Waves and Vibrations PDF - FEE 3101 Engineering Physics 1A

WAVES AND VIBRATIONS FEE 3101: ENGINEERING PHYSICS 1A  Wave motion is a means of transferring energy from one point to another without there being any transfer of matter between the points. Waves are classified into two categories namely, 1) Mechanical waves e.g. water waves, waves in str...

WAVES AND VIBRATIONS FEE 3101: ENGINEERING PHYSICS 1A  Wave motion is a means of transferring energy from one point to another without there being any transfer of matter between the points. Waves are classified into two categories namely, 1) Mechanical waves e.g. water waves, waves in string, sound etc. 2) Electromagnetic waves e.g. light, radio, X – rays etc. Differences between oscillations and waves. 1) An oscillation is confined to a body while a wave extends through space. 2) A wave is a means by which energy is transferred from one place to another while an oscillation is a means by which energy can be stored in a confined mass. 3) Any oscillation can be resolved into a number of simple harmonic motions of different amplitudes and frequencies while a wave can be resolved into a continuous set of oscillations in the medium. FEE 3101: PHYSICS 1A 2  Waves obey the following properties: Reflection; refraction, diffraction, interference and polarization. Waves and Their Characterization Waves are further are classified into two classes depending on their mode of propagation, i.e. 1) Transverse waves – It is a wave in which the particles of the medium move in the direction perpendicular to the direction of wave motion, e.g water waves, waves in a string (rope), all electromagnetic waves etc. 2) Longitudinal Waves – are waves in which the particles of the medium move in a direction parallel to the direction of the wave motion e.g. sound wave, waves in a slingy spring Rarefactions Crest Compressions TroughFEE 3101: PHYSICS 1A 3 Definition of Terms: Amplitude (𝒂): It is the maximum displacement from the mean position, measured in metres (𝑚). Wavelength 𝝀 : It is the length of a complete cycle of a wave, measured in metres (𝑚). Frequency (𝒇): It is the number of cycles made by a wave in a second, measured in cycles per second 𝑐/𝑠 𝑜𝑟 𝑐𝑠 −1 where: 1 𝑐𝑠 −1 = 1 𝐻𝑒𝑟𝑡𝑧 (𝐻𝑧) Period (𝑻): It is the time taken for wave to make a complete oscillation. It is measured in seconds (s) 1 1 𝑇= 𝑓= 𝑣 = 𝜆𝑓 𝑓 𝑇 Phase difference FEE 3101: PHYSICS 1A 4  Consider the wave below: An equation can be formed to represent generally the displacement y, of a vibrating particle in a medium in which a wave passes. Suppose the wave moves from left to right their amplitude at the origin 𝑂 then vibrates according to the equation 𝑦 = 𝑎 sin 𝜔𝑡. 