: a) Diameter = 2𝑟 2.4 𝑟= = 1.2 𝑚𝑚 2 𝑟 = 1.2 x10−3 𝑚 𝜇 = 𝜌 𝜋𝑟 2 = 8920 x 3.14 x (1.2 x 10−3 )2 = 0.04 kg/m 𝑇 10 b) v = = = 15.8 m/s 𝜇 0.04 24 Extra Exercise When the horn on a ship is sounded, the passengers hear an echo from a cliff after 4s. If the speed of sound is 340 m/s, how far away is the cliff? A) 170 m B) 340 m C) 680 m D) 1360 m 25 Extra Exercise When the horn on a ship is sounded, the passengers hear an echo from a cliff after 4s. If the speed of sound is 340 m/s, how far away is the cliff? A) 170 m B) 340 m C) 680 m D) 1360 m SOLUTION: As its an Echo, it is meaning the sound wave has traveled double the distance: 2𝑑 v= 𝑡 𝑉𝑡 340 𝑋 4 𝑑= = = 680 m 2 2 26  Equation summary (1) 𝟏 𝝎 = 𝟐𝝅𝒇 (2) 𝒇= 𝑻 (3) 𝒌 𝝎= 𝒗 𝟐𝝅 𝝎= 𝑻 𝑤 𝑣 = λ𝑓 = (4) 𝑘 (5) 𝑀 𝐹 𝜇= 𝑣= 𝐿 𝜇 𝜇 = 𝜌(𝜋𝑟 2 ) 27 The END OF CHAPTER (15) 28