Analyzing Mass on a Vertical Spring

[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!

 

[HallsofIvy]

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

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.

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!

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.

 

[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..

 

[gramentz]

I have the general solution to equal c₁e^4x + 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?

 

[HallsofIvy]

I have the general solution to equal c₁e^4x + c₂e^-4x

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 e^{4i x} and e^-4i or, if you want to avoid complex numbers, sin(4x) and cos(4x). Do you see why those are equivalent?

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

 

[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=$$\omega$$): y_p(x) = Asin(wt) + Bcos(wt)

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

Plugging back in, I get:

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

 

[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...