# Velocity eqn

1. Dec 5, 2009

### gramentz

Mass on vertical spring with force applied.

dy/dt = v

dv/dt = 5sin(wt) - 16y

I am trying to figure out the velocity of this object at any given time when w= 2 rad/ sec. (Would ideally like to have it for any w and any t)

Did I do this correctly? (In calc 2, haven't had diff eq yet)

dv/dt = 5sin(2t)- 16y

y= v2/2

dv/dt = 5sin(2t) - 8v2

dv/dt +8v2 = 5sin(2t)

8v2dv = 5sin (2t)dt

Integrated both sides:

(8/3)v3 = (-5/2)cos(2t)

v3 = (-15/16) cos(2t)

v= $$\sqrt[3]{ ((-15/16) cos (2t))}$$

Did I do this correctly or even approach this the right way? Thanks for any help!

2. Dec 5, 2009

### HallsofIvy

Staff Emeritus
How do you arrive at this? If your first equation were "dy/dv= v" then that would be a first integral, but you don't. You have dy/dt= v.

Here's what I would do: differentiate that second equation again:
Differentiating $dv/dt = 5sin(\omega t) - 16y$
$$\frac{d^2v}{dt^2}= 5\omega cos(\omega t)- 16\frac{dv}{dt}$$
$$\frac{d^2v}{dt^2}= 5\omega cos(\omega t)- 16v$$
so
$$\frac{d^2v}{dt^2}+ 16v= 5\omega cos(\omega t)$$
A "linear non-homogenous differential equation with constant coefficients" that should be easy to solve. Once you have found v (which, because this is a second order equation, will have two undetermined constants), to avoid introducing a third constant, use $16y= 5 sin(\omega t)- dv/t$ to find y.

3. Dec 5, 2009

### gramentz

I arrived at that first "solution" because I made the mistake of thinking it was in fact dy/dv. Thanks for clearing that up for me. I'm going to work on this new setup now..

4. Dec 6, 2009

### gramentz

I have the general solution to equal c1e4x + c2e-4x

And the particular solution (so far) to be c2cos(ax+$$\beta$$) + c2sin(ax+$$\beta$$) Does this look like I'm on the right track?

5. Dec 6, 2009

### HallsofIvy

Staff Emeritus
No, that would satisfy y"- 16y= 0, not y"+ 16y= 0. y"+ 16y= 0 has "characteristic equation" $r^2+ 16= 0$ which has imaginary roots 4i and -4i, not 4 and -4. Try exponentials e4i x and e-4i or, if you want to avoid complex numbers, sin(4x) and cos(4x). Do you see why those are equivalent?

6. Dec 7, 2009

### gramentz

Ohh ok right, somewhere along the line I copied down the equation to be a "-" and not a "+" for the gen. solution. I had to look it up, but I believe that sin(4x) and cos(4x) are equivalent because of imaginary numbers "rotating" between x and y, and the shifts between +i and -i correlate to the differences in graphs of sin x and cos x?

I set the particular solution up to be (w=$$\omega$$): yp(x) = Asin(wt) + Bcos(wt)

yp'(x) = Awcos(wt) - Bwsin(wt)
yp''(x) = -Aw2sin(wt) - Bw2cos(wt)

Plugging back in, I get:

-Aw2sin(wt) - Bw2cos(wt) + 16Awcos(wt) - 16Bwsin(wt) = 5wcos(wt)?

7. Dec 8, 2009

### gramentz

Then, I believe the complete solution will equal sin(4x) + cos(4x) + 5/16 cos(wt). Again, I'm not positive, but I think I did everything correctly...