In Exercises 67–68, an object is attached to a coiled spring. ...

Question

In Exercises 67–68, an object is attached to a coiled spring. In Exercise 67, the object is pulled down (negative direction from the rest position) and then released. In Exercise 68, the object is propelled downward from its rest position. Write an equation for the distance of the object from its rest position after t seconds.

Accepted Answer

Welcome back. Everyone in this problem. A bunch of jumpers preparing to jump off a bridge exciting, right? The bungee card is attached to a platform above the jumper and has been stretched and coiled when the jumper jumps off the platform, they are propelled downwards by gravity. And we want to write an equation for the distance of the jumper from the platform. After three seconds, we are already told that the initial distance from the position at rest equals zero. The amplitude is 1/6 of a centimeter and the period is seconds for our answer choices. A equals Y equal negative six multiplied by the cosine of six divided by pi B is Y equals negative 1/6 multiplied by the cosine of six D divided by pi C is Y equals 1/ multiplied by the cosine of pi divided by six, multiplied by T and D is Y equals six multiplied by the cosine of pi divided by six R multiplied by T. Now let's sketch it to, to make sense of what we already know. So this is a bungee jump, right? And this is a bungee jump off the vertical distance. Let's call it Y over time. No, they're jumping from the top of a platform. So this jump is going to model a cosine equation because if you think about it, they're going to jump and then the bungee will pull them back up and then it's going to repeat until it gets smaller and smaller. Now, generally, for this problem recall that it can be modeled by the trigonometric equation Y equals plus or minus a multiplied by the cosine of omega multiplied by T right, where A is our amplitude, OK. And Omega is our angular frequency and our angular frequency is equal to two pi divided by or period. Now, what do we know from our question? Well, we know that the amplitude is 1/6 of a centimeter that's given. So we can rewrite A as 1/6. Let me put that in red to show that that's our, that's our answer. So A is 1/6 that requires our angular frequency is two pi divided by the period, the period is 12 seconds. So our angular frequency is going to be two pi divided by 12, which simplified is equal to pi divided by six. So alar frequency is pi divided by six. So know that we have our angular frequency and we have our amplitude, we can write our equation. So that means Y is going to be equal to plus or minus A plus plus R minus 1/ multiplied by the cosine of the angular frequency pi divided by six multiplied by T. So our final question is, is this going to be positive or negative? Well, since the jumper is propelled downwards by gravity, it is going in a positive direction. Therefore, Y is going to be equal to positive 1/6 multiplied by the cosine of 1/6 of pi almost brought by tea. And that tells us then that our answer choice will be answer. C Thanks for watching guys. I hope this video helped.

Key Concepts

Simple Harmonic Motion (SHM)

Trigonometric Functions in Oscillations

Initial Conditions and Phase Shift

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