A skyscraper sways 55 cm back and forth from “the vertical” during high winds. At t = 5 s, the building is 55 cm to the right of vertical. The building sways back to the vertical and, at t = 35 s, the building sways 55 cm to the left of the vertical. a) What is the length of the period? b) Sketch a graph of this function for 0 75 s t s   . c) Write an equation that models the motion of the building in terms of time.

视频信息

答案文本 复制

视频字幕 复制

In this problem, a skyscraper sways back and forth during high winds. The maximum displacement is 55 centimeters from the vertical position. At time equals 5 seconds, the building is at its maximum displacement to the right. At time equals 35 seconds, the building is at its maximum displacement to the left. Let's find the period of this oscillation. The time it takes to go from maximum right position to maximum left position is 35 minus 5, which equals 30 seconds. This represents half of the period. Therefore, the full period is 2 times 30 seconds, which equals 60 seconds. So the answer to part a is that the period is 60 seconds. Now, let's sketch the graph of this function for time between 0 and 75 seconds. Since we know the period is 60 seconds and the amplitude is 55 centimeters, we can identify key points on the graph. At time equals 5 seconds, we have a peak with displacement 55 centimeters to the right. At time equals 20 seconds, the building crosses the vertical position with zero displacement. At time equals 35 seconds, we have a trough with displacement 55 centimeters to the left. At time equals 50 seconds, the building crosses the vertical position again. And at time equals 65 seconds, we have another peak with displacement 55 centimeters to the right. The graph shows a complete cycle from 5 to 65 seconds, which confirms our period of 60 seconds. At time equals 0, the displacement is approximately 47.6 centimeters, and at time equals 75 seconds, the displacement is about 27.5 centimeters. Now, let's find the equation that models the motion of the building. We know the amplitude is 55 centimeters and the period is 60 seconds. From the period, we can calculate the angular frequency omega as 2π divided by T, which equals π/30 radians per second. We'll use the general form of a cosine function: y equals A times cosine of omega t plus phi, where phi is the phase shift. At time equals 5 seconds, we know the displacement is 55 centimeters, which is a peak of the function. Substituting these values, we get 55 equals 55 times cosine of π/30 times 5 plus phi. Simplifying, we get 1 equals cosine of π/6 plus phi. For cosine to equal 1, its argument must be 0, so π/6 plus phi equals 0, which means phi equals negative π/6. Therefore, our final equation is y of t equals 55 times cosine of π/30 times t minus π/6. This equation correctly models the building's motion, with the peak occurring at t equals 5 seconds and the period being 60 seconds. Let's verify that our equation correctly models the building's motion. We derived the equation y of t equals 55 times cosine of π over 30 times t minus π over 6. First, let's check at time equals 0. Substituting, we get y of 0 equals 55 times cosine of negative π over 6, which equals 55 times square root of 3 over 2, approximately 47.6 centimeters. At time equals 5 seconds, which should be our first peak, we get y of 5 equals 55 times cosine of π over 30 times 5 minus π over 6, which simplifies to 55 times cosine of 0, which equals 55 centimeters. Perfect! At time equals 35 seconds, which should be our trough, we get y of 35 equals 55 times cosine of π over 30 times 35 minus π over 6, which equals 55 times cosine of π, which equals negative 55 centimeters. Again, this matches our expected value. Finally, at time equals 65 seconds, which should be our second peak, we get y of 65 equals 55 times cosine of π over 30 times 65 minus π over 6, which equals 55 times cosine of 2π, which equals 55 centimeters. All our verification points match the expected values, confirming that our equation correctly models the building's motion. Let's summarize what we've learned about the skyscraper swaying problem. The skyscraper sways with simple harmonic motion during high winds, which can be modeled using trigonometric functions. We found that the period of oscillation is 60 seconds, calculated from the time between extreme positions. The amplitude of the motion is 55 centimeters, representing the maximum displacement from the vertical position. We derived an equation that models this motion: y of t equals 55 times cosine of pi over 30 times t minus pi over 6. This equation includes the appropriate phase shift to ensure that the building is at its maximum right displacement at time equals 5 seconds. We verified that this equation accurately predicts the building's position at any time t by checking several key points. This problem demonstrates how trigonometric functions can be used to model real-world periodic phenomena, such as the swaying of buildings during high winds.