Differential equations: period of a mass on a hanging spring

In summary: Oh well. In summary, the conversation discusses a homework problem involving finding the period of simple harmonic motion of a mass hanging from a spring. The conversation also touches on using the differential equation my'' + ky = mg to solve the problem, with a particular focus on the use of the method of undetermined coefficients and variation of parameters. The correct solution is ultimately found to be y_p = mg/k.
  • #1
OEP
6
0
Hey guys. This was a homework problem (I could never figure it out) and I asked my instructor for help as well as researching on the Internet, but still no dice. It seems basic enough but it bothers me that I still don't know how to solve it.

Homework Statement



Show that the period of the simple harmonic motion of a mass hanging from a spring is [tex]2 \pi \sqrt{l/g}[/tex], where [tex]l[/tex] denotes the distance beyond the spring's natural length when the mass is at equilibrium.


Homework Equations



I figured the differential equation we should solve is a simple mass-on-a-spring case but with the external force being mass-gravity. That is:
[tex]my'' + ky = mg[/tex].

The Attempt at a Solution



The general solution looks something like:

[tex]c_1 cos(\sqrt{\frac{k}{m}}t) + c_2 sin(\sqrt{\frac{k}{m}}t) + y_p[/tex]

For now, let's say:

[tex]\omega := \sqrt{\frac{k}{m}}[/tex]

Now we need to solve for [tex]y_p[/tex].

I did this in the two ways I know how, that being method of undetermined coefficients and variation of parameters. Both of them ended in some pretty nonsensical solutions (0 = mg??).

For the sake of comparison, after using the method of variation of parameters, I ended up with this particular solution and resulting position function:

[tex]y_p = \frac{g}{\omega(1+\omega)}[/tex]

(I know that can't be right since it's a constant.)

[tex]y(t) = (L - \frac{g}{\omega(1+\omega)}) cos(\omega t) + \frac{g}{\omega(1+\omega)}[/tex]

Where L is the natural length of the spring (I used the initial conditions y(0) = L, y'(0) = 0).

Anyone want to take a whack at it or point me to where someone already has?
 
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  • #2
You usually want to do this by working in a variable where u=0 is the equilibrium position. Then mu''+ku=0. I.e. y=u+mg/k. What's wrong with that? A particular solution in the variable y is y=mg/k, isn't it? No call for variation of parameters or anything.
 
  • #3
OEP said:
Hey guys. This was a homework problem (I could never figure it out) and I asked my instructor for help as well as researching on the Internet, but still no dice. It seems basic enough but it bothers me that I still don't know how to solve it.

Homework Statement



Show that the period of the simple harmonic motion of a mass hanging from a spring is [tex]2 \pi \sqrt{l/g}[/tex], where [tex]l[/tex] denotes the distance beyond the spring's natural length when the mass is at equilibrium.


Homework Equations



I figured the differential equation we should solve is a simple mass-on-a-spring case but with the external force being mass-gravity. That is:
[tex]my'' + ky = mg[/tex].

The Attempt at a Solution



The general solution looks something like:

[tex]c_1 cos(\sqrt{\frac{k}{m}}t) + c_2 sin(\sqrt{\frac{k}{m}}t) + y_p[/tex]

For now, let's say:

[tex]\omega := \sqrt{\frac{k}{m}}[/tex]

Now we need to solve for [tex]y_p[/tex].

I did this in the two ways I know how, that being method of undetermined coefficients and variation of parameters. Both of them ended in some pretty nonsensical solutions (0 = mg??).

For the sake of comparison, after using the method of variation of parameters, I ended up with this particular solution and resulting position function:

[tex]y_p = \frac{g}{\omega(1+\omega)}[/tex]

(I know that can't be right since it's a constant.)

[tex]y(t) = (L - \frac{g}{\omega(1+\omega)}) cos(\omega t) + \frac{g}{\omega(1+\omega)}[/tex]

Where L is the natural length of the spring (I used the initial conditions y(0) = L, y'(0) = 0).

Anyone want to take a whack at it or point me to where someone already has?
"where [tex]l[/tex] denotes the distance beyond the spring's natural length when the mass is at equilibrium."
The initial conditions "y(0)= L, y'(0)= 0" give no motion at all! The spring is sitting motionless at its natural length so there is no force on it. You want y(0)= L+ l, y'(0)= 0.
 
  • #4
Dick said:
You usually want to do this by working in a variable where u=0 is the equilibrium position. Then mu''+ku=0. I.e. y=u+mg/k. What's wrong with that? A particular solution in the variable y is y=mg/k, isn't it? No call for variation of parameters or anything.

I get the feeling there's some sort of progression of logic that does involve variation of parameters given that this is from a math book and is from a chapter that is trying to teach how to use MUC/VoP.

HallsofIvy said:
The initial conditions "y(0)= L, y'(0)= 0" give no motion at all! The spring is sitting motionless at its natural length so there is no force on it. You want y(0)= L+ l, y'(0)= 0.

There wouldn't be? I thought of it like this:

y(0) = L would be like if the spring were hanging with no mass on it. The spring, under its own weight, wouldn't stretch noticeably beyond its natural length. As soon as you attach a mass to it and let the oscillations settle, the mass would be at length L + l.
 
  • #5
A particular solution IS a constant. If you used VOP and got yp=g/(w*(w+1)) you came pretty close but you made a mistake. Not surprising, VOP is complicated. You should have gotten g/w^2=g/sqrt(k/m)^2=g/(k/m)=mg/k. But VOP is massive overkill for that. You can tell a constant is a solution by inspection.
 
Last edited:
  • #6
Thanks :)

Since you mentioned that, I remembered another time where I solved for a particular solution mg/k with just algebra, but it didn't occur to me to use it as y_p.
 

What is a differential equation?

A differential equation is a type of mathematical equation that involves an unknown function and its derivatives. It describes how the value of the function changes as its input (usually time or space) changes.

What is the period of a mass on a hanging spring?

The period of a mass on a hanging spring refers to the amount of time it takes for the mass to complete one full oscillation (back and forth motion) on the spring. It is affected by the mass of the object, the stiffness of the spring, and the force of gravity.

How can differential equations be used to model the period of a mass on a hanging spring?

Differential equations can be used to model the period of a mass on a hanging spring by describing the relationship between the displacement of the mass and the forces acting on it, such as gravity and the spring force. By solving the differential equation, we can determine the period of the mass on the spring.

What are the assumptions made when using a differential equation to model the period of a mass on a hanging spring?

Some common assumptions made when using a differential equation to model the period of a mass on a hanging spring include assuming the spring is ideal (massless and frictionless), ignoring air resistance, and assuming the mass is small enough to not affect the spring's behavior.

What are some real-life applications of using differential equations to model the period of a mass on a hanging spring?

Differential equations are commonly used in engineering and physics to model and predict the behavior of systems, such as the motion of a mass on a hanging spring. This can be applied to real-life scenarios such as the design of suspension systems for vehicles, studying the movement of pendulum clocks, and analyzing the motion of objects in space.

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