PHYS 132 - University of Maryland _ Standing and Traveling Waves: Solutions

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Standing and traveling waves: Solution In the figure at the right is shown three graphs of the shapes of the same taut elastic string in three different circumstances, labelled A, B, and C. In so... me of the problems below, the string represents a guitar string, tied down at both ends, while in others, it is part of a very long telephone wire whose ends are not shown. Be careful to note which is which for each problem! The string is light enough that the effects of gravity can be ignored. In each case, the bit of string at the position x = 23 cm is painted blue. For each of the 5 problems below, put all the correct answers in the box at the right of the problem. 1. Suppose the three graphs all represent the same string, tied down at x = 0 and x = 70 cm, and at the same tension. What can you say about the three frequencies of oscillation? a. Situation A has the highest frequency. b. Situation B has the highest frequency. c. Situation C has the highest frequency. d. The three frequencies are all the same. e. You can’t say anything from the information given.For any wave, v = fλ, where v is the wave speed, f is the frequency, and λ is the wavelength. Because these three graphs all represent the same string at the same tension, v must be the same for all of them. Therefore, the highest frequency mmust occur when there is the shortest wavelength, which we can see in graph C. 2. Suppose the three graphs all represent the same string, tied down at x = 0 and x = 70 cm, but their tensions have been adjusted so that all three situations are all oscillating at the same frequency. What can you say about the tensions in the three cases? a. You can’t adjust the tensions to make the frequencies the same since the wavelengths are different. b. The tension in string C is the greatest. c. The tension is string A is the greatest. d. All three tensions are the same. e. You can’t say anything from the information given. Once again, v = fλ. But this time around, f is the same in all cases. Therefore, the wavelength λ is directly proportional to the wave speed v. For a given string, the wave speed is faster when the string is held at greater tension. Therefore, greater tension implies longer wavelength, so the tension must be greatest in string A. 3. Suppose that graph C represents the string, tied down at x = 0 and x = 70 cm, under tension. What can you say about the acceleration of the blue bead in case C at the instant shown? a. It is upward. b. It is downward. c. It is 0. d. You can’t say anything from the information given. This is a standing wave. The blue bead is at a node of the wave, so this bead never moves, so its acceleration is zero. 4. Suppose that graph C represents part of a long wire, under tension, and is showing a left traveling wave at a particular instant. What can you say about the velocity of the blue bead in case C at the instant shown? a. It is upward. b. It is downward. c. It is 0. d. You can’t say anything from the information given. The wave is moving to the left, and the part of the wave that is about to reach the blue bead (i.e. the part that is to the right of the blue bead at this point in time) is below the horizontal axis. Therefore, the blue bead is moving downward.5. Suppose that graph C represents part of a long wire, under tension, and is showing a left traveling wave at a particular instant. What can you say about the acceleration of the blue bead in case C at the instant shown? a. It is upward. b. It is downward. c. It is 0. d. You can’t say anything from the information given. You can think of the individual beads as masses attached to springs, so they experience forces proportional to their displacements away from equilibrium (as in Hooke’s Law). The blue bead is currently at the equilibrium position, so it experiences zero net force, and therefore zero acceleration. [Show More]

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