Chapter 17 - Superposition (Physics)

Physics Cutnell & Johnson Chapter 17 The Principle of Linear Superposition and Interference Phenomena Copyright ©2022 John Wiley & Sons, Inc.  Physics Cutnell & Johnson Chapter 17 The Principle of Line...

Physics Cutnell & Johnson Chapter 17 The Principle of Linear Superposition and Interference Phenomena Copyright ©2022 John Wiley & Sons, Inc.  Physics Cutnell & Johnson Chapter 17 The Principle of Linear Superposition and Interference Phenomena Copyright ©2022 John Wiley & Sons, Inc. 17.1 The Principle of Linear Superposition (1 of 3) Copyright ©2022 John Wiley & Sons, Inc. 3 17.1 The Principle of Linear Superposition (1 of 3) When the pulses merge, the Slinky assumes a shape that is the sum of the shapes of the individual pulses. Copyright ©2022 John Wiley & Sons, Inc. 4 17.1 The Principle of Linear Superposition (2 of 3) When the pulses merge, the Slinky assumes a shape that is the sum of the shapes of the individual pulses. Copyright ©2022 John Wiley & Sons, Inc. 5 17.1 The Principle of Linear Superposition (3 of 3) The Principle of Linear Superposition When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves. Copyright ©2022 John Wiley & Sons, Inc. 6 17.1 The Principle of Linear Superposition (3 of 3) The Principle of Linear Superposition When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves. ‘Reinforcement’ A1 A1+A 2 + = A2 Constructive Superposition (Phase) Copyright ©2022 John Wiley & Sons, Inc. 7 17.1 The Principle of Linear Superposition (3 of 3) The Principle of Linear Superposition When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves. A1 A1 - A1 = 0 + = Destructive Superposition A1 (Antiphase) ‘Cancellation’ Copyright ©2022 John Wiley & Sons, Inc. 8 17.1 The Principle of Linear Superposition (3 of 3) The Principle of Linear Superposition When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves. Copyright ©2022 John Wiley & Sons, Inc. 9 17.2 Constructive and Destructive Interference of Sound Waves (3 of 8) WHY SHOULD WE CARE? Copyright ©2022 John Wiley & Sons, Inc. 10 17.2 Constructive and Destructive Interference of Sound Waves (Noise cancelling headphones) Copyright ©2022 John Wiley & Sons, Inc. 11 17.2 Constructive and Destructive Interference of Sound Waves (1 of 8) When two waves always meet condensation-to-condensation and rarefaction-to-rarefaction, they are said to be exactly in phase and to exhibit constructive interference. Copyright ©2022 John Wiley & Sons, Inc. 12 17.2 Constructive and Destructive Interference of Sound Waves (1 of 8) When two waves always meet condensation-to-condensation and rarefaction-to-rarefaction, they are said to be exactly in phase and to exhibit constructive interference. Copyright ©2022 John Wiley & Sons, Inc. 13 17.2 Constructive and Destructive Interference of Sound Waves (2 of 8) When two waves always meet condensation-to-rarefaction, they are said to be exactly out of phase and to exhibit destructive interference. Copyright ©2022 John Wiley & Sons, Inc. 14 Practice problem Two pulses on a string, travelling