{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>
b. Find the time and the displacement when the object reaches its lowest point.

b. Find the time and the displacement when the object reaches its lowest point.

Accepted Answer

A damped harmonic oscillator's displacement from equilibrium is described by the equation X of T equals 4 E to the negative 0.5 T cosine 4T, where T is time in seconds. Find the first time at which the oscillator displacement is at a maximum. We have four possible answers, being 2.754, 0.754, 1.754, or 0.457 all in seconds. Now, to solve this, we want the maximum of the displacement function. We can find the maximum by finding the critical point. Critical point means that we will take the first derivative of our equation and set it equals to zero. So we have X of T equals 4 E rates is negative 0.5 T. Cosine 40 And we first want to find this derivative. This is a product rule derivative. Pro rule states that if you have two functions multiplying each other, In our case, we'll call it U and V. The derivative is U multiplied by V plus U multiplied by V. U in this case is 4E 0.5T. V is cosine 4 T. You prime then Based on this exponent derivative. would be -2 to 0.5 T. And would be 4 of 40. We can then write out our derivatives. We have U Native to E To the negative 0.5 T. Multiplied by V Plus Our U ore, the negative 0.5 T. Multiplied By V -4 4 T. We can now simplify You have -2 E to the 0.5T. To assign 40 -16 E to the negative 0.5 T. Signed 40 All of it equaling 0. This will help us find our critical point, and we do notice that E to the negative 0.5 T can never equal 0. Because of this, we can actually divide out our E to the negative 0.5 T term. From our equation. We end up getting negative too. Gos sign a 40 -16 Sign of 40 equals 0. Now, we just need to solve this routine. We divide everything first by -2. We get cosine 4 T. Plus 840. Equals 0. Now, let's move one of these terms over. If we subtract. 84 T. Both sites We end up getting Cosine 4 T equals -8, of 4 T. Now if we divide both sides. By cosine of 4 T. We get 1 equals negative 8s. Tangent F 4 T. We know sine divided by cosine is tangent. Divide both sides by negative 8. To get tangent or T equals -1 divided by 8. Now, we can get rid of tangent by taking the inverse tangent. We get 40 equals the tangent in verse. Of negative 1/8. Now tangent inverse of 1/8 is a decimal value that we have to approximate. We get 40, it equals 2. The approximation 0 0.12. 44 And because tangent has multiple values, we will add In pi to account for those values. No. Let's solve for tea. We can use example values for N, for instance, N. Equals 1 That would give us 40 equals -0.1244 plus 1 multiplied by pi. If we simplify this, we get approximately 40 equals. 3.0172. Dividing both sides by 4 gives us T. is approximately 0.754 seconds. If we look at our possible answers. You can see the answer to our problem. I answer B. OK, hope to help you solve the problem. Thank you for watching. Goodbye.

Key Concepts

Damped Oscillator

Cosine Function in Oscillations

Finding Extrema

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