1. Transverse Waves: the particles in the medium move perpendicular to the direction of the wave motion

Transcription

1 Mechanical Waves Represents the periodic motion of matter e.g. water, sound Energy can be transferred from one point to another by waves Waves are cyclical in nature and display simple harmonic motion There are many types of waves, we will look at two: 1. Transverse Waves: the particles in the medium move perpendicular to the direction of the wave motion eg) earth during earthquake, water, snapping a rope, light (special case) crest of wave = highest point rest position trough of wave = lowest point

2 2. Longitudinal Waves: the particles in the medium move parallel to the direction of the wave eg) earth during earthquake, sound waves wave direction source of wave compression = particles together rarefaction = particles apart Definitions 1. Amplitude = maximum displacement of wave from rest position the greater the amplitude, the greater the amount of energy being transferred and the greater the work that can be done Rest position

3 2. Frequency (f) = number of cycles (waves) per second f = # cycles time unit = cycles per second or the Hertz (Hz) 3. Period (T) = time for one cycle T = 1/f T = time # cycles f = 1/T Unit of period (T) is second (s) Example A wave emits 20 cycles in 10 seconds. Calculate the frequency and the period. t = 10 s f = # of cycles time

4 =20/10 =2.0 cycles/s or 2.0 Hz T = time = 10 = 0.50 s # of cycles 20 ***Note: to calculate period, you could also use T = 1/f = 1/2.0 = 0.50 s 4. Wavelength (λ) = shortest distance between two points where the wave pattern repeats itself the unit is the meter (m) 1 λ 1 λ 1 λ Two points on a wave are said to be in phase if they are travelling in the same direction and are the same distance from the rest position

5 G A F B H L K M rest position C J E D Points in phase: B&H, D&I, E&J, F&K, G&L Wavelength is the distance between two adjacent points in phase. 5. Speed (v) = distance travelled by the wave in a certain time period v = d/t =λ/1/f v =fλ This leads us to the universal wave equation: v = fλ I where: v = speed of wave in m/s

6 f = frequency of wave in Hz (cycles per second) λ = wavelength in m Example 1 A wave has a speed of 340 m/s. If the wavelength is 20 m, what is the frequency? v = 340 m/s λ = 20 m v=fλ 340=f (20) f = 17 Hz Example 2 The distance between successive crests in a series of water waves is 1.5 m and the crests travel 6.5 m in 2.8 s. Calculate the frequency of a surfer bobbing up and down in the water. d = 6.5 m v = d/t t= 2.8 s =6.5/2.8 =2.32 m/s v=fλ 2.32=f(1.5) f=1.5 Hz

7 p. 435 key terms

8 Properties of Waves There are 4 fundamental properties of waves: 1. Reflection 2. Refraction 3. Diffraction 4. Interference 1. Reflection of Waves waves are reflected when they hit a barrier if waves hit a barrier straight on, they are reflected in the opposite direction barrier reflected waves incoming (incident) waves if waves hit a barrier at an angle, they will be reflected at an angle

9 barrier incident wave angle of incidence angle of reflection reflected wave normal Normal = a line perpendicular to the barrier angle of incidence = angle between the normal and the incident wave angle of reflection = angle between the normal and the reflected wave Law of Reflection = the angle of incidence is always equal to the angle of reflection Example If the angle between the incident wave and the reflected wave is 120, what is the angle of reflection?

10 barrier reflected wave incident wave = Refraction of Waves Refraction is the change in direction of waves at the boundary between two different media (substances). the frequency and period remain constant the wavelength, speed and direction of the waves change eg) waves from deep to shallow water, light through a prism, your legs in a pool normal i r

11 Angle of incidence (i) = angle between the normal and the incident wave Angle of refraction (r) = angle between the normal and the refracted wave angle i angle r Example If the speed of waves in deep water is 2.5 m/s and in shallow water is 1.0 m/s, and the wavelength is 10 m in deep water, what is the wavelength in shallow water? Deep Shallow v = 2.5 m/s v = 1.0 m/s λ = 10 m λ =? λ = 4 m v 1 = v 2 λ 1 λ 2 3. Diffraction of Waves Diffraction is the spreading of waves around the edge of a barrier

12 If there is a hole in the barrier, the wave will go through and bend the smaller the hole, the more the waves bend (eventually producing a circular wave) The longer the wavelength of the incident wave, the greater the diffraction eg) sound diffracts more than light, red light diffracts more than blue light (this is one reason why the sky is blue) diffracted wave reflected wave incident wave 4. Wave Interference Waves can combine to form larger or smaller waves when they interfere Constructive interference = when two or more waves in phase (same direction above or below rest position) combine to form a larger wave wave 1 combination of wave 2 waves 1 and 2

