Answer 1: Top answer: compression. A compression is an area where the particles are closest together. Bottom answer: rarefaction. A rarefaction is an area where the particles are furthest apart.
Answer 2: Largest volume: blue. This is because the amplitude is the largest. Smallest volume: orange. This is because the amplitude is the smallest.
Answer 3: The frequency remains constant, while the amplitude decreases. The amplitude decreases because the volume decreases.
Answer 4: y = A * sin(2 * pi * frq * x). A is the new value added in from the last video.
Answer 5: N/m^2, or Pascals.
Answer 6: A = 2.5. We first plug the point given into the parent function. Also, it is apparent on the graph that the frequency is 1, so we plug that in too: -1.177 = A * sin(2 * pi * 1 * 1.578). Isolate A by dividing both sides by sin(2 * pi * 1 * 1.578), getting 2.5.
Answer 7: A = 3.7. To go about solving this problem, simply reference the problem above. The only real difference is that the frequency is not 1; it is 5.
Answer 8: Pressure difference = 0. When asking for pressure difference, we must find the y value. In our equation, have 3 variables: A, frq, and x. X is given; it's 4.2. A is visible; it's 4. Frq is also visible, to an extent; in .2 seconds, there are 3 periods, so the frequency is 15 Hz. Plug these in and solve for y.
Answer 9: Pressure difference = -1.705. This one is a bit more tricky. We still need to solve for y, but the variables A and frq are not as clear. To find A, just look at the point given; it's at the peak, so A is the y value, 2.9. Now we have to find frq. There are two ways we can go about find it. First, we notice that the x value is given at halfway through the crest. By multiplying it by 4, we know the x value at the end of the trough, or the complete period. There is 1 period in .02632 seconds, so there are 38 periods in 1 second (that's the frequency). The second method
† is plugging the point into the function. 2.9 = 2.9 * sin(2 * pi * frq * .00658), so divide both sides by 2.9. 1 = sin(2 * pi * frq * .00658), now inverse-sine both sides (it looks like sin^-1 on your calculator). This cancels out the sin part, giving us 1.5708 = 2 * pi * frq * .00658. Solve for frq, getting 38. Now with our frequency, we plug it into our function with our desired x value (4.2) in mind: y = 2.9 * sin(2 * pi * 38 * 4.2).
Answer 10: .4122, .4628, and .4955 seconds. This problem is the toughest yet. As you know, the sine function is not one-to-one, meaning that for every y value, there may be multiple x values (in this case, infinitely many). We are asked to limit them to just three. To begin this problem, first find an x value that produces the desired y value—with this, we can find 3 others that happen to fall into the range we want. We need to solve for x, but we need the other variables first; the y is given, -1.909. The amplitude is observable, 6. The frequency is also observable, 12. -1.909 = 6 * sin(2 * pi * 12 * x), solving for x and getting roughly -.0043. What other x values can give us the same value? Well, a period over can; add the length of a period to this number to find a number in our range. If the frequency (cycles/second) is 12, then a period length is 1/12, or .0833. We add this multiple times to our calculated x value getting .0790, .1623, .2456... and eventually .4122 (let's call this point 1) and .4955 (point 2)! Both of those work, but the question asks for
three x values, not two. If we look at the graph, we can see there's another point mirroring the trough on which .4955 is placed. We have to think using common sense here; a period consists of a trough and a crest. Divide that by two and we get the length of just one. So then, the length of a trough is .0417, using our previously calculated period length. Subtract that from .5 and we get the left side of the trough's x value, .4583. The difference between the right side of the trough (.5) and point 2 is .0045, so the difference between the left side of the trough and point 3 must be the same, making point 3 .4583 + .0045, or .4628.
†Note that this method does not always work. If the point were given on a different crest—say, the second or third one—then another wave with a different frequency would be able to produce the same point, only it would be the first crest for that said wave.
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