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It’s your turn on harmonic motion
solutions to the exercises
Hello. It’s Your Turn on harmonic motion. In this exercise, we have to compute the amplitude, the period, the frequency, and the phase of the harmonic motion represented by this function. You see this function is the sum of two sinusoidal functions. You have seen in the Francis’ video that because, you see, you have the same coefficient in front of x here and here, then you can easily transform this sum into what? Into something like k times the cosine of 3 times x minus phi. We will see in a few seconds how to compute k and phi.
But just knowing that it’s possible to transform our function into a function like this one, we can immediately answer to a couple of questions of the exercise. Because we know that the period of this function– we can read here the period is equal to 2 pi over 3– and the frequency, which is the inverse of the period, is 3 over 2 pi. And then when we will know the constant k and the constant phi, then what we will have? That the k will be the amplitude of this function, and phi will be the phase. OK. Let us compute now k and phi. Let us manipulate this sum of functions in such a way to get something of this form. Good.
Let us write. We apply immediately the formula for the sine of a sum and the sine of a difference of angles. And what do we get? We get 2 times– OK, I have the sine of the first angle times the cosine of the second angle, plus the cosine of the first angle times the sine of the second the angle. Good. Plus 3 times– and now I have the sine of the first angle times the cosine of the second angle, minus the cosine of the first angle times the sine of the second angle. Good. And now let us collect sine of 3 times x. What do we get?
Sine of 3 times x, which multiplies 2 times cosine of 4 plus 3 times the cosine of 2. OK. And then we collect cosine of 3 times x, which multiplies 2 times the sine of 4 minus 3 times the sine of 2.
Good. And now let us call this quantity A, and this quantity B. You see, A and B are just numbers.
Good. Now let us rewrite this expression in this form. Using the fact that this is A and this is B, we can write A times the sine of 3 x plus B times the cosine of 3 times x. Good. And now let us rewrite this expression in this apparently more complicated form. You see, I write the square root of A squared plus B squared times A over the square root of A squared plus B squared, times the sine of 3 x, plus B over the square root of A squared plus B squared times the cosine of 3 x.
Good. You see, it’s very easy that– we have an equality between these two expressions because we have multiplied and divided by the square root of A plus B squared. OK. And now look at the this coefficient here and this coefficient here. They have a precise meaning. You see, if you take the square root of this and you add the square of this, what do you get? You get 1. Therefore, these two coefficients are the coordinates of a point in the circumference of a circle of radius 1. Then there exists an angle phi such that this is equal to the sine of phi, and this here is equal to the cosine of phi. OK.
And we call now this constant here k. OK. Using just a calculator, you can easily compute k, sine of phi, and cosine of phi. Indeed, they are– k is approximately equal to 4.95. The sine of phi is approximately equal to minus 0.52. And the cosine of phi is approximately equal to minus 0.85. Now, immediately you’ll see from these two informations the fact that the sine and the cosine of phi are both negative. We know that the angle phi is an angle between what? Between pi and 3 over 2 times pi. Again, using a calculator, from these two informations you get that precisely the angle phi is equal to 3.69. Good. And now let us continue our computations.
What do we get from here? We can continue writing. OK. This is k, and here we have the sine of phi times the sine of 3 times x, plus the cosine of phi times the cosine of 3 times x. OK, but you immediately realize that this sum is exactly what? The cosine of 3 times x minus phi.
Cosine of 3 times x minus phi. Good. Therefore, we obtained what we were looking for– an expression of this shape for our sum of these two functions. Then what we can conclude? We can conclude that the amplitude of this harmonic motion is k– that is, 4.95– and that the phase of our harmonic motion is 3.69. OK, but exactly what we have done? You see, here we have the sum of these two sinusoidal functions, and we have obtained that this sum is another sinusoidal function. And looking at the graphs, we can see what we have done. You see, here in blue you have this function here. In red, you have this sinusoidal function here.
And in green, you have this sum that is this function. You see, again, the sum of these two sinusoidal functions has a sinusoidal behaviour because it is a sinusoidal function. OK. Thank you very much for your attention.
Do your best in trying to solve the following problems. In any case some of them are solved in the video and all of them are solved in the pdf file below.
Exercise 1.
Determine the amplitude, the period, the frequency and the phase of the harmonic motion represented by the function [xmapsto 2sin(3x+4)+3sin(3x-2).]
Exercise 2.
Determine the amplitude, the period, the frequency and the phase of the harmonic motion represented by the function [xmapsto 2sin(2x-frac pi 6)+frac 34cos(2x+frac{2pi}3).]
Exercise 3.
John’s blood pressure is modeled by the function [p:tmapsto 115+25sin(160pi t)] where (t) is the time measured in minutes.
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Find the period of (p);
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Find the number of heartbeats per minute;
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Graph the function (p);
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Compute the maximum and the minimum blood pressures.
Exercise 4.
Graph the functions (f:xmapsto x^2), (g:xmapsto -x^2), and (h:xmapsto x^2cos(6pi x)).
This article is from the online course:
Advanced Precalculus: Geometry, Trigonometry and Exponentials
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Advanced Precalculus: Geometry, Trigonometry and Exponentials
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