at 10 m/s, are approaching each other. Draw a snapshot graph of the string 1.0 seconds later and explain the principles applied to create the snapshot. Copyright ©2022 John Wiley & Sons, Inc. 15 Practice problem Two pulses on a string, travelling at 10 m/s, are approaching each other. Draw a snapshot graph of the string 1.0 seconds later and explain the principles applied to create the snapshot. +1 0 0 Copyright ©2022 John Wiley & Sons, Inc. 16 Practice problem Two pulses on a string, travelling at 10 m/s, are approaching each other. Draw a snapshot graph of the string 1.0 seconds later and explain the principles applied to create the snapshot. +1 0 0 0 -1 -1 Copyright ©2022 John Wiley & Sons, Inc. 17 Practice problem Two pulses on a string, travelling at 10 m/s, are approaching each other. Draw a snapshot graph of the string 1.0 seconds later and explain the principles applied to create the snapshot. The Principle of Linear Superposition When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves. +1 0 0 0 00 -1 -1 -1 Copyright ©2022 John Wiley & Sons, Inc. 18 17.2 Constructive and Destructive Interference of Sound Waves (4 of 8) If the wave patters do not shift relative to one another as time passes, the sources are said to be coherent. For two wave sources vibrating in phase, a difference in path lengths that is zero or an integer number (1, 2, 3,.. ) of wavelengths leads to constructive interference; 1 1 1  a difference in path lengths that is a half-integer number  2 ,1 2 , 2 2 ,...  of wavelengths leads to destructive interference. Copyright ©2022 John Wiley & Sons, Inc. 19 17.2 Constructive and Destructive Interference of Sound Waves (8 of 8) constructive interference: destructive interference: Copyright ©2022 John Wiley & Sons, Inc. 20 17.2 Constructive and Destructive Interference of Sound Waves (8 of 8) constructive interference: destructive interference: Copyright ©2022 John Wiley & Sons, Inc. 21 17.2 Constructive and Destructive Interference of Sound Waves (8 of 8) constructive interference: destructive interference: Copyright ©2022 John Wiley & Sons, Inc. 22  Review 16.5 The Nature of Sound Waves (1 of 5) Click picture to try simulation PhET: Waves intro PhET: Sound Waves Copyright ©2022 John Wiley & Sons, Inc. 23 17.2 Constructive and Destructive Interference of Sound Waves (5 of 8) Example 1 What Does a Listener Hear? Two in-phase loudspeakers, A and B, are separated by 3.20 m. A listener is stationed at C, which is 2.40 m in front of speaker B. Both speakers are playing identical 214-Hz tones, and the speed of sound is 343 m/s. Does the listener hear a loud sound, or no sound? Copyright ©2022 John Wiley & Sons, Inc. 24  17.2 Constructive and Destructive Interference of Sound Waves (6 of 8) Calculate the path length difference. 2 2 = 3.20 m   2.40 m   2.40 m 1.60 m Calculate the wavelength. −1 =343 𝑚. 𝑠 v 343m s −1  f 214 Hz 1.60 m = =214 𝑠 For two wave sources vibrating in phase, a difference in path lengths that is zero or an integer number (1, 2, 3,.. ) of wavelengths leads to constructive interference; LOUD SOUND Copyright ©2022 John Wiley & Sons, Inc. 25  17.2 Constructive and Destructive Interference of Sound Waves (6 of 8) Calculate the path length difference. 