solve-the-given-problems-by-finding-the-appropriate-derivative-an-object-on-the-end-of-a-spring-is-3

Question

Solve the given problems by finding the appropriate derivative.
An object on the end of a spring is moving so that its displacement (in $\mathrm{cm}$ ) from the equilibrium position is given by $y=e^{-0.5 t}(0.4 \cos 6 t-0.2 \sin 6 t) .$ Find the expression for the
velocity of the object. What is the velocity when $t=0.26 \mathrm{s} ?$ The motion described by this equation is called damped harmonic motion.

Stanley Enemuo · Accepted Answer

we need to find the expression for the velocity of the object. This movement of the end of the spring. Well given to displacement and centimeters from the equilibrium position. The equilibrium position is given by the function Y equal to eat the negative zero point five T. Times 0.4 Co sign coastline 16 minus 0.2 time 16. Also we have to find the velocity see when the time 0.26 seconds. So we know the velocity of an object. To find that we have to find the first derivative of the displacement with respect to time. So we know the function. Why? So now we have to find revenue, the white cheese. And we can use integration by parts to find the former by explaining this. So D. Y. T. Is equal to E. to the negative 0.5 T. Times -2.3. Science signed 16 -1.4 co sign 16. And remember that he is 0.26 seconds. Told you plug in 0.26 seconds and sort of derivative function. We have the uh the U. I. D. C. To be equal to negative two points. There are 32 um centimeters per second.

Joseph Lentino · Answer

This is another application. In this case, we&#x27;re looking at harmonic motion. Why of T is the displacement and feet and one quarter cosine of 60 and T is going to be time in seconds. Okay, so in this case that we wanna be able to do three problems, we want to know what&#x27;s the displacement when time is zero, when time is a quarter and when time is half. So I&#x27;m going to evaluate y of zero, which is one quarter cozied of six times zero. Well, the cosine six times zero is zero. The cosine at zero is one. So the value here would be the displacement would be one quarter. When I substitute one quarter in. I have one quarter cosine of six times a quarter, and that&#x27;s gonna end up being a displacement of about 0.18 Um, feet, these were all in feet. And then when I plug in one half, I&#x27;m gonna have one quarter cosine of six times a half, so 0.25 times the cosine of three, which is going to give us negative 0.25 ft

Evan Leonard · Answer

all right, so we have our object on a spring. Cool it down on we let it go. It has simple harmonic motion with amplitude six and period high. So we want to figure out which what function represents the motion of this object because it&#x27;s simple harmonic. We have two choices. It will either be a sign function or will be a coast sonic function. If we can now need to figure out which one it is well looking at our objects. It&#x27;s resting. Position is here, and we&#x27;re gonna pull it down so that it&#x27;s starting position is somewhere down here. So thinking about a graphically this is it&#x27;s resting position, so it&#x27;s gonna start somewhere down here. When we release it, it&#x27;s gonna go up and then down and then up and then down. So if it were a sign function, it would start at the Orange because there&#x27;s no horizontal or vertical transformations on the sine function. So we know that it&#x27;s not a sign function because it starts down here. So it&#x27;s co sign. However, we know amplitude is positive, so if it were just this, we&#x27;d be starting somewhere up here. But since we&#x27;re starting down here. We need to flip it. So to flip it, we just put a negative out front. Okay, so we&#x27;ve got started. Now we need to do is figure out what our amplitude is and figure out what are making its well, we were given our amplitude are amplitude is six. So we can go ahead employing that. And so six. Now, I need to figure out what Omega is. Um, if you remember on figuring out how to find periods, we know that two pi divided by, uh, Omega is the period of a, um, co sign or sine function. So that means that this equals pot because our period is pie. So to solve for omega will first multiply both sides by omega. We get too high, equals omega pi. We&#x27;ll divide both sides by pie to get a megabyte. So that leaves us with two equals omega. Okay, so we just pulling into Let&#x27;s see that. Quick. Bring in two and we are done. This is our answer. D equals negative six co sign of two t

Ankit Gupta · Answer

even a question. ISTEA off is equal to six people on my 0.8 cause. Six flighty plus four here these the distance and in just off read from the point it was the string is attached to the ceiling. How far do you think the spring this from the ceiling? When the spring jobs moving so and he is equal to zero, we have to find the distance.

Doruk Isik · Answer

in this problem, we&#x27;re gonna function that displacement of an object. And there were Castillo. Find a position in the war. Still after fire range. Seconds such final position. First, we&#x27;re just gonna plot fire. Eight to depression. One bird co sign for pie. Old eight minus four. Sign full five or eight and we find a position for this money. Seconds to be one board. Now the whole city is the ritual displacements of white parents only Wipe our first. Why, Colonel t as one. Oh, negative sign for tea terms. 12 minus work forward of co starring Faulty turn full So we can write. This one is, um negative four. Science. Lt my instated minus three co sign. Well, t they&#x27;re fine people all see, after this man a second working I ate into this aggression and finally a lot to be thankful or sign. Full 58 minus three. Co sign for file eight. And this is for be for a second

Evan Leonard · Answer

All right, So we have an object. We pulled it down is going up and down continuously with simple harmonic motion amplitude 10 and period three and at time, zero. It&#x27;s at the resting position and it&#x27;s going down. So we need to figure out which function for a simple harmonic motion we&#x27;re gonna use. We have two choices. Is either gonna be signed or co sign? We need to figure out which one we need. So let&#x27;s think about what this looks like graphically. Let&#x27;s use read. Okay, So time zero or at the resting position and it&#x27;s moving down. So it&#x27;s gonna look something like this is gonna go down that&#x27;s gonna go up that&#x27;s gonna get down and continue on forever. Well, because it starts at the origin, we know that it&#x27;s sign because come sign. Mileva started here down here. Um, so we know it&#x27;s sign, however, with a positive coefficient. We know that amplitude is positive with a positive coefficient. The sine curves starts at the origin, but it goes up and then it goes, Damn. So we need to flip this, so to flip it, we just put a negative there, okay? And that&#x27;s a good starting place. Now we need to find the amplitude and omega, or we were given the ample too. So we can just go ahead and plug that straight in. So a equals 10. Okay, now we need to find Omega. Think back to when we find the period of a sign graph. We take too high over Omega, and I gives us the period here. The period is three. So that equals three. Okay, now we just need to solve for making well, most flying by omega on both sides. We get two pi equals three omega. Okay, now we just need to divide by three on both sides to get omega by itself. So I&#x27;ll give us 2/3. Pi equals amega so we can just plug in 2/3 pi for Omega. So let&#x27;s do that, Uh, my recovery Erase all that. So 2/3 hi times t Okay. And that is our finalists. D equals negative 10. Sign of 2/3 pi t

Problem

Solve the given problems by finding the appropria…

Problem 39 Easy Difficulty

Answer

$$v=-e^{-0.5 t}(1.4 \cos 6 t+2.3 \sin 6 t),-2.03 \mathrm{cm} / \mathrm{s}$$

Related Courses

Basic Technical Mathematics with Calculus

Related Topics

Discussion

Top Calculus 1 / AB Educators

Catherine R.

Kristen K.

Samuel H.

Michael J.

Calculus 1 / AB Courses

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Watch More Solved Questions in Chapter 27

Video Transcript

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Catherine R.

Kristen K.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$$v=-e^{-0.5 t}(1.4 \cos 6 t+2.3 \sin 6 t),-2.03 \mathrm{cm} / \mathrm{s}$$

Basic Technical Mathematics with Calculus

Catherine R.

Kristen K.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Catherine R.

Kristen K.

Samuel H.

Michael J.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus