A mass hanging from a spring is set in motion, and its...

A mass hanging from a spring is set in motion, and its ensuing velocity is given by $v(t)=2 \pi$ cos $\pi t$ for $t \geq 0 .$ Assume the positive direction is upward and $s(0)=0$. a. Determine the position function, for $t \geq 0$ b. Graph the position function on the interval [0,4] c. At what times does the mass reach its low point the first three times? d. At what times does the mass reach its high point the first three times?

Transcript

00:01 B of t equals 2 pi cosine pi t gives the velocity of a spring hanging from a mass at time t with t greater than equal to 0 the positive direction is upward and s of 0 equals 0 where s of t gives the position of the mass at time t so first we want to find the position function of the mass for t is greater than or equal to zero so we have the velocity function is 2 pi cosine of pi t so the integral of the velocity function will give our position function.

00:33 So s of t is going to be the integral of 2 pi, cosine of pi t, d t, and then we have to have the plus c, of course.

00:49 So we could take the 2 out, the constant of 2 out of that integrand, and get 2 times the integral of pi, cosine of pi t d t plus c and then if you need to do a u substitution for this you could you could say you could let u be um pi over t and then d u would be pi d t and then we could substitute we could say two we could say cosine of pi t is now um that would just be the cosine of u of u and then pi d t we have pi and d t there so that's right there's going to be our d u so d u and have the plus c so this just becomes two the integral of cosine u d u that's easy that's just going to be sign of u and they have the plus c and then we go back and re -s substitute u that gives us two times a sign u is pi over t and then we have the plus c and that is our position function we just need to find the c since we're going to given s of 0 equals 0 this is an initial value problem so we know this is going to be 0 whenever our t is 0 so that we give um 2 sine of pi times 0 plus c and that give sign of pi times 0 is 0 this just become 0 plus c that shows us that c equals 0 so our position function is just s of t equals um 2 sine pi t so if we want to graph that um so now i want to graph this position function on the interval from zero to four so there's our axis so the amplitude is given by this it's two so we're going to have a max of two and a minimum of negative two sign function starts the origin so it's going to go up to its maximum come back down to its equilibrium come down and come right back up there's one cycle um so if we find the period the period is given by 2 pi divided by that b value of pi and that gives us 2.

03:05 That means it completes this cycle in two units of time.

03:09 That means this would happen at 1.

03:11 This would happen at 1 half of a second.

03:14 And this would happen at 1 .5 seconds.

03:18 We'll just change that into decimals there.

03:19 So this would be 0 .5 seconds.

03:23 And we want to graph this from 0 to 4.

03:25 So that just means we need one more cycle here.

03:28 So we're going to go up, down, and then back up...

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