2 2 = 3.20 m   2.40 m   2.40 m 1.60 m Calculate the wavelength. −1 =343 𝑚. 𝑠 v 343m s −1  f 214 Hz 1.60 m = =214 𝑠 For two wave sources vibrating in phase, a difference in path lengths that is zero or an integer number (1, 2, 3,.. ) of wavelengths leads to constructive interference; LOUD SOUND Copyright ©2022 John Wiley & Sons, Inc. 26 17.3 Diffraction (1 of 3) The bending of a wave around an obstacle or the edges of an opening is called diffraction. Copyright ©2022 John Wiley & Sons, Inc. 27 17.3 Diffraction (2 of 3) longer wavelength more spreadout wider slit (normal door) less spread out  single slit – first minimum sin   D Copyright ©2022 John Wiley & Sons, Inc. 28 17.3 Diffraction (2 of 3) 𝝀  single slit – first minimum sin   D Copyright ©2022 John Wiley & Sons, Inc. 29 17.3 Diffraction (2 of 3) longer wavelength more spreadout short wavelength long wavelength (red) (blue) less spreadout more sinspreadout 𝜃= 𝜆 single slit – first minimum 𝐷 Copyright ©2022 John Wiley & Sons, Inc. 30 17.3 Diffraction (2 of 3) wider slit less spread out large slit (normal door) small slit (slit in lab) less spreadout more spreadout 𝜆 sin 𝜃 = single slit – first minimum 𝐷 Copyright ©2022 John Wiley & Sons, Inc. 31 17.3 Diffraction (3 of 3) samething, but for circular openings(add 1.22) longer wavelength more spreadout wider slit (normal door) less spread out 𝜆 Circular opening – first minimum sin 𝜃 =1.22 𝐷 Copyright ©2022 John Wiley & Sons, Inc. 32 17.3 Diffraction (3 of 3) longer wavelength more spreadout short wavelength long wavelength (red) (blue) less spreadout Circular opening – first minimum more spreadout sin 𝜃 =1.22 𝜆 𝐷 Copyright ©2022 John Wiley & Sons, Inc. 33  17.3 Diffraction (3 of 3) longer wavelength more spreadout short wavelength long wavelength (red) (blue) less spreadout Circular opening – first minimum more spreadout sin 𝜃 =1.22 𝜆 𝐷 distance from center to first dark line Copyright ©2022 John Wiley & Sons, Inc. 34  17.3 Diffraction (3 of 3) wider slit less spread out large slit (normal door) small slit (slit in lab) less spreadout more spreadout 𝜆 Circular opening – first minimum sin 𝜃 =1.22 𝐷 distance from center to first dark line Copyright ©2022 John Wiley & Sons, Inc. 35  17.3 Diffraction (3 of 3) A 3000-Hz sound and a 8000-Hz sound emerge from a loudspeaker through a circular opening that has a diameter of 0.35 m (see Figure 17.11). Assuming that the speed of sound in air is 343 m/s, find the diffraction angle θ for each sound. 𝑓 1 =3000 𝐻𝑧 𝑓 2 =8000 𝐻𝑧 𝐷=0.35 𝑚 𝑣 𝑎𝑖𝑟 =343 𝑚/ 𝑠 𝜃? 𝜃 1= 2= ? Copyright ©2022 John Wiley & Sons, Inc. 36  17.3 Diffraction (3 of 3) A 3000-Hz sound and a 8000-Hz sound emerge from a loudspeaker through a circular opening that has a diameter of 0.35 m (see Figure 17.11). Assuming that the speed of sound in air is 343 m/s, find the diffraction angle θ for each sound. 𝜆 sin 𝜃 =1.22 𝑓 1 =3000 𝐻𝑧 𝐷 𝑓 2 =8000 𝐻𝑧 𝐷=0.35 𝑚 𝑣 𝑎𝑖𝑟 =343 𝑚/ 𝑠 𝜃? 