{Use of Tech} A damped oscillator The displacement of an ...

Question

{Use of Tech} A damped oscillator The displacement of an object as it bounces vertically up and down on a spring is given by y(t) = 2.5e⁻ᵗ cos 2t, where the initial displacement is y(0) = 2.5 and y = 0 corresponds to the rest position (see figure). <IMAGE>

d. Find the time and the displacement when the object reaches its high point for the second time.

d. Find the time and the displacement when the object reaches its high point for the second time.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says a wave on a string is described by the equation YFT is equal to 6 multiplied by E raised to the quantity of minus T and quantity multiplied by sine of 5 T. Where YFT is the displacement in meters from the equilibrium position and T is the time in seconds. Calculated time when the wave reaches a local maximum for the second time. Round the answer to three decimal places. We're given four possible choices as our answers. For choice A, we have T is equal to 0.903 seconds. For choice B, we have T is equal to 1.531 seconds. For choice C, we have T is equal to 0.275 seconds. And for choice D, we have T is equal to 1.275 seconds. Now, we're asked to calculate the time at which the wave reaches a local maximum for the 2nd time. So, to determine the time at which our local maxima occur, we need to determine the critical points, and for that, we need to take a derivative with respect to T of our function YFT. So, we'll calculate YT, and that's going to be equal to the derivative with respect to T. Of the quantity of 6 multiplied by E raises the quantity of minus T in quantity multiplied by the sign of 5 T in quantity. So taking this derivative, this is going to be equal to, and here we're taking the derivative of a product, so we need to use our product rule. So we call for the product rule that we're gonna have the derivative of the first term of a product. So we'll have the derivative with respect to T of the quantity of 6 multiplied by E raised to the quantity of minus T in quantity. And so when I take that derivative, I get -6 multiplied by E raised to the minus T. And this gets multiplied by the second term in our product, which is sign of 5T. Then we'll have plus the first term of products, we have 6 multiplied by Eras of the quantity of minus T in quantity multiplied by the derivative of the second term in our product. So we'll have the derivative with respect to T of sine of 5T. So recall that when you take the derivative of a sine function, you'll get a cosine function. So I'll multiply this by the cosine of 5T. And however, we need to apply our chain rule here, so we'll need to multiply this by the derivative with respect to T of 5T, which is 5. So we can simplify this expression by factoring out. A 6 multiplied by E raised to the quantity of minus T and quantity from both terms. So this is going to be equal to 6 multiplied by E raised to the quantity of minus T in quantity, multiplied by the quantity of 5, multiplied by cosine of 5 T. Minus sign of 5 T in quantity. So now we need to determine our critical points. So, recall that for critical points, that's going to be the values of T at which either our first derivative is undefined. Here we have exponentials and sine and cosine functions, and all of those are defined for all real numbers. So, our first derivative here is defined for all real numbers. But the other condition that we need to look for is when our first derivative is equal to 0, and since the exponential term can never be equal to 0, we're going to set 5 multiplied by cosine of 5 T. Minus sine of 5 T equal to 0. And now we just need to solve for T. So the first step is to add sine of 5T to both sides. We'll have 5 multiplied by cosine. Of 5T is equal to sine of 5T. Dividing both sides by cosine of 5T, we will get 5 is equal to the tangent. Of 5T. And now we just need to take the inverse tangent of both sides, so we'll have the inverse tangent of 5. is equal to 5, and then to solve for T, just divide both sides by 5. So we'll have T is equal to 1 divided by 5, multiplied by the inverse tangent of 5. And we can evaluate this expression error calculator, and this is approximately equal to 0.275. So this is the time for our first critical point. And if we look at a function YFT, we notice that we have sine of 5T in it. So that's going to be the part that's going to determine whether we have a local maximum or a local minimum. And recall that the sine function is initially increasing. So our time here is the first time we reach a critical point, which is going to be a local maximum. However, the question was just define when our wave reaches a local maximum for a second time. So that means that we need the sine function here, the sine of 5T, to have completed another period. And so the period of sin of 5T is equal to 2 pi divided by 5. And so to find the time, or our second local maximum, we'll call that T2. This is just going to be equal to the time that we found for the first local maximum, which is 0.275. Plus the period of our sign function, which is 2 pi divided by 5. And when we evaluate this expression in our calculator, this is approximately equal to 1.531. That means that the time at which a wave reaches a local maximum for the second time is 1.531 seconds. So that is the answer that we were looking for. And if we compare the answer that we calculated to the answer choices, we see that the correct answer choice corresponds to answer choice B. So, I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Damped Oscillator

Cosine Function in Oscillations

Finding Extrema in Functions

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