Suppose we have a spring which has a uniform cross sectional area $A$, and has a uniform linear mass density $J$, and has a total unstretched length $L$. Let the spring also obey Hooke's Law.

Further, suppose we hang this spring in a room from the ceiling and hold the spring in unstretched state. Then we slowly bring the spring to its equilibrium state under gravity (if I just release the spring from initial state, I think the Motion of the spring will be really complex, so I assume we bring it to a state of equilibrium very slowly)

Finally, we are to find the extension in the spring.

I'm at a loss as to how to approach this question. I've never encountered massive Springs in my course of study and am at a loss of how to approach problems related to them.

I'm looking for a hint to approach the problem.

Any Help would be appreciated, Thank You.


1 Answer 1


The trick is to evaluate the load situation of a slice of lenght $\Delta h$. As it is at rest, the downward force at the bottom of the slice $(F_h)$, plus the weight of the slice must be equal to the upward force at the top of the slice $(F_{h+\Delta h})$.

After converting to tensions, applying Hooke's law, and going to limit where $\Delta h \to 0$ we get a differential equation. Using the appropriate boundary conditions, it is possible to find the displacement as a function of $h$.

  • $\begingroup$ Thank You, I arrived at the answer JLg/2k (where k is the spring constant) and I'm pretty sure it's correct. $\endgroup$
    – neofyt
    Commented May 23, 2022 at 8:43
  • $\begingroup$ I got the same result for the total stretch of the spring. This does assume that no force is required to separate two adjacent coils which are in contact. $\endgroup$
    – R.W. Bird
    Commented May 27, 2022 at 16:24

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