Mathematical Recreations

MathRec Home
Last Month's Problem
Introductory Problem
Older Problems
and Links
Educational Resources
Steve's Personal Page

Harmonic Oscillator With a Massive Spring: Solutions

In order to attack this problem, we need to consider how a small segment of the spring will behave. If the entire spring has spring constant k, then what is the spring constant of a portion of the spring? To answer this, we realize that a fraction u of the spring will deform by u x if the entire spring deforms by x (if the spring is horizontal, so that we may ignore its own weight). Since all parts of the spring are under the same force, the spring constant of a fraction u of the spring must be k/u.

Now let's consider the extension of the spring under its own weight m and the attached weight M. We need a coordinate to represent position along the spring. Let the fraction of the spring which is above a particular point be represented by u, so that u = 0 at the top of the spring where it is rigidly attached and u = 1 at the bottom where it is attached to the weight M. At any given point u, the tension in the spring is given by

τ  = Mg + (1 – u)mg.

A small section of the spring du with spring constant k/du is extended by

dx  =  τ du/k.

Integrating from zero to u gives

x(u)  =  [Mu + (uu2/2)m] g/k.

Here I have written x(u) to indicate that we are talking about the displacement of the spring at u. The displacement x is measured relative to the corresponding location if there were no load on the spring. So, if the spring has a natural length l, then the position relative to the anchor point is x + ul. We now have the answer to the first question. At rest, the spring is stretched by (M + m/2)g/k.

Now let's consider the spring in motion. If we look at a small section of the spring du, then the force on that section is its own weight mg du plus any difference in the tension in the spring below the section and the tension in the spring above the section. We find

F  =  mg du + dτ.

We know that the tension in the spring is proportional to its extension

τ  =  k
 dx 

 du 

Now, we're almost there. We can now write the governing differential equation

F
  =  mg du + k
 d2x 

 du2 
du
  =  m
 d2x 

 dt2 
du

The solution must also satisfy the boundary conditions at u = 0 and u = 1.

x(0)  =  0
τ(1)  =  k
 dx 

 du 
|u=1   =  MgM
 d2x 

 dt2 
|u=1

The solution is additive. x = xs + xd, where xs is the static portion given above, and xd is the dynamic part. Using

xs(u)  =  [Mu + (uu2/2)m] g/k,

we can remove the static components from the differential equation and the boundary conditions. We find

k
 d2xd 

 du2 
  =  m
 d2xd 

 dt2 

xd(0)  =  0
k
 dxd 

 du 
|u=1   =  – M
 d2xd 

 dt2 
|u=1

Let's start by solving the differential equation. We expect to have a solution which is oscillatory, so we try xd = f(u)eiωt. For any who might not be accustomed to physicists' shorthand, eiωt is traditionally used to carry through calculations involving oscillatory terms. Any linear combination of eiωt and –iωt, or equivalently, sin ωt and cos ωt will have the same properties. We find

k
 d2f 

 du2 
  =  –2f

This equation is easily solved by

f  =  A sin pu,

where p2 = 2/k and A is an arbitrary amplitude. (We have chosen the sine function for the solution, recognizing that f(0)=0 is a boundary condition.) At this point, we have no restrictions on ω and one boundary condition left to apply. The boundary condition requires that

kp cos p  =  2 sin p.

This condition restricts the allowed values of p and ω. Since p is the natural argument of the trig functions, it makes the most sense to eliminate ω from the equation, and we get

p tan p  = m/M.

This condition has infinitely many solutions, spaced roughly π apart. (There are both positive and negative solutions, but the negative solutions give the same answers as the positive solutions, since we have undetermined amplitudes and phases in our equations. We'll restrict our discussion to positive values of p.) The lowest-order solution is between zero and π. Let's look at two limits.

First, consider the case where M = 0. In this case, p = π/2, and ω = (π/2)(k/m)1/2. There are no nodes of vibration (except the attachment point at u = 0), and there is never any tension in the spring at u = 1.

Second, consider the case where M >>m. In this case, we can expand tan p = p – p3/3 + o(p5), where o(p5) indicates terms of order p5 and higher. From here, we can solve for p in terms of m/M.

p  =  (m/M)1/2 (1 – (m/M)/6 + o(m/M)2)

and

ω  = [k/(M + m/3 + o(m2/M) )]1/2.

Since p is small, the spatial function f is nearly proportional to u, which is just what you would expect from a massless spring. That is, df/du is constant and the spring expands uniformly. Interestingly, the static extension of the spring is proportional to M + m/2, whereas the vibration is more nearly dependent on M + m/3 (for small m). (Would anyone care to try for a concise explanation of why this is so?)

Finally, let's consider the higher-order modes. The second mode occurs for p between π and 3π/2. If M >>m, then p is approximately π. The mass doesn't move very much as the spring oscilates with a half wave between the fixed point and the suspended mass. (Just over a half wave, actually. There is a node near, but not at, the mass M.) For higher order modes, there are additional nodes of vibration.

If M = 0, then the second mode occurs for p = 3π/2. There is a node 2/3 of the way down the spring. (The 2/3 refers to the unstretched spring. If you actually suspend a Slinky®, you'll find that it's near the end of the spring as measured in linear units like centimeters, since the top is stretched more than the bottom.) There is 3/4 of a wave from top to bottom. Again, higher order modes have additional nodes.

If you haven't already done so, I encourage you to get your hands on a Slinky® or some equivalent toy. (Stay away from the plastic ones, since they don't have enough mass.) You can easily hang the spring off any balcony and look at the vibrational modes. (I wouldn't put any weight on the end. The M = 0 case seems to be the most fun.)

After you've played with the vibrational modes, you can try the rotational modes. I haven't tried the math on this one, but it is very interesting to look at the way the spring stretches. It might be necessary to make some approximations to get a solution... I think you can do the same thing with a rope. The math might be simpler if you just eliminate the stretching entirely. But then you don't get to consider the vibrational modes in the rotating spring. (But it is fun to do with a Slinky®!)

Send all responses to my email address is mathrec at this domain.

Thanks,
Steve