homework14 - This print-out should have 12 questions....

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Unformatted text preview: This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points For a transverse wave on a string the string displacement is described by y ( x, t ) = f ( x- at ) where f is a given function and a is a positive constant. Which of the following does NOT neces- sarily follow from this statement? 1. The speed of the waveform is x t . correct 2. The waveform moves in the positive x direction. 3. The shape of the waveform does not change as it moves along the string. 4. The speed of the waveform is a . 5. The shape of the string at time t = 0 is given by f ( x ). Explanation: It is obvious that the shape at time t = 0 is f ( x ), and since the displacement is f ( x- at ), it does not change shape as it moves along (since the amplitude is independent of time). Furthermore, since the argument of f is x- at , as time increases we need to move to larger x to keep the phase the same, i.e. the waveform moves to the right. Since the general form for a traveling wave is f ( x- v t ), we see that the speed of the waveform is a , not x t . 002 (part 1 of 1) 10 points The lowest A on a piano has a frequency of 27 . 5 Hz. Assume: The tension in the A piano wire (of length 2 m) is 300 N, and one-half wave- length occupies the wire. What is the mass of m the wire? Correct answer: 0 . 0495868 kg. Explanation: Basic Concepts: v = s F for waves on wire . v = f L Solution: If one half-wavelength occupies the wire, then the wavelength = 2 L , where L is the length of the wire. Then the velocity of waves on the wire is v = f . But this velocity is also given by v = s F , where is the mass per unit length of the wire. The total mass of the wire is m = L . Thus we have m = L F v 2 = L F [2 Lf ] 2 = (2 m) 300 N [2(2 m)(27...
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homework14 - This print-out should have 12 questions....

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