𝜃 1= 2= ? Copyright ©2022 John Wiley & Sons, Inc. 37  17.3 Diffraction (3 of 3) A 3000-Hz sound and a 8000-Hz sound emerge from a loudspeaker through a circular opening that has a diameter of 0.35 m (see Figure 17.11). Assuming that the speed of sound in air is 343 m/s, find the diffraction angle θ for each sound. 𝜆 sin 𝜃 =1.22 𝑓 1 =3000 𝐻𝑧 𝐷 𝑓 2 =8000 𝐻𝑧 → 𝜃 =sin − 1 ( 1.22 𝜆 𝐷 ) 𝐷=0.35 𝑚 𝑣 𝑎𝑖𝑟 =343 𝑚/ 𝑠 𝑣=𝑓 𝜆 𝜃? 𝜃 1= ? 𝜆= 2=→ 𝑣 𝑓 𝜆 =? Copyright ©2022 John Wiley & Sons, Inc. 38  17.3 Diffraction (3 of 3) A 3000-Hz sound and a 8000-Hz sound emerge from a loudspeaker through a circular opening that has a diameter of 0.35 m (see Figure 17.11). Assuming that the speed of sound in air is 343 m/s, find the diffraction angle θ for each sound. 𝜆 sin 𝜃 =1.22 𝑓 1 =3000 𝐻𝑧 𝐷 𝑓 2 =8000 𝐻𝑧 → 𝜃 =sin − 1 ( 1.22 𝜆 𝐷 ) 𝐷=0.35 𝑚 𝑣 𝑎𝑖𝑟 =343 𝑚/ 𝑠 𝑣=𝑓 𝜆 𝜃? 𝜃 1= ? 𝜆= 2=→ 𝑣 𝑓 ¿ sin − 1 ( 1.22 𝑣/ 𝑓 𝐷 ) 𝜆 =? Copyright ©2022 John Wiley & Sons, Inc. 39  17.3 Diffraction (3 of 3) A 3000-Hz sound and a 8000-Hz sound emerge from a loudspeaker through a circular opening that has a diameter of 0.35 m (see Figure 17.11). Assuming that the speed of sound in air is 343 m/s, find the diffraction angle θ for each sound. 𝜃 1= sin − 1 ( 1.22 𝑣 𝑎𝑖𝑟 / 𝑓 𝐷 1 ) 𝑓 1 =3000 𝐻𝑧 𝑓 2 =8000 𝐻𝑧 ( ) −1 −1 −1 343 𝑚. 𝑠 / 3000 𝑠 𝜃 1= sin 1.22 0.35 𝑚 𝐷=0.35 𝑚 𝑣 𝑎𝑖𝑟 =343 𝑚/ 𝑠 𝜃 𝜃 1 =1 ?= 23.47 ¿ sin °1.22 𝑣 / 𝑓 −1 ( 𝐷 ) 𝜃 2= ? Copyright ©2022 John Wiley & Sons, Inc. 40 17.3 Diffraction (3 of 3) A 3000-Hz sound and a 8000-Hz sound emerge from a loudspeaker through a circular opening that has a diameter of 0.35 m (see Figure 17.11). Assuming that the speed of sound in air is 343 m/s, find the diffraction angle θ for each sound. − 1( ( 𝑣 𝑎𝑖𝑟 / 𝑓 2 ) ) 𝑣 𝑎𝑖𝑟 / 𝑓 𝜃 1= sin − 1 1.22 1 𝑓 1 =3000 𝐻𝑧 𝜃2 = sin 1.22 𝐷 𝐷 𝑓 2 =8000 𝐻𝑧 ( ) −1 −1 −1 343 𝑚. 𝑠 / 8000 𝑠 𝜃 2= sin 1.22 0.35 𝑚 𝐷=0.35 𝑚 𝑣 𝑎𝑖𝑟 =343 𝑚/ 𝑠 𝜃 2= 8.60 ° 𝜃1=23.47° 𝜃 2= ? Copyright ©2022 John Wiley & Sons, Inc. 41  17.3 Diffraction (3 of 3) A 3000-Hz sound and a 8000-Hz sound emerge from a loudspeaker through a circular opening that has a diameter of 0.35 m (see Figure 17.11). Assuming that the speed of sound in air is 343 m/s, find the diffraction angle θ for each sound. 𝑓 1 =3000 𝐻𝑧 𝑓 2 =8000 𝐻𝑧 𝐷=0.35 𝑚 𝑣 𝑎𝑖𝑟 =343 𝑚/ 𝑠 𝜃1=23.47° longer wavelength more spreadout 𝜃 2= 8.60 ° 1 𝜆∝ 𝑓 Copyright ©2022 John Wiley & Sons, Inc. 42  17.3 Diffraction (3 of 3) A 3000-Hz sound and a 8000-Hz sound emerge from a loudspeaker through a circular opening that has a diameter of 0.35 m (see Figure 17.11). Assuming that the speed of sound in air is 343 m/s, find the diffraction angle θ for each sound. 𝑓 1 =3000 𝐻𝑧 𝑓 2 =8000 𝐻𝑧 𝐷=0.35 𝑚 𝑣 𝑎𝑖𝑟 =343 𝑚/ 𝑠 𝜃1=23.47° 𝜃 2= 8.60 ° smaller frequency more spreadout Copyright ©2022 John Wiley & Sons, Inc. 43 17.4 Beats (1 of 2) Two overlapping waves with slightly different frequencies gives rise to the phenomena of beats. Copyright ©2022 John Wiley & Sons, Inc. 44 17.4 Beats (2 of 2) The beat frequency is the difference between the two sound frequencies. 