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PROBLEM 7 & 8 April 5 2010

PROBLEM 7 & 8 April 5 2010 - f = v = 10 Hz The...

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SOLVING PROBLEM 7 April 5, 2010 (like 13.11): A sinusoidal traveling wave is propagating to the right (in the + x direction) on a taught piece of string. At a certain instant the part of the string at the origin ( x =0) is at rest with amplitude y = 0.01 m. If the wave velocity is 10 m/s and the wavelength is 1 m, what is the velocity of the same part of the string when it later crosses the x -axis? a) 0.2 m/s; b) 2 m/s; c) 20 m/s; d) 0.02 m/s; e) 0.002 m/s. What do you know? If the portion of the string is at rest, the amplitude must be a maximum. Thus the amplitude of the wave is A = 0.01 m. If the wave velocity is v and the wavelength is , then the frequency is
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Unformatted text preview: f = v = 10 Hz. The velocity of the portion of string as it crosses the x-axis is Af A t kx A dt t kx dA dt t x dy 2 ) cos( ) sin( ) , ( 0.2 m/s. SOLVING PROBLEM 8 (like 13.15): What is the wave speed on a thin string of mass per unit length 0.001 kg/m suspending a mass of 1 kg? a) 100 m/s; b) 10 m/s; c) 1 m/s; d) 0.1 m/s; e) 0.01 m/s. What do you know? The speed of transverse traveling waves on a string is / T v , where T is the tension and is the mass per unit length. The tension is mg = 10 N. Thus / T v = (10 N/0.001 kg/m) 1/2 = 100 m/s...
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