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The motion of a particle performing damped oscillations is given by the formula y =...

Question:

The motion of a particle performing damped oscillations is given by the formula {eq}y = e^{-t}sin(2t) {/eq}, where y is the displacement from its mean position and t is time in seconds.

(a) Determine the time at which the velocity of the particle is 0.

(b) Determine if the displacement of the particle reaches a maximum or a minimum at the time when the velocity is 0.

(c) Hence, find the displacement.

Velocity and Displacement

If we have a position function, its derivative tells us the velocity of that moving object at any time. If we solve for when the velocity is zero, we may find the values for which the displacement is maximized or minimized, depending on the analysis of these points.

Answer and Explanation:

a) In order to find when the velocity is zero, we first need to define a velocity function. We can do this by differentiating the given function, as it's defined as the displacement from its mean position. This derivative will require us to use the product rule, as we cannot simplify this product of an exponential and sine function.

{eq}\begin{align*} &f = e^{-t} && f'= -e^{-t}\\ &g = \sin 2t && g' = 2\cos 2t \end{align*} {/eq}


Putting these pieces together will yield the derivative of this displacement function, which is the velocity of this particle.

{eq}\begin{align*} y' &= f'g+fg'\\ &= -e^{-t}\sin 2t + e^{-t}2\cos 2t\\ &= e^{-t}(2\cos 2t - \sin 2t) \end{align*} {/eq}


We can set this equal to zero to find when the velocity is zero. An exponential expression is never zero, as it always approaches zero asymptotically, so we only need to consider the trigonometric piece. Unfortunately, we can't use trigonometric identities to rewrite this into an easily solvable expression without some intense creativity, so it's best to turn to a computer to find our solutions. Since we're setting a trigonometric expression equal to a constant, there are an infinite amount of solutions, so let's use n as an integer placeholder.

{eq}e^{-t}(2\cos 2t - \sin 2t)=0\\ 2\cos 2t - \sin 2t=0\\ t = \pi n - 1.0172\\ t = \pi n + 0.55357 {/eq}


b) We can determine whether the displacement reaches a maximum or minimum at the points we found in part a, which are the critical points of this displacement function, by applying the first or second derivative test. As a note, since this is a periodic function, we only need to evaluate one instance of each of the repeating critical points (thus for one value of n), as the rest will behave in the same way.

Let's use the first derivative test, which requires us to evaluate the first derivative on each side of each critical point, and let's use {eq}n=1 {/eq} so that we are considering positive critical points.

{eq}y'(2) = e^{-2}(2\cos(2(2)) - \sin (2(2))) = -0.0745\\ y'(3) = e^{-3}(2\cos(2(3)) - \sin (2(3))) = 0.109519\\ y'(4) = e^{-4}(2\cos(2(4)) - \sin (2(4))) = -0.0234506 {/eq}


As the derivative does change sign on each side of each critical point, we have identified both a local minimum and a local maximum.

Local maximum: {eq}t = \pi n + 0.55357 {/eq}

Local minimum: {eq}t = \pi n - 1.0172 {/eq}


c) Normally, when we work with a trigonometric function, all of the maxima have the same value and all of the minima have the same value. This is not the case for our function. Since the amplitude can be considered to be the exponential expression in front of the sine function, the values of the maxima are actually decreasing and the values of the minima are actually increasing. As there are an infinite amount of local maxima or minima, we can't evaluate each point, as we would have an infinite amount of calculations here!

However, if we want only an absolute maxima or minima, and we consider only positive values of time, we can indeed find a single maximum and a single minimum. The maximum will occur at the critical point indicated when we let n be zero, and the minimum will occur at the critical point indicated if we let n be one.

{eq}y(0.55357) = e^{-0.55357}\sin (2(0.55357)) = 0.514198\qquad\text{absolute maximum}\\ y(\pi - 1.0172) = e^{-(\pi - 1.0172)}\sin (2(\pi- 1.0172)) = -0.10689\qquad\text{absolute minimum} {/eq}


Learn more about this topic:

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Using Differentiation to Find Maximum and Minimum Values

from Math 104: Calculus

Chapter 9 / Lesson 4
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