Problem

A Cepheid variable star is a star whose brightnes…

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Problem 80 Hard Difficulty

If the equation of motion of a particle is given by $ s = A \cos (wt + 8), $ the particle is said to undergo simple harmonic motion.
(a) Find the velocity of the particle at time $ t. $
(b) When is the velocity 0?

Answer

a. $v=-A \omega \sin (\omega t+\delta)$
b. $t=\frac{n \pi-\delta}{\omega}$ Where $n$ is an integer

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Video Transcript

here. We have an equation of motion, and, uh, some of the letters in here are Greek letters. So if you're not familiar with, um, the one that looks kind of like a W is Omega, and the one that looks kind of like a D is Delta. And those air constants in this equation and a is also constant. So are variable is teeth. So to find the velocity, we start by finding the derivative of position. Velocity is a derivative of position. Okay, so we're going to leave our constant a and then we notice that we have a composite function. So we're going to use the chain rule. The outer function is co sign, and it's derivative is negative sign. So we have negative sign of Omega T plus Delta. Now we need to multiply that by the derivative of the inside and the inside is Omega T plus Delta. And so it's derivative would be omega. Okay, let's simplify this a little bit. What we could do is multiply the a the negative and the Omega together, and we have the opposite of a Omega Times, a sign of omega T plus delta. So that's part A We have our velocity now for part B were interested in finding when the velocity is zero. Okay, that doesn't mean the time is zero. That means the velocity is zero and we're solving for the time. So we're going to take our our velocity negative eight times Omega sign of Omega T plus Delta. We're gonna set it equal to zero and solve this equation so that solving a trigger the metric equation, it's going to involve lots of trick. The first thing we can do is divide both sides of the equation by negative a Omega. And we have the sign of Delta T plus Excuse me, Omega T plus Delta equal zero. Okay, The next step is going to be to take the inverse sign of both sides. And so we have Omega t plus Delta equals Enver Sign of zero. That means the angle who signed is zero. And it just so happens lucky us that one of the angles who signed a zero is just zero radiance, but unlucky us. It also just so happens that they're infinitely many more answers. So let's take a look at that. So here we have our unit circle. We have zero radiance. This is the 00.10 So the sign is zero. So sign of zero is zero. Okay. However, you could go around the circle again and get to the same spot. Get to the same sign value and you have gone to pie Radiance. Or you could go again and you've gone four Pi radiance or again and again and again. You could go infinitely many times around the circle that way or back the other way so you could add two pi radiance infinitely many times. Similarly, you have another angle over here that has assigned value of zero, and you could come back to it infinitely many times by adding two pi radiance to it. Now let's consolidate those together. We have two answers at zero and at pi. And how far apart are they? They are pie radiance apart. So if you have one answer, you can go pie units and get to the next answer. Go pie radiance and get to the next answer and so on and so on infinitely many times. So how do we express that we can take one of her answers zero and say they were adding pi times n where n is an integer is an integer so that means you can add pie one time you can add pie two times you can add pine 99 times you can subtract pi Ah, 142 times You get all of the possibilities. So we're going to incorporate that into our answer back here. We're going to say that Omega T plus Delta is not just zero but zero plus pi en where n is an integer Now remember, the goal is to solve her time. So we're going to isolate T from here. So we're going to subtract Delta from both sides and we have Omega t equals Negative Delta plus pi n. We could write that as pie n minus Delta just to make it look a little bit better. And then finally will divide both sides by omega. So we have tea is equal to pi times n minus delta over omega. What a strange looking answer

Heather Z.

Oregon State University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 4

The Chain Rule

Problem

A Cepheid variable star is a star whose brightnes…

Problem 80 Hard Difficulty

If the equation of motion of a particle is given by $ s = A \cos (wt + 8), $ the particle is said to undergo simple harmonic motion.
(a) Find the velocity of the particle at time $ t. $
(b) When is the velocity 0?

Answer

a. $v=-A \omega \sin (\omega t+\delta)$
b. $t=\frac{n \pi-\delta}{\omega}$ Where $n$ is an integer

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Calculus: Early Transcendentals

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Calculus: Early Transcendentals

Grace H.

Kayleah T.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

a. $v=-A \omega \sin (\omega t+\delta)$b. $t=\frac{n \pi-\delta}{\omega}$ Where $n$ is an integer

Calculus: Early Transcendentals

Grace H.

Kayleah T.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Grace H.

Kayleah T.

Kristen K.

Joseph L.

a. $v=-A \omega \sin (\omega t+\delta)$
b. $t=\frac{n \pi-\delta}{\omega}$ Where $n$ is an integer