𝑓 𝑏𝑒𝑎𝑡 =¿ 𝑓 2 − 𝑓 1∨¿ Copyright ©2022 John Wiley & Sons, Inc. 45 17.4 Beats (2 of 2) The beat frequency is the difference between the two sound frequencies. 𝑓 𝑏𝑒𝑎𝑡 =¿ 𝑓 2 − 𝑓 1∨¿ Copyright ©2022 John Wiley & Sons, Inc. 46 17.5 Transverse Standing Waves (1 of 5) Transverse standing wave patters. Copyright ©2022 John Wiley & Sons, Inc. 47 17.5 Transverse Standing Waves (2 of 5) In reflecting from the wall, a forward-traveling half-cycle becomes a backward-traveling half-cycle that is inverted. Unless the timing is right, the newly formed and reflected cycles tend to offset one another. Repeated reinforcement between newly created and reflected cycles causes a large amplitude standing wave to develop. Copyright ©2022 John Wiley & Sons, Inc. 48 17.5 Transverse Standing Waves (2 of 5) Standing / stationary waves are created by the superposition of two progressive (moving) waves of equal frequency and amplitude in opposite directions Copyright ©2022 John Wiley & Sons, Inc. 49 17.5 Transverse Standing Waves (2 of 5) The points of zero amplitude are called nodes and the maxima are called antinodes  No displacement at the nodes N A N A N A N A N Pink = instantaneous displacement of particles Blue = displacement of particles T/2 later Copyright ©2022 John Wiley & Sons, Inc. 50 17.5 Transverse Standing Waves (2 of 5) Copyright ©2022 John Wiley & Sons, Inc. 51  17.5 Transverse Standing Waves (3 of 5) 𝑛 =1 𝑛 =2 𝑛 =3 Standing wave frequency → 𝑓 𝑛 =𝑛 ( 𝑣 ) 𝑛=1 2 𝐿 , 2 , 3 ,… (String fixed at both ends) Copyright ©2022 John Wiley & Sons, Inc. 52  17.5 Transverse Standing Waves (3 of 5) 𝑣=𝑓 𝜆 𝑣 → 𝑓 = 𝜆 Standing wave frequency → 𝑓 𝑛 =𝑛 ( 𝑣 ) 𝑛=1 2 𝐿 , 2 , 3 ,… (String fixed at both ends) Copyright ©2022 John Wiley & Sons, Inc. 53 The heaviest string on an electric guitar has a linear density of kg/m and is stretched with a tension of 284 N. This string produces the musical note E when vibrating along its entire length in a standing wave at the fundamental frequency of 164.8 Hz. (a) Find the length L of the string between its two fixed ends (see Figure 17.17a). −3 𝑚/ 𝐿=6.24 × 10 kg / m 𝐹 =284 𝑁 𝑓 𝑛 =164.8 𝐻𝑧 Copyright ©2022 John Wiley & Sons, Inc. 54 The heaviest string on an electric guitar has a linear density of kg/m and is stretched with a tension of 284 N. This string produces the musical note E when vibrating along its entire length in a standing wave at the fundamental frequency of 164.8 Hz. (a) Find the length L of the string between its two fixed ends (see Figure 17.17a). −3 𝑚/ 𝐿=6.24 × 10 kg / m 𝐹 =284 𝑁 𝑓 𝑛 =164.8 𝐻𝑧 𝑛 =1 Copyright ©2022 John Wiley & Sons, Inc. 55  The heaviest string on an electric guitar has a linear density of kg/m and is stretched with a tension of 284 N. This string produces the musical note E when vibrating along its entire length in a standing wave at the fundamental frequency of 164.8 Hz. (a) Find the length L of the string between its two fixed ends (see Figure 17.17a). −3 𝑚/ 𝐿=6.24 × 10 kg / m 𝐹 =284 𝑁 𝑓 𝑛 =164.8 𝐻𝑧 𝑛 =1 𝐿=? 