PHYS_2014_Lecture_28

PHYS_2014_Lecture_28 - Lecture 28: Shock Waves (u = v) A B...

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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 28, Slide 1 Lecture 28: Shock Waves ( u = v ) B A u=v shockwave front Wave crests pile up and compress along the shockwave front, concentrating the energy of the waves in a small space. λ
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 28, Slide 2 Shock Waves ( u > v ) shockwave front θ u v sin v u = The ratio u / v is referred to as the Mach Number.
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 28, Slide 3
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 28, Slide 4 Boats or ships moving faster than the speed of waves in water produce bow shocks analogous to the shock waves produced when objects like bullets or aircrat break the sound barrier.
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 28, Slide 5 U.S. Navy F-14 Breaking the Sound Barrier
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 28, Slide 6 θ = 60 o sin m 316 s m 365 s sin sin60 v u v u = = = = r m 365 s 1.15 speed of sound m 316 s = × speed of sound at 20,000 ft altitude: 316 m/s
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 28, Slide 7 The fundamental frequency of the A 4 key on my baby grand piano is 440 Hz. I measured the length of the string to be 38 cm and the diameter of the string to be 0.3 cm. Piano strings in this frequency range are typically made of high quality carbon steel of density ρ = 7.8 g/cm 3 . What is the tension of the A 4 string? Solution: The frequencies for standing waves on strings fixed at both ends is: 2 v f n L = where L is the length of the string, v is the speed of the wave on the string, and n is an integer. ..1 in the case of the fundamental frequency.
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 28, Slide 8 2 v f L = The velocity of waves on a string is given by: T v μ = where T is the tension in the string and is the mass per unit length. Substitute for v in the first equation and solve for T . 2 2 2 2 1 2 4 4 T f L T L f T L f = = =
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 28, Slide 9 We still need μ , the mass per unit length. We know the string is 38 cm long, has a diameter of 0.3 cm and is made of steel with a density of ρ = 7.8 g/cm 3 . We can find the volume, V , of the string and multiply that by to get the mass. We can then divide the mass by the length of the string to obtain . ( ) ( ) ( ) 2
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This note was uploaded on 11/28/2011 for the course PSYCH 2014 taught by Professor Staff during the Fall '10 term at Oklahoma State.

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PHYS_2014_Lecture_28 - Lecture 28: Shock Waves (u = v) A B...

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