Traveling Waves: Refraction

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Presentation transcript:

Traveling Waves: Refraction Refraction: Zero Incident Angle (speed of propagation is the frequency times the wavelength) Crest B Crest A (time for crest B to reach medium 2) (distance traveled by crest A in time Dt) (Law of Refraction) (frequency is the same) R. Field 11/19/2013 University of Florida PHY 2053

Traveling Waves: Refraction Refraction: Incident Angle q1 (speed of propagation is the frequency times the wavelength) (incident angle) (refracted angle) (Law of Refraction) (Law of Refraction) (Law of Refraction) R. Field 11/19/2013 University of Florida PHY 2053

Standing Waves: Superposition (Right Moving) (Left Moving) (Speed of the travelling waves) Nodes fixed in position! 2A Standing Wave (not rotating) X L Add the two travelling waves together (right moving plus left moving) as follows: (Standing Wave) (Allowed wavelengths of the Standing Wave) (Allowed frequencies of the Standing Wave) (Fundamental Frequency is n = 1 ) R. Field 11/19/2013 University of Florida PHY 2053

Example Problem: Standing Waves A string in a grand piano is 2 m long and has a mass density of 1 g/m. If the fundamental frequency of oscillations of the string is 440 Hz, what is the tension in the string (in N)? Answer: 3097.6 R. Field 11/19/2013 University of Florida PHY 2053

Example Problem: Standing Waves A nylon guitar string has a linear density of 5 g/m and is under a tension of 200 N. The fixed supports are D = 60 cm apart. The string is oscillating in the standing wave pattern shown in the figure. What is the frequency of the traveling waves whose superposition gives this standing wave? Answer: 500 Hz R. Field 11/19/2013 University of Florida PHY 2053

Example Problem: Standing Waves A string, which is tied to a sinusoidal oscillator at P and which runs over a support Q, is stretched by a block of mass m. The distance L = 2.0 m, the linear mass density of the string m = 4.9 g/m, and the oscillator frequency f = 100 Hz. The motion at P is in the vertical direction, and its amplitude is small enough for that point to be considered a node. A node also exists at Q. What mass allows the oscillator to set up the second harmonic on the string? Answer: 20 kg (Second harmonic is n = 2) R. Field 11/19/2013 University of Florida PHY 2053