𝑛 =𝑛 ( 𝑣 2 𝐿 ) Copyright ©2022 John Wiley & Sons, Inc. 56  The heaviest string on an electric guitar has a linear density of kg/m and is stretched with a tension of 284 N. This string produces the musical note E when vibrating along its entire length in a standing wave at the fundamental frequency of 164.8 Hz. (a) Find the length L of the string between its two fixed ends (see Figure 17.17a). −3 𝑚/ 𝐿=6.24 × 10 kg / m 𝐹 =284 𝑁 𝑓 𝑛 =164.8 𝐻𝑧 𝑛 =1 √ 𝐹 𝑣 = 𝑚 / 𝐿 𝐿=? 𝑣 =? =𝑛 ( 𝑛 2 𝐿 ) 𝑣→ 𝐿 =𝑛 (2 𝑓 ) 𝑣 𝑛 Copyright ©2022 John Wiley & Sons, Inc. 57  The heaviest string on an electric guitar has a linear density of kg/m and is stretched with a tension of 284 N. This string produces the musical note E when vibrating along its entire length in a standing wave at the fundamental frequency of 164.8 Hz. (a) Find the length L of the string between its two fixed ends (see Figure 17.17a). −3 𝑚/ 𝐿=6.24 × 10 kg / m 𝐹 =284 𝑁 𝑓 𝑛 =164.8 𝐻𝑧 𝑛 =1 √ 𝐹 𝑣 = 𝑚 / 𝐿 𝐿=? (√ ) 𝐹 𝑣 =? → 𝐿 =𝑛 𝑚 / 𝐿 2 𝑓 𝑛 =𝑛 ( 𝑛 2 𝐿 ) 𝑣→ 𝐿 =𝑛 (2 𝑓 ) 𝑣 𝑛 Copyright ©2022 John Wiley & Sons, Inc. 58 The heaviest string on an electric guitar has a linear density of kg/m and is stretched with a tension of 284 N. This string produces the musical note E when vibrating along its entire length in a standing wave at the fundamental frequency of 164.8 Hz. (a) Find the length L of the string between its two fixed ends (see Figure 17.17a). −3 𝑚/ 𝐿=6.24 × 10 kg / m √ ( ) 𝐹 𝑚 / 𝐿 → 𝐿 =𝑛 𝐹 =284 𝑁 2 𝑓 𝑛 𝑓 𝑛 =164.8 𝐻𝑧 𝑛 =1 (√ ) 284 𝑁 − 3 6.24 × 10 kg / m 𝐿=? → 𝐿 =1 2 ( 16 5. 0 𝐻𝑧 ) → 𝐿 =0.65 𝑚 Copyright ©2022 John Wiley & Sons, Inc. 59 The heaviest string on an electric guitar has a linear density of kg/m and is stretched with a tension of 284 N. This string produces the musical note E when vibrating along its entire length in a standing wave at the fundamental frequency of 164.8 Hz. (b) A guitar player wants the string to vibrate at a fundamental frequency of 2 × 164.8 Hz = 329.6 Hz. What length must she press? −3 𝑚/ 𝐿=6.24 × 10 kg / m √ ( ) 𝐹 𝑚 / 𝐿 → 𝐿 =𝑛 𝐹 =284 𝑁 2 𝑓 𝑛 𝑓 𝑛 =329.6 𝐻𝑧 𝑛 =1 (√ ) 284 𝑁 − 3 6.24 × 10 kg / m 𝐿=? → 𝐿 =1 2 ( 329.6 𝐻𝑧 ) → 𝐿 =0.32 𝑚 Copyright ©2022 John Wiley & Sons, Inc. 60 17.5 Transverse Standing Waves (4 of 5) →𝑚 𝐿 =0.65 𝐿 =0.32 𝑚 𝑓 𝑛 =164.8 𝐻𝑧 𝑓 𝑛 =329.6 𝐻𝑧  v  f n n   n 1,2,3,4,  2L  Copyright ©2022 John Wiley & Sons, Inc. 61 17.5 Transverse Standing Waves (5 of 5) Conceptual Example 5 The Physics of the Frets on a Guitar Look at the figure, which shows the distances where the frequency doubles. Explain why pressing at these distances would double the frequency  v  f n n   n 1,2,3,4,  2𝑓L ∝ 1 𝐿 Frequency increases with smaller length 1 Frequency doubles, 2 𝑓 ∝ 1 𝐿 2 If lengths halves Copyright ©2022 John Wiley & Sons, Inc. 62 17.6 Longitudinal Standing Waves (1 of 5) A longitudinal standing wave pattern on a slinky. Copyright ©2022 John Wiley & Sons, Inc. 63 17.6 Longitudinal Standing Waves (2 of 5)  v  Tube open at both ends f n n   n 1,2,3,4,  2L  Copyright ©2022 John Wiley & Sons, Inc. 64  17.6 Longitudinal Standing Waves (3 of 5) When all the holes are closed on one type of flute, the lowest note it can sound is middle C (261.6 Hz). If the speed of sound is 343 m/s, and the flute is assumed to be a cylinder open at both ends, determine the distance L. 