13 wave 1 Amplitude increases waves return to original size and position relative to rest position after they pass each other Destructive interference = when two or more waves out of phase (opposite direction above or below rest position) combine to form a smaller wave or cancel each other out combination of waves 1 and 2 wave 2 wave 1 combination of waves 1 and 2..cancels out wave 2 amplitude decreases Principle of Superposition = when two or more waves combine, the amplitude

14 of the resultant wave is the sum of the amplitudes of the individual waves Example Two waves of amplitude 5.0 cm and 8.0 cm interfere. Calculate the amplitude of the resultant if the waves are: a) in phase b) out of phase a) in phase 8.0 cm cm = 13 cm 13 cm 5.0 cm 8.0 cm b) out of phase cm cm = cm 5.0 cm 3.0 cm 8.0 cm

15 ys/mmedia/waves/swf.html Nodes and Antinodes Node = point of zero amplitude (caused by destructive interference) Antinode = point of maximum amplitude (caused by constructive interference) if you increase the frequency, you increase the number of nodes the distance between two adjacent nodes = ½ wavelength

16 node antinode Example The distance between adjacent nodes on a standing wave is 1.50 m. The frequency is 50.0 Hz. a) Calculate the wavelength Distance between 2 adjacent nodes = ½ wavelength λ = 1.50 m x 2 = 3.00 m b) Calculate the speed of the wave λ = 3.00 m v = fλ f = 50.0 Hz = (50.0)(3.00) = 150 m/s Example The distance between the second node and the sixth node is 50 cm. Find the wavelength of the wave.

17 2λ=50 cm 1λ=25 cm Waves at Boundaries Speed of a wave depends on the medium that it is passing through eg) water depends on depth sound depends on air temperature rope depends on thickness, tension (force) When a wave reaches a boundary between two media, part of the energy is transmitted into the new medium as a wave of the same frequency the wavelength and speed change once in the new medium the rest of the energy is reflected back as a wave of the same frequency in the original medium reflected wave transmitted wave medium 1 medium 2 incident wave

18 the amount of energy transmitted depends on the difference between the two media ie) small difference = most transmitted large difference = most reflected the relative densities of the two media affect the reflected and transmitted waves: Situation #1: Less Dense to More Dense reflected transmitted less dense incident more dense ia/waves/fix.html characteristics: 1. reflected wave is inverted

19 2. transmitted wave (in more dense medium) has shorter wavelength and smaller speed 3. frequency (and period) remain constant Situation #2: More Dense to Less Dense reflected wave transmitted wave more dense medium incident wave less dense medium characteristics: 4. reflected wave is upright(erect) 5. transmitted wave (in less dense medium) has longer wavelength and faster speed 6. frequency (and period) remain constant p. 698 #1-11 (Ch. 14)

20 Nodes and Antinodes Node = point of zero amplitude (caused by destructive interference) Antinode = point of maximum amplitude (caused by constructive interference) if you increase the frequency, you increase the number of nodes the distance between two adjacent nodes = ½ wavelength Sound Sound waves are longitudinal which means they consist of a series of compressions and rarefactions direction source of sound wave rarefaction compression

21 Sound requires a medium for transmission ie) it will not travel in a vacuum Travels through solids, liquids and gases Travels fastest through solids (particles closer together), slowest through gases Speed of sound in air is affected by temperature Changes by 0.60 m/s per C Speed of sound at 0 C in air = 332 m/s Example 1 Find the speed of sound at: a) 18 C v = 332 m/s + (18 x 0.60 m/s) = 332 m/s m/s = m/s = 3.4 x 10 2 m/s

22 b) -30 C v = 332 m/s + (-30 x 0.60 m/s) = 332 m/s + (-18 m/s) = 314 m/s = 3.1 x 10 2 m/s Example Calculate the time taken for an explosion at Syncrude, 50 km away, to reach Comp when the air temperature is -15 C. d = 50 km t = d = m v v = 332 m/s + (-15 x 0.60 m/s) = 323 m/s t=50,000/323 = s = 2.6 min Mach number = the number of times the speed of an object is greater than the speed of sound

23 mach # = v object v sound Example Find the mach number of a plane travelling at 1650 km/h at 12 C. v object = 1650 km/h = m/s v sound = (12 x 0.60 m/s) = m/s Mach number = /339.2 = = 1.4 Doppler Shift = change in frequency of sound caused by the motion of either the source or the detector When a source generating waves approaches an observer, the frequency increases When the source moves away, the frequency decreases You can t tell the direction the source is coming from but from the change

