{Use of Tech} A damped oscillator The displacement of a mass o...

Question

{Use of Tech} A damped oscillator The displacement of a mass on a spring suspended from the ceiling is given by $y = 10 e^{- \frac{t}{2}} \cos (\frac{π t}{8})$ .

a. Graph the displacement function.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says the amplitude of a sound wave produced by a speaker decreases over time due to air resistance. The amplitudey in decibels of the sound wave at time T in seconds is given by the equation Y is equal to 5 multiplied by E rates to the quantity of minus T divided by 3 in quantity multiplied by the cosine of the quantity of pi T divided by 6 in quantity. Draw the graph of the amplitude function on the closed interval from 0 to 9. And below the problem we're given an empty graph on which to draw our function. Now, in order to draw our function here, we need to determine a couple of properties. The first thing we're going to look at are our points of interest. So, we're gonna look at the Y intercept, which occurs when T is equal to 0. And when T is equal to 0, we will have Y is equal to 5, multiplied by E raised to the quantity of minus 0 divided by 3 in quantity multiplied by the cosine of quantity of pi multiplied by 0 divided by 6 in quantity. And when we simplify this expression, this is going to be equal to 5. That it's going to be our white intercept. The next point of interest that we need to look at are our X intercepts. So we're gonna set Y equal to 0. That means we'll have 0 is equal to 5 multiplied by E raised to the quantity of minus T divided by 3 in quantity multiplied by cosine of the quantity of pi T divided by 6 in quantity. Now, our exponential um term there can never be equal to 0, so the only term that can actually be equal to 0 is our cosine function. And recall that our cosine function is equal to 0 when its argument is equal to an odd multiple of pi divided by 2. That means that we're going to have pi multiplied by T divided by 6, is equal to. Pi divided by 2, plus an integer, which we'll call K multiplied by pi. So K here is an integer. So we need to solve this for T by multiplying both sides of this equation. By 6 divided by pi, and doing so, we will get T is equal to 3 plus. 6K. So now we need to find the values of T that fall within the close interval from 0 to 9, and that only occurs twice. We'll have T equal to 3 when K is equal to 0, and we'll have T equal to 9 when K is equal to 1. And for other values to K, the value of T will fall outside of our interval. So those are the values for T at which our function crosses the X axis. So now we want to do is locate our critical points. For this we'll need to take a derivative of our function. So we'll calculate Y and that's going to be equal to the derivative with respect to T. Of the quantity of 5 multiplied by E raised to the quantity of minus T divided by 3 in quantity multiplied by cosine of the quantity of pi T divided by 6 in quantity. So this is going to be equal to, and in order to take this derivative, I need to apply our product rules. So recall that for our product rule. Product rule, we'll need to first take the derivative of the first term in our product, which is going to be the derivative of expected T of 5 multiplied by E raised to the quantity of minus T divided by 3. And so when I take that derivative, I get -5 divided by 3, multiplied by E raised to the quantity of minus T divided by 3. And this gets multiplied by the second term in our product, which is just going to be cosine of the quantity of pi T divided by 6. Then we'll have plus the first term in our product, which is 5 multiplied by E raised to the quantity of minus T divided by 3 in quantity, and this gets multiplied by the derivative of the second term in our product. So we'll have the derivatives with respect to T of the cosine of the quantity of pi T divided by 6 in quantity. So recall that when I take the derivative of the cosine function, I get a minus sine function. So I'll have minus sign. Of the quantity of pi T divided by 6 in quantity. But I, I'm going to have to apply our chain rule here and take the derivative of the argument. And so I have the derivative with respect to tea. Of the quantity of pi T divided by 6, and so when I take that derivative, I get pi divided by 6, and so I'll multiply our assign function here by pi divided by 6. And so now we can simplify this expression. I noticed that I can factor out a minus 5 divided by 3, multiplied by E rates to the quantity of minus T divided by 3 in quantity out of both terms. So doing so, this is going to be equal to. -5 divided by 3, multiplied by E rates to the quantity of minus T divided by 3 in quantity, and that gets multiplied by the quantity of cosine of the quantity of pi T divided by 6 in quantity. Plus, Pi divided by 2, multiplied by sine of quantity of pi T divided by 6 in quantity. So, in order to find a critical points, we need to um determine the T values which our first derivative here is either undefined or equal to zero. Now, our The first derivative here is defined everywhere within the interval from 0 to 9. So we need to just find when our first derivative here is equal to 0, and that only occurs when the term multiplying the exponential that contains the sine and cosine is going to be equal to 0, since our exponential can never equal 0. That means we're going to have cosine of the quantity of pit divided by 6 in quantity, plus pi divided by 2. Multiplied by sign of the quantity of pi T divided by 6 in quantity is equal to 0. So now I need to solve this for T. So, to do so, the first step I'm going to do is subtract. Pi divided by 2 multiplied by the sine of the quantity of pi T divided by 6 in quantity from both sides. That means I will have cosine of the quantity of pi T divided by 6. And quantity is equal to minus pi divided by 2 multiplied by sign of the quantity of pit divided by 6. So now what I'm going to do is divide both sides of the equation by cosine of the quantity of pi T divided by 6 in quantity, and I'm also going to divide both sides by minus pi divided by 2. Doing so, I will get Minus 2 divided by pi. is equal to the tangent of the quantity of pit divided by 6. So now need to take the inverse tangent of both sides of this equation as well as Multiplying both sides by 6 divided by pi to solve for T. So doing so, I will get T is equal to. 6 divided by pi, multiplied by the inverse tangent. Of the quantity of minus 2 divided by pi. Now, here we do have to be careful, because our inverse tangent function is only defined. On the 4th and 1st quadrants. However, we could have values that are in the 2nd and 3rd quadrants. So, what we're going to do to account for that is solve for T is equal to. 6 divided by pi, multiplying the quantity, and here in for our inverse tangent function, we'll now add to we will add pi to it. We'll have the inverse tangent. Of the quantity of minus 2 divided by pi in quantity plus pi. So this will give us the value that's in the 2nd and 3rd quadrant. So, evaluating these two values for T, we'll have T is equal to minus 1.08. And we'll have T is equal to. 4.92. Now the first term, the first t value is outside of our interval, so the only value that we're going to really look at is the T equal to 4.92. So now we have enough information to actually draw our graph. So coming back up to our graph, we're going to plot a couple of points. First, we'll plot our Y intercept when T is equal to 0. And that occurred when Y is equal to 5, so we'll plot that point on the graph. The next point that we're going two points that we're gonna plot are are X intercepts, and those occurred when X is equal to 3, or T is equal to 3, and T is equal to 9. And now we're going to plot. Our critical point. And so, if we calculate Y when T is equal to 4.92. We get -0.82. So we're going to plot that point. On our graph as well. So now we need to do is just connect these points with a smooth curve. So our curve is going to decrease. Starting at Y equal to 5, want to decrease crossing the x axis at T equals to 3. Go down to the minimum point that we just calculated, which was our critical point, and then it's going to increase up to T equal to 9 at the X axis. So, the graph that we have drawn on the screen is the answer to this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

{Use of Tech} A damped oscillator The displacement of a mass on a spring suspended from the ceiling is given by y=10e−t2cos⁡(πt8)y=10e^{-\frac{t}{2}}\cos\left(\frac{\pi t}{8}\right).a. Graph the displacement function.

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{Use of Tech} A damped oscillator The displacement of a mass on a spring suspended from the ceiling is given by $y = 10 e^{- \frac{t}{2}} \cos (\frac{π t}{8})$ .
a. Graph the displacement function.