𝑓 𝑛 =261.6 𝐻𝑧 𝑓 𝑛 =𝑛 ( 𝑣 2 𝐿 ) 43 𝑛 =1 𝐿=? Copyright ©2022 John Wiley & Sons, Inc. 65  17.6 Longitudinal Standing Waves (3 of 5) When all the holes are closed on one type of flute, the lowest note it can sound is middle C (261.6 Hz). If the speed of sound is 343 m/s, and the flute is assumed to be a cylinder open at both ends, determine the distance L. 𝑓 𝑛 =261.6 𝐻𝑧 𝑓( 𝑛 =𝑛 𝑣 2 𝐿 ) 43 𝑛 =1 → 𝐿 =𝑛 ( 𝑣 2 𝑓 𝑛 ) 𝐿=? ( ) −1 343 𝑚. 𝑠 𝐿=(1 ) 2 (261.6 𝐻𝑧 ) 𝐿=0.656 𝑚 Copyright ©2022 John Wiley & Sons, Inc. 66 17.6 Longitudinal Standing Waves (5 of 5)  v  Tube open at one end f n n   n 1,3,5,  4L  Copyright ©2022 John Wiley & Sons, Inc. 67 17.6 Standing Waves (5 of 5) Both ends closed 𝑓 𝑛 =𝑛 ( 2𝑣𝐿 ) Both ends open 𝑓 𝑛 =𝑛 ( 𝑣 2 𝐿 ) 𝑣 Tube open at one end 𝑓 𝑛 =𝑛 4 𝐿 closed ( ) 4 looks like a tube, 1 side closed 1 side open open Copyright ©2022 John Wiley & Sons, Inc. 68  The human ear canal (Figure 17.23) essentially acts like a tube closed at one end. If the ear canal has a length of 2.3 cm, what are the fundamental wavelength and frequency for standing waves in the ear? Take the speed of sound to be 343 m/s. 𝐿=2.3 ¿ 𝑐𝑚0.023 𝑚 𝜆 𝑛=? 𝑛 =? 𝑣 =343 𝑚/ 𝑠 𝑛 =1 Copyright ©2022 John Wiley & Sons, Inc. 69  The human ear canal (Figure 17.23) essentially acts like a tube closed at one end. If the ear canal has a length of 2.3 cm, what are the fundamental wavelength and frequency for standing waves in the ear? Take the speed of sound to be 343 m/s. 𝐿=0.02 3𝑚 𝜆 𝑛=? 𝑛 =? 𝑣 =343 𝑚/ 𝑠 𝑛 =1 𝑓 𝑛 =𝑛 ( 𝑣 4 𝐿 ) Copyright ©2022 John Wiley & Sons, Inc. 70  The human ear canal (Figure 17.23) essentially acts like a tube closed at one end. If the ear canal has a length of 2.3 cm, what are the fundamental wavelength and frequency for standing waves in the ear? Take the speed of sound to be 343 m/s. 𝐿=0.02 3𝑚 𝑣=𝑓 𝜆 𝜆 𝑛=? 𝑛 =? 𝑣 =343 𝑚/ 𝑠 𝑛 =1 ( 4𝑣𝐿 ) ( ) −1 343 𝑚. 𝑠 ¿ (1 ) 𝑓 =𝑛 ¿ 3728.26 𝐻𝑧 𝑛 4 (0.023 𝑚 ) Copyright ©2022 John Wiley & Sons, Inc. 71  The human ear canal (Figure 17.23) essentially acts like a tube closed at one end. If the ear canal has a length of 2.3 cm, what are the fundamental wavelength and frequency for standing waves in the ear? Take the speed of sound to be 343 m/s. 𝑣 → 𝜆= 𝐿=0.02 3𝑚 𝑣=𝑓 𝜆 𝑓 𝜆 𝑛=? 343 𝑚/ 𝑠 → 𝜆= 𝑛 =? 3728.26 𝐻𝑧 𝑣 =343 𝑚/ 𝑠 → 𝜆=0.092 𝑚 𝑛 =1 ( 4𝑣𝐿 ) ( ) −1 343 𝑚. 𝑠 ¿ (1 ) 𝑓 =𝑛 ¿ 3728.26 𝐻𝑧 𝑛 4 (0.023 𝑚 ) Copyright ©2022 John Wiley & Sons, Inc. 72 17.7 Complex Sound Waves (1 of 2) SKIP Copyright ©2022 John Wiley & Sons, Inc. 73 17.7 Complex Sound Waves (2 of 2) SKIP Copyright ©2022 John Wiley & Sons, Inc. 74 Copyright Copyright © 2022 John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that permitted in Section 117 of the 1976 United States Act without the express written permission of the copyright owner is unlawful. Request for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. The purchaser may make back-up copies for his/her own use only and not for distribution or resale. 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