Analyzing Mass on a Vertical Spring

In summary, the mass on a vertical spring experiences a force that causes the velocity to decrease over time.
  • #1
gramentz
16
0
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= [tex]\sqrt[3]{ ((-15/16) cos (2t))}[/tex]

Did I do this correctly or even approach this the right way? Thanks for any help!
 
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  • #2
gramentz said:
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= [tex]\sqrt[3]{ ((-15/16) cos (2t))}[/tex]

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 [itex]dv/dt = 5sin(\omega t) - 16y[/itex]
[tex]\frac{d^2v}{dt^2}= 5\omega cos(\omega t)- 16\frac{dv}{dt}[/tex]
[tex]\frac{d^2v}{dt^2}= 5\omega cos(\omega t)- 16v[/tex]
so
[tex]\frac{d^2v}{dt^2}+ 16v= 5\omega cos(\omega t)[/tex]
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 [itex]16y= 5 sin(\omega t)- dv/t[/itex] to find y.
 
  • #3
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
I have the general solution to equal c1e4x + c2e-4x

And the particular solution (so far) to be c2cos(ax+[tex]\beta[/tex]) + c2sin(ax+[tex]\beta[/tex]) Does this look like I'm on the right track?
 
  • #5
gramentz said:
I have the general solution to equal c1e4x + c2e-4x
No, that would satisfy y"- 16y= 0, not y"+ 16y= 0. y"+ 16y= 0 has "characteristic equation" [itex]r^2+ 16= 0[/itex] 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?

And the particular solution (so far) to be c2cos(ax+[tex]\beta[/tex]) + c2sin(ax+[tex]\beta[/tex]) Does this look like I'm on the right track?
 
  • #6
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=[tex]\omega[/tex]): 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
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...
 

Related to Analyzing Mass on a Vertical Spring

1. What is the purpose of analyzing mass on a vertical spring?

The purpose of analyzing mass on a vertical spring is to understand and study the relationship between the mass of an object and the displacement of a vertical spring under the force of gravity. This can provide valuable insights into the properties of the spring and the gravitational force acting on the mass.

2. How is the mass on a vertical spring calculated?

The mass on a vertical spring can be calculated by using the equation m = kx/g, where m is the mass of the object, k is the spring constant, x is the displacement of the spring, and g is the acceleration due to gravity. This equation is based on Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement.

3. What factors affect the displacement of a vertical spring?

The displacement of a vertical spring is affected by several factors, including the mass of the object, the spring constant, and the strength of the gravitational force. Other factors that can influence the displacement include the length and stiffness of the spring, as well as any external forces acting on the system.

4. How does the mass on a vertical spring affect its period of oscillation?

The mass on a vertical spring has a direct impact on its period of oscillation, or the time it takes for the spring to complete one full cycle. As the mass increases, the period of oscillation also increases. This is because a heavier mass requires more force to move, causing the spring to oscillate at a slower rate.

5. Can the analysis of mass on a vertical spring be applied to real-world scenarios?

Yes, the analysis of mass on a vertical spring has many practical applications in various fields such as engineering, physics, and biomechanics. For example, it can be used to study the behavior of suspension systems in vehicles, the movement of buildings during earthquakes, and the impact of body weight on the human musculoskeletal system.

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