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. $\endgroup$
    – Raj Shukla
    May 23 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
    May 27 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.