𝑤ℎ𝑒𝑟𝑒 𝑡 𝑖𝑠 𝑡𝑖𝑚𝑒 𝑎𝑛𝑑 𝜔 = 2𝜋𝑓 A particle 𝑃 at a distance 𝑥 from 𝑂 to the right the phase of the vibration will be different from that at 𝑂. A distance 𝜆 from 𝑂 corresponds to a phase difference of 2𝜋. So the phase difference 𝜙 at 𝑃 is given by 𝑥 Τ𝜆 × 2𝜋. 𝑂𝑟 2𝜋𝑥 Τ𝜆 FEE 3101: PHYSICS 1A 5  Then the displacement of any particle at a distance x from the origin is given by 𝑦 = 𝑎 sin 𝜔𝑡 − 𝜙 2𝜋𝑥 𝑦 = 𝑎 sin 𝜔𝑡 − 𝜆 2𝜋𝑣 2𝜋𝑥 2𝜋𝑣 𝑦 = 𝑎 sin 𝑡− 𝜔 = 2𝜋𝑓 = 𝜆 𝜆 𝜆 2𝜋 𝑦 = 𝑎 sin 𝑣𝑡 − 𝑥 𝜆 2𝜋 𝑡 𝑥 Also, 𝜔 = , then 𝑦 = 𝑎 sin 2𝜋 − 𝑇 𝑇 𝜆 The negative sign shows the wave is moving from left to right. If a wave travels in opposite direction, then, 𝑡 𝑥 𝑦 = 𝑎 sin 2𝜋 + 𝑇 𝜆 FEE 3101: PHYSICS 1A 6 Equation of waves Waves can be expressed using Sine or Cosine function, i.e. a wave in one dimension can be expressed as 2𝜋 𝑦 = 𝐴 sin 𝑥 − 𝑣𝑡 𝜆 𝑦 = 𝐴 sin 𝑘𝑥 − 𝜔𝑡 Example 1: A generator at one end of a very long string makes a wave governed by the equation, 𝜋 𝑦 = 6.0 𝑐𝑚 cos 2.0𝑚−1 𝑥 + 8.0𝑠 −1 𝑡 2 Calculate the frequency, period, amplitude, wavelength and speed of the wave. Amplitude = 6.0 cm 𝑦 = 6.0 𝑐𝑚 cos 𝜋𝑥𝑚−1 + 4𝜋𝑡𝑠 −1 2𝜋 2 𝐾 = = 𝜋𝑚−1 = 1𝑚−1 , 𝜆 = 2 𝑚 𝜆 𝜆 𝜔 = 2𝜋𝑓, 4𝜋𝑠 −1 = 2𝜋𝑓 𝑓 = 2𝑠 −1 𝑇 = 1ൗ𝑓 = 1Τ2𝑠−1 = 0.5 𝑠 𝑣 = 𝜆𝑓 = 2 × 2 = 4 𝑚𝑠 −1 FEE 3101: PHYSICS 1A 7 Example 2: Suppose a wave is represented by 𝜋𝑥 𝑦 = 𝑎 sin 2000𝜋𝑡 − 0.17 Find frequency, period, wavelength, speed and phase difference, given that x = 0.17. 2𝜋 𝑦 = 𝑎 sin 𝑣𝑡 − 𝑥 𝜆 2𝜋𝑡 2𝜋 𝜋 = 2000𝜋 𝑎𝑛𝑑 = 𝜆 = 2 × 0.17 = 0.34 𝑚 𝜆 𝜆 0.17 𝑣 = 1000𝜆 = 1000 × 0.34 = 340 𝑚/𝑠 𝑣 340 𝑓= = = 1000 𝐻𝑧 𝜆 0.34 1 1 𝑇= = = 1 × 10−3 𝑠 𝑓 1000 2𝜋×0.17 If the waves are separated by 0.17, then 𝜙 = = 2𝜋 0.17 FEE 3101: PHYSICS 1A 8 Exercise 1: The equation of a certain sound wave (simple harmonic progressive wave) is given by y = 0.05 sin 10π(t/0.025 – x/8.5), where x and y are in meters and t is in seconds. What are the (1) amplitude (2) frequency (3) wavelength of the wave? What is the velocity and direction of propagation of the wave? Exercise 2: The equation of a wave can be represented by y = 0.02 sin 2π /0.5 (320t – x) where x and y are in metres and t is in seconds. Find the amplitude, frequency, wavelength, and