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I want to know the way to derive the time period equation of a spring mass system accounting for the mass of the spring but not using the energy analysis method but by proceeding in the same way as we do by ignoring the mass of the spring. Please help. I did not find any texts at my level. Any links would suffice gratefully.

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  • $\begingroup$ If you account for the mass of the spring, you end up with a wave equation coupled to a mass at the end of the elastic medium of the spring. The result of that is a system that does not just have one period, but a whole continuum of solutions. If you don't want that, you have to place the mass of the spring somewhere along the spring. If you place it on the fixed end, the spring mass doesn't matter. If you place it on the mass end, it simply adds to the mass. If you place it anywhere else, then you have a system with two massless springs and two masses, which has two frequencies, again. $\endgroup$ – CuriousOne Sep 18 '14 at 9:08
  • $\begingroup$ Placing the mass of the spring any where along the spring as an assumption in the beginning yields an incorrect expression. I know from energy analysis that the effective mass of the whole spring mass system is M+(m/3), where M is the mass of the block attached and m is that of the spring. But I want to use the method of integration and differential equations rather than taking the kinetic energy and potential energy to see the effect. $\endgroup$ – BibThePhysicist Sep 18 '14 at 9:15
  • $\begingroup$ If the mass of the spring is homogeneously distributed along the spring, then the system is not a simple harmonic oscillator with an effective mass but it's an elastic medium that follows a wave equation. As far as I can see, it does not matter what you do, the effective mass treatment is plain wrong with either assumption. $\endgroup$ – CuriousOne Sep 18 '14 at 9:19
  • $\begingroup$ I think every infinitely small element on the spring taken as an individual mass undergoes simple harmonic motion. en.wikipedia.org/wiki/… and there must be a way to integrate those small shms to find the equation of motion of the whole system. $\endgroup$ – BibThePhysicist Sep 18 '14 at 9:33
  • $\begingroup$ Only if you additionally assume that M>>m AND that none of the degrees of freedom of the spring are excited. Then you can imagine replacing each small mass element $dm$ with a massless rod and an equivalent mass element $d\tilde{m}$ moving with the main mass M. Since the individual mass elements $dm$ that are closer to the wall are acting on a shorter (and therefor stiffer spring), the effective mass of their virtual counterpart $d\tilde{m}$ moving in parallel with M must be smaller than those $dm$ that are closer to M. If you integrate the weight factors, you should end up with the 1/3. $\endgroup$ – CuriousOne Sep 18 '14 at 9:41
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I needed to solve the same question. Here is what i did:-

Suppose that we have a spring of mass $M$ with a mass $m$ attached to it, suspended against gravity. Assume that the mass of the spring is distributed evenly through its length. So, we can imagine our spring to be divided into infinitesimally small springs. Each small spring will have its own rate of oscillation depending upon the effective mass felt by it. Suppose the length of the spring lies along $x$ axis with origin at the bottom tip of the spring.

Now, assume that we have completely compressed the spring such that it can no more be compressed against its solid self. Let the length of the spring in this state be $L$. Now, for a small spring element located at a distance $x$ from the bottom in this state, time period $T$ of oscillation will be given by-

$T=2\pi \sqrt{\frac{\frac{M.x}{L} +m}{k} } $

Here, $k$ is the spring constant of the spring element.

So, frequency $f$ of simple harmonic motion can be given by-

$f=\sqrt {\frac{k} {\frac{M.x}{L} +m} } $

We realize that every spring element will have different frequency of oscillation. To simplify and approximate, we can imagine the entire spring to have the frequency of the spring element located at the middle of the spring.

So, for the time period of the entire spring, we get-

$T=2\pi \sqrt{\frac{\frac{M}{2} +m}{k} } $

Problem solved...!

Note that this solution will be valid only for a small time duration after the spring is allowed to oscillation. For example, if we let the spring oscillate for several minutes, we will start observing discrepancy. But for small time duration since the beginning of the experiment, the results will be quite accurate.

Edit: I just checked my physics lab manual, which has following equation for $T$:

$T=2\pi \sqrt{\frac{\frac{M}{3} +m}{k} } $

So, my answer was close. My approximation turns out to be too crude, and instead of assuming the frequency of the whole system to be close to that of the spring element halfway, we can get a better approximation if we assume it to be one-third from bottom, for the sake of calculus...

The equation in the book is also approximate, one must remember.

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