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1 30 Class Average = Counts Scores
2 Chapter 12 Mechanical Waves and Sound To describe mechanical waves. To study superposition, standing waves, and interference. To understand sound as a longitudinal wave. To study sound intensity and beats. To understand the Doppler effect and frequency shifts.
3 12.1 Mechanical Wave Mechanical waves are produced by mechanical disturbance and there are two types. Transverse Waves The wave disturbance is perpendicular to the direction of wave propagation. crest trough Longitudinal Waves The wave disturbance is parallel to the direction of wave propagation.
4 12.2 Periodic Mechanical Wave y(x,t) λ o x Wave speed v = λ T = λf Periodic in time----at a given x, the displacement from equilibrium changes periodically in a time period T, which is 1/f where f is the frequency of the oscillations. Periodic in space---at a given time (taking a snap shot), the displacement from equilibrium repeats itself after a certain distance known as the wavelength λ. Periodic in propagation---in a time period T, the wave appears to propagate forward by a distance that is equal to the wavelength λ.
5 12.3 Waves Speed Speed of a Transverse Wave (in a rope or string) is v = F T : Tension force in N, μ: The linear mass density, mass per unit length, in kg/m. Example 12.2 on page 358 Waves on a Long Rope Under Tension Given: See the diagram. Ignore the weight of the rope for F T. Find: (a) the speed of the wave. (b) if f = 20 Hz, what is λ? Solution: (a) F T = m sample g = (20 kg)(9.8 m/s 2 ) = 196 N μ = m/l = (2.00 kg)/(80.0 m) = kg/m v = F T /μ = 88.5 m/s (b) λ = v/f = 4.43 m F T μ
6 12.4 Mathematical Description of a Wave How is the displacement from equilibrium related to position and time? y = f(x, t) Consider x = 0 where the wave is generated: y = A sin ωt = A sin 2πft Now, consider x 0. It takes a time t = x for the wave generated at x = 0 to v propagate to x. Or, the oscillations at x is delayed by a time t = x compared with the oscillations at x = 0. Therefore, v the oscillations at x is given by x = 0 t = 0 y x, t = A sin ω t x = A sin 2πf t x v v = A sin 2π t fx = A sin 2π t x T v T λ
7 12.5 Reflections and Superposition Reflections
8 Superposition: Vector addition of displacements of all the waves
9 12.6 Standing Waves and Normal Modes N: Nodes, points at which the string does not move, or, displacement is zero. A: Antinodes, points at which the string has the largest displacements.
10 Normal Modes: Allowed modes of wave oscillation L = n λ 2 (n = 1, 2, 3,.) Allowed wavelengths: λ n = 2L n (n = 1, 2, 3, ) Allowed frequencies: f n = v λ n = n v 2L = nf 1 (n = 1, 2, 3, ) Fundamental frequency: f 1 = v 2L The fundamental frequency of a vibrating string of length L, fixed at both ends, is f 1 = v 2L = 1 2L F T μ
11 Example 12.3 on page 367, a bass string Given: f 1 = 41.0 Hz, L = 0.86 m, μ = kg/m Find: (a) F T (b) f 2 and λ 2 (second harmonic) (c) f 3 and λ 3 (third harmonic) Solution: (a) Since f 1 = v 2L = 1 2L F T μ F T = 4μL 2 f 2 1 = 76 N (b) f 2 = 2f 1 = 82.0 Hz λ 2 = 2L = 0.86 m 2 (a) f 3 = 3f 1 = 123 Hz λ 3 = 2L = 0.57 m 3
12 Quantitative Analysis 12.4 on page 367 Given: The vibrating string has fundamental frequency f when a block of mass m is hung as shown. Find: (A) (B) (C) To increase the fundamental frequency of the vibration to 2f, what mass should be hung on the string? 2m 2m 4m
13 12.7 Longitudinal Standing Waves
14 Open Pipe and Normal Modes L = n λ n (n = 1, 2, 3,.) 2 Allowed wavelengths: λ n = 2L n (n = 1, 2, 3, ) Allowed frequencies: f n = v λ n = n v 2L = nf 1 (n = 1, 2, 3, ) Fundamental frequency: f 1 = v 2L
15 Stopped Pipe and Normal Modes L = n λ n (n = 1, 3, 5, 7.) 4 Allowed wavelengths: λ n = 4L n Allowed frequencies: (n = 1, 3, 5, 7 ) f n = v λ n = n v 4L = nf 1 (n = 1, 3, 5, 7 ) Fundamental frequency: f 1 = v 4L
16 Example 12.4 on page 369 Speed of Sound in Hydrogen Given: f = 25 khz Find: Speed of the sound wave Solution: λ/2 = m λ = m v = λf = (0.052 m)( Hz) = 1300 m/s
17 Example 12.5 on page 371: Harmonics of an organ pipe Given: Speed of sound v = 345 m/s; Open pipe f 1 = 690 Hz Question: If the n = 2 mode of the open pipe has the same wavelength as the n = 5 mode of a stopped pipe, what is the length of each pipe? Solution: (a) For the open pipe f 1 = Therefore, (b) λ 2 open = λ 5 stopped L open = v Given λ n open = 2L open we have Therefore, n 2f 1 = and 2L open 2 v 2L open 345 m/s 2(690 Hz) = m λ n stopped = 4L stopped n = 4L stopped 5 L stopped = 5L open 4 = m,
18 12.8 Interference S 1 S 2 S 1 S 2 P d 1 d 2 Constructive Interference: d = d 1 d 2 = nλ = n v f (n = 0, 1, 2, 3, ) Destructive Interference: d = d 1 d 2 = n λ = n (n = 0, 1, 2, 3, ) v f
19 12.9 Sound and Hearing Sound Intensity Intensity: I = P 4πr 2 Decibels: β = 10 db log I I 0 with I 0 = W/m 2.
20 12.11 Beats in phase out of phase Let T 1 < T 2. Beat occurs when nt 1 = T beat = (n - 1)T 2 Or, T beat = T 1T 2 T 2 T 1, f beat = 1 T beat = T 2 T 1 T 1 T 2 = 1 T 1 1 T 2 = f 1 f 2
21 12.12 The Doppler Effect Consider a stationary source: emitting sound at frequency f s and wavelength λ = v/f s, where v is the speed of sound in air. Moving listener What is the frequency of the sound that a moving listener hear? To a listener moving at a speed v L toward the stationary sound source, the speed of sound is v + v L. Therefore, f L = v+v L λ = v+v L v f s.
22 The Doppler Effect Continued Consider a source moving at a speed v S, emitting sound wave at a frequency f s. What is the frequency of the sound that a stationary listener hear? To a listener, the speed of the sound is not affected by the motion of the source. It is the speed of sound in air, v. However, the wavelength of the sound is affected by the speed of the source: λ = v f s + v S T = v f s + v S f s = v+v S f s. Therefore, the frequency the listnere hears is: f L = v = λ v v+v S f s Moving Source
23 The Doppler Effect Continued What if both the source and the listener are moving? What is the frequency of the sound that the listener hear? Moving source and moving listener f L = v+v L λ = v+v L v+v S f s, since λ = v+v S f s. How to determine the signs of v L and v S? General rule: moving closer leads to an increasing frequency, vice versa. Therefore: If listener moving toward source, v L is positive, and, vice versa. If source moving toward listener, v S is negative, and, vice versa.
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