velocity of propagation of the wave. Exercise 3: A plane progressive wave is represented by the equation y=0.1 sin (200πt−20πx/17) where y is displacement in m, t in second and x is distance from a fixed origin in meter. The frequency, wavelength and speed of the wave respectively are? FEE 3101: PHYSICS 1A 9 Principle of Superposition When two ends of a string are jerked as shown below, then the waves interfere. The above observation shows that in I the waves interfere constructively and in II, they interfere destructively. The observation can be explained by applying the principle of superposition which states that, ‘whenever two waves travelling in the same medium, the total displacement at any point is the sum of the separate displacements due to the waves at that point.” FEE 3101: PHYSICS 1A 10  Consider two waves 𝑦1 and 𝑦2 travelling in the same direction and phase difference between them is 𝜙. 𝑦1 𝑥, 𝑡 = 𝑎 sin 𝑘𝑥 − 𝜔𝑡 𝑎𝑛𝑑 𝑦2 𝑥, 𝑡 = 𝑎 sin 𝑘𝑥 − 𝜔𝑡 + 𝜙 𝑦 ′ 𝑥, 𝑡 = 𝑦1 𝑥, 𝑡 + 𝑦2 𝑥, 𝑡 = 𝑎 sin 𝑘𝑥 − 𝜔𝑡 + 𝑎 sin(𝑘𝑥 − 𝜔𝑡 + 𝜙) 1 1 Using sin 𝛼 + sin 𝛽 = 2 sin 𝛼 + 𝛽 cos 𝛼−𝛽 2 2 1 𝑦 ′ 𝑥, 𝑡 = 𝑎 ൜ 2 sin 𝑘𝑥 − 𝜔𝑡 + 𝑘𝑥 − 𝜔𝑡 + 𝜙 + 2 1 2 cos 𝑘𝑥 − 𝜔𝑡 − 𝑘𝑥 − 𝜔𝑡 + 𝜙 ൠ 2 = 𝑎 2 sin 1ൗ2 2𝑘𝑥 − 2𝜔𝑡 + 𝜙 cos 1ൗ2 𝜙 The amplitude = 2a cos 1Τ2 𝜙 and phase difference = 1Τ2 𝜙, then the amplitude doubles. Task: Consider two waves 𝑦1 𝑥, 𝑡 = 𝑎 sin 𝑘𝑥 + 𝜔𝑡 and 𝑦2 𝑥, 𝑡 = 𝑎 sin 𝑘𝑥 − 𝜔𝑡 travelling in opposite direction, find 𝑦 ′ 𝑥, 𝑡 = 𝑦1 𝑥, 𝑡 + 𝑦2 𝑥, 𝑡 FEE 3101: PHYSICS 1A 11 Exercise: Two waves of the same frequency, velocity and amplitude travelling in opposite directions on a string fixed at both ends are represented as: 𝑦1 𝑥, 𝑡 = 𝑎 sin 𝑘𝑥 − 𝜔𝑡 𝑦2 𝑥, 𝑡 = 𝑎 sin 𝑘𝑥 + 𝜔𝑡 On superposition, these give rise to a standing wave given by 𝜋𝑥 𝑦 𝑥, 𝑡 = 2 sin cos 40𝜋𝑡 𝑐𝑚 4 a) Determine the equations representing the component waves 𝑦1 𝑥, 𝑡 and 𝑦2 𝑥, 𝑡. b) Calculate the distance between two successive nodes. FEE 3101: PHYSICS 1A 12 Stationary/Standing Waves A stationary (standing) wave is formed when two equal progressive (travelling) waves travelling in opposite directions are superposed on each other. Stationary waves have nodes at points of zero displacement and antinodes at points of maximum displacement. In a stationary wave, vibrations of particles at points between successive nodes are in phase. Between successive nodes, particles have different amplitudes of vibrations. The distance between successive