24 in pitch of the sound you can tell when it passes you away λ longer v same f decrease source towards λ- shorter v-same f increase eg) radar, Doppler radar for weather, ultrasound used by physicians, bats, ambulance traveling toward an observer, train approaching or moving away from a railway station Properties of Sound 1. Reflection sound waves can be reflected off of hard surfaces angle of incidence = angle of reflection 2. Refraction

25 the bending of sound waves occurs when they move through air at different temperatures 3. Diffraction sound waves can go around objects they diffract more than light because of longer wavelength 4. Interference Constructive and destructive interference can make sounds louder or softer Beats are produced at regular intervals due to the interference of two sound waves Beat frequency is simply the difference in frequency of two sources emitting a sound wave and is the number of beats heard per second B f = f 1 f 2 Where: B f = beat frequency in Hz or beats

26 per second f 1 = frequency of source 1 in Hz f 2 = frequency of source 2 in Hz Example : Calculate the beat frequency for two tuning forks. One is 256 Hz and the other is 250 Hz. f 1 = 256 Hz f 2 = 250 Hz B f = f 1 f 2 = = 6.00 Hz Example Two tuning forks produce 5.00 beats per second. How many beats are heard in 1.00 minute? 5.00 beats/s x 60.0 s = 300 beats

27 p. 309 #1-4 p. 324 (key terms) Example 3 A tuning fork with a frequency of 300 Hz is sounded with another tuning fork to produce 5.00 beats per second. What are the two possible frequencies for the second tuning fork? B f = 5.00 beats per second B f = f 1 f 2 f 1 = 300 Hz beats per second f 2 = f 1 ± B f = 300 Hz ± 5.00 = 305 Hz or 295 Hz Example 4 Some plasticine is placed on the 300 Hz tuning fork. Now only 3 beats are heard. What is the actual frequency of second tuning fork?

28 The plasticine decreases frequency. Since only fewer beats are heard, the frequency of the tuning fork is 295 Hz. Example A 200 Hz tuning fork is sounded with another tuning fork and 7 beats are heard. If some plasticine is placed on the 200 Hz tuning fork and 9 beats are heard, what is the actual frequency of the second tuning fork? Your Assignment: 34-38,40 Vibrating Strings 1. 1 st harmonic (Fundamental frequency) = lowest frequency making up a sound. The string vibrates as a whole segment

29 2. 2 nd harmonic (1 st overtone) = string vibrating in two segments 3. 3 rd harmonic (2 nd overtone) = string vibrating in three segments Mode of Vibration 1 st harmonic (fundamental frequency) # of frequency wavelength loops 1 f ½ λ

30 2 nd harmonic (1 st overtone) 3 rd harmonic (2 nd overtone) 2 2f 1 λ 3 3f 1 ½ λ Example Find the frequency of the second overtone if the fundamental frequency of the vibrating string is 150 Hz. f = 150 Hz 2 nd overtone = 3f = 3(150 Hz) = 450 Hz Factors Affecting the Frequency of Vibrating Strings 1. Tension of String (T): frequency is directly proportional to the square root of the tension f 1 = T 1 Equation #40 f 2 T 2 where: f 1, f 2 = frequencies in Hz T 1, T 2 = tensions in N

31 Example The frequency of a string is 200 Hz. What is the frequency if the tension is tripled? f 1 = 200 Hz f 1 = T 1 T 1 = 1 f 2 T 2 T 2 = 3 = (200 Hz)( 3) 1 = 346 Hz 2. Length of String (l) frequency is inversely proportional to the length of the string. fα1/l f 1 = l 2 Equation #39 f 2 l 1 where: f 1, f 2 = frequencies in Hz l 1, l 2 = lengths in m

32 Example The frequency of a 2.0 m long vibrating string is 400 Hz. If the length is increased to 2.5 m, what is the new frequency? f 1 = 400 Hz f 1 = l 2 l 1 = 2.0 m f 2 l 1 l 2 = 2.5 m f 2 = f 1 l 1 l 2 = (400 Hz)(2.0 m) = 2.5 m =3.2x10 2 Hz 3. Diameter of String (D) frequency is inversely proportional to the diameter of the string f 1 = D 2 Equation #41 f 2 D 1 where: f 1, f 2 = frequencies in Hz D 1, D 2 = diameters in m 4. Density of String (d)

33 frequency is inversely proportional to the square root of the density f 1 = d 2 f 2 d 1 where: f 1, f 2 = frequencies in Hz d 1, d 2 = densities(kg/m 3 ) Resonance two objects are said to be in resonance if they vibrate at the same frequency eg) cordless phones, remote and TV, garage door openers all objects have a natural frequency when an object is affected by a wave of the same frequency (sound, wind etc.), resonance causes an increase in the amplitude of the wave increase in amount of energy eg) louder sound, higher swing

34 Soldiers are asked to break step when they march on bridges. The bridge will otherwise reach its natural frequency. Characteristics of Musical Sound pitch: depends on frequency ie) frequency = pitch loudness: depends on amplitude (energy) ie) amplitude = loudness quality of sound = depends on overtone Vibrating Air Columns if a tuning fork is held above a column of air of a particular length, the column will vibrate at the same frequency as the fork ie) resonance occurs

35 the lengths of the air columns where resonance occurs are called the resonant lengths changing the length changes the pitch (frequency) 1. Closed Pipe Resonator a cylindrical tube of air with one end closed and a sound source at the other end a standing wave is set up with nodes and antinodes the shortest resonant length is ¼ λ l 1= 1/4λ Equation #33 where: l 1 = first resonant length in m λ = wavelength in m each additional resonant length is ½ λ further down the air column

36 1st resonant length: l 1 = λ 4 2nd resonant length: l 2 = 3λ 4 3rd resonant length: l 3 = 5λ 4 Example 1 If the shortest resonant length for a closed pipe resonator is 16.5 cm when a 500 Hz tuning fork is used, what is the speed of sound? f = 500 Hz λ = 4l l 1 = 16.5 cm = (4)(0.165 m)

37 = m = m v = fλ = (500 Hz)(0.660 m) = 330 m/s Example 2 If the first resonant length in a closed air column is 18.5 cm when sounded with a 480 Hz tuning fork, what is the speed of sound? v=355 m/s. Example 3 The shortest resonant length for a 470 Hz tuning fork over a closed pipe resonator is 18 cm. Find the speed of sound. f = 470 Hz λ = 4l l 1 = 18 cm = (4)(0.18 m) = 0.18 m = 0.72 m v = fλ = (470 Hz)(0.72 m) = m/s = 3.4 x 10 2 m/s

38 Example 3 Find the first three resonant lengths of a closed air column that will resonate with a 300 Hz tuning fork at 20 C v = 332 m/s + (0.60 m/s x 20) = 344 m/s f f = 300 Hz v=fλ 344=300λ λ = m l 1 =1/4λ=1/4x1.1466=0.29 m l 2= 3/4λ=3/4x1.146=0.86 m l 3 =5/4λ=5/4x1.1466=1.4 m 2. Open Pipe Resonator cylindrical column of air open at both ends and a sound source at one end a standing wave is set up with nodes and antinodes the shortest resonant length is ½ λ l 1 = λ

39 2 where: l 1 = first resonant length in m λ = wavelength in m each additional resonant length is ½ λ further down the air column 1st resonant length: l 1 = λ 2 2nd resonant length: l 2 = 2λ 2 3rd resonant length: l 3 = 3λ 2 Example 1 Find the shortest length of an open tube air column that will resonate with a 400 Hz tuning fork at 15 C.

40 v = 332 m/s + (0.60 m/s x 15) = 341 m/s f f = 400 Hz λ = v = 341 m/s 400 Hz = m 1. l 1 = λ = m = m = m Simple Harmonic Motion simple harmonic motion is motion where there is a restoring force that varies linearly with displacement objects undergo a change in velocity, acceleration, and force equilibrium position Equilibrium: mass at rest F = max a = max v = 0 F = 0 a = 0 v = max Inertia makes the mass continue to move up F = max a = max v = 0 Spring now compressed then gravity pulls it down

41 to calculate the period of a mass suspended on a spring: T = 2π m/k where: T = period in s m = mass in kg k = spring constant in N/m π = 3.14 Example Calculate the period for a spring with a force constant of 15 N/m if the mass suspended is 1.0 kg. k = 15 N/m m = 1.0 kg

42 π = 3.14 = (2)(3.14) (1.0 kg) (15 N/m) = s = 1.6 s F = 0 a = 0 v = max F = max a = max v = 0 F = max a = max v = 0 F = 0 a = 0 v = max rest position Assignment 1.A 0.23 kg object vibrates at the end of a horizontal spring (k=32 N/m) along

43 a frictionless surface. What is the period of the vibration? 2. An object vibrates at the end of a horizontal spring(k=115 N/m) along a frictionless surface. If the frequency of vibration is 1.50 Hz, what is the mass of the object?