nodes or antinodes is 𝜆Τ2. The distance between a node and the next antinodes is 𝜆Τ4. Task: Give differences between a stationary and progressive wave. FEE 3101: PHYSICS 1A 13 Equation of a standing wave Consider two travelling waves traveling in opposite directions; 𝐴1 = 𝐴0 sin 𝑘𝑥 − 𝜔𝑡 𝑎𝑛𝑑 𝐴2 = 𝐴0 sin 𝑘𝑥 + 𝜔𝑡 Adding the two waves A = 𝐴0 sin 𝑘𝑥 − 𝜔𝑡 + 𝐴0 sin 𝑘𝑥 + 𝜔𝑡 Now using trigonometric identities sin 𝑥 − 𝑦 = sin 𝑥 cos 𝑦 − cos 𝑥 sin 𝑦 sin 𝑥 + 𝑦 = sin 𝑥 cos 𝑦 + cos 𝑥 sin 𝑦 Thus, 𝐴 = 𝐴0 ሾsin 𝑘𝑥 cos 𝜔𝑡 − cos 𝑘𝑥 sin 𝜔𝑡 + sin 𝑘𝑥 cos 𝜔𝑡 + cos 𝑘𝑥 sin 𝜔𝑡ሿ 𝐴 = 2𝐴0 sin 𝑘𝑥 cos 𝜔𝑡 Comparing (3.04) with linear wave equation 𝑦 = 𝑦0 cos 𝜔𝑡 amplitude of resultant wave is therefore 𝐴𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 = 2𝐴0 sin 𝑘𝑥 FEE 3101: PHYSICS 1A 14  Minimum displacement (node) occurs when sin 𝑘𝑥 = 0 (since maximum amplitude 𝐴0 cannot be equal to zero) 𝑘𝑥 = 𝑛𝜋, 𝑛 = 0, 1, 2, ….. 𝑥 = 𝑛Τ𝑘 𝜋 Putting 𝑘 = 2𝜋Τ𝜆 ⇒ 𝑥 = 𝑛Τ2 𝜆 𝑛 = 0, 1, 2, … Thus, nodes occur at 1 3 𝑥 = 0, 𝜆, 1𝜆, 𝜆, … … 𝑛 = 0, 1, 2, 3, … …. 2 2 Maximum displacement (antinode) equal ±2𝐴0 which occurs when sin 𝑘𝑥 = ±1 which occurs when; 2𝑛 + 1 𝑘𝑥 = 𝜋, 𝑛 = 0,1,2, … 2 2𝑛 + 1 𝜋 𝑥= 𝑛=0,1,2,… 2 𝑘 2𝜋 Putting 𝑘 = Τ𝜆 2𝑛 + 1 𝑛=0,1,2,… 𝑥= 𝜆 4 1 3 5 Thus, antinodes occur when 𝑥 = 𝜆, 𝜆, 𝜆 … …. 4 4 4 FEE 3101: PHYSICS 1A 15 Standing Waves on Strings Musical instruments like guitar produce their sound as a result of stretched strings being made to vibrate by plucking them. A transverse wave of a given frequency travels along the string and is reflected back by a fixed end causing the incident and reflected waves to interfere and form a stationary wave. A vibrating string exhibits different stationary waves (modes of vibrations) depending on where it has been plucked. The frequency of the transverse wave depends on the tension (T) and mass per unit length (𝜇) of the vibrating string. It can be shown that the velocity of the transverse wave along the string is given by: FEE 3101: PHYSICS 1A 16  𝑇 𝑚 𝑇 𝑣= 𝜇= 𝑣= 𝑚Τ𝑙 𝑙 𝜇 Fundamental Frequency 𝑓0 This is the lowest frequency that can be obtained when a musical instruments is played. A stationary wave in its simplest form possible, produces a fundamental note which gives sound its basic pitch. Overtone: It is a fundamental note accompanied by other notes smaller in amplitude but of higher frequencies than the fundamental frequency, these notes are called overtones. Harmonics: This is the name given to a note whose frequency is a whole number multiple of the fundamental frequency. Frequencies 𝑓0 , 2𝑓0 , 3𝑓0 , 𝑎𝑛𝑑 4𝑓0 are the first, second, third and fourth harmonics respectively. FEE 3101: PHYSICS 1A 17 Modes of vibration Fundamental Frequency (First Harmonics) The string is plucked in the middle. This produces the simplest possible stationary wave as shown below. 𝜆 𝑙= ,𝜆 = 2𝑙 2 1 𝑇 𝑓= = 𝑣 𝑣 for 𝑣 = 𝑇 𝑓= 𝜆 2𝑙 𝜇 2𝑙 𝜇 This is the frequency of the fundamental or lowest note obtained from a string. 1 𝑇 𝑓0 = 2𝑙 𝜇 FEE 3101: PHYSICS 1A 18 First Overtone (Second Harmonic) This is obtained by holding the midpoint of the vibrating string and plucking the string at a point a quarter of its length from one end. If 𝜆1 is the wavelength and 𝑓1 the frequency, then 𝑣 𝑣 𝑙 = 𝜆1 , 𝑓1 = = … … … … … … … … … … …. (1) 𝜆1 𝑙 𝑣 But 𝑓0 = … … … … … … … … … … … … … ….. ….. (2) 2𝑙 Dividing equation (2) by (1) 𝑓0 𝑣Τ2𝑙 1 = = 𝑓1 𝑣 Τ𝑙 2 1 𝑇 𝑓1 = 2𝑓0 = 𝑙 𝜇 FEE 3101: PHYSICS 1A 19 Second Overtone (Third Harmonic) It is obtained by plucking the string in the middle while touching the string one-third from end. Let the wavelength be 𝜆2 and the frequency be 𝑓2 𝑣 3 2 𝑓2 = , 𝑙 = 𝜆2 , 𝜆2 = 𝑙 𝜆2 2 3 3𝑣 𝑓2 = … … … … … … … … … … … …. (1) 2𝑙 𝑣 𝑓0 = … … … … … … … … … … … ….. (2) 2𝑙 Dividing (2) by (1) 3 𝑇 𝑓0 = 𝑣Τ2𝑙 = 1 𝑓2 = 3𝑓0 = 𝑓1 3𝑣Τ2𝑙 3 2𝑙 𝜇 FEE 3101: PHYSICS 1A 20 It therefore follows that, 𝑛𝑡ℎ overtone, 𝑓𝑛 = 𝑛 + 1 𝑓0 Note: waves in vibrating strings give both odd and even harmonics. Example: The length of a stretched string is 0.4 𝑚. Its mass is 1 × 10−4 𝑘𝑔. If the tension in the string is 10 N, calculate: a) The velocity of the transverse wave in the string. b) The fundamental frequency. c) The frequency of the first overtone. 𝑚 1×10−4 𝜇= = = 0.00025 𝑘𝑔/𝑚3 𝑙 0.4 a) 𝑇 10 𝑣= = = 200 𝑚/𝑠 𝑓1 = 2𝑓0 𝜇 0.00025 c) = 2 × 250 = 500 𝐻𝑧 1 𝑇 200 b) 𝑓0 = = = 250 𝐻𝑧 2𝑙 𝜇 2×0.4 FEE 3101: PHYSICS 1A 21 Example: A wire of uniform cross-section area has a tension of 20 𝑁 and produces a note of 100 𝐻𝑧 when plucked in the middle. Find its diameter if it is 1m long and has a density of 6000 𝑘𝑔/𝑚3. State the frequency of the third overtone and the wavelength of the third harmonic. Solution: This is the first harmonic 1 𝑇 𝑇 𝑓0 = 𝑣= 2𝑙 𝜇 𝜇 𝑣 = 2𝑙𝑓0 = 2 × 1 × 100 = 200 𝑚/𝑠 𝑇 20 𝑣2 = 𝜇= = 0.0005 𝑘𝑔/𝑚 𝜇 200 × 200 𝑚 𝜇 = , 𝑚 = 𝜇𝑙 = 0.0005 × 1 = 0.0005 𝑘𝑔 𝑙 𝑚 5 × 10−4 𝑉= = = 8.33 × 10−8 𝑚3 𝜌 6000 FEE 3101: PHYSICS 1A 22