If we assume the slinky to have a uniform mass (mass per unit length around the circumference of slinky to be constant, or simply slinky is made of same material and has uniform thickness) and that the slinky follows Hooke's law. What is the total vertical length of the slinky, when it is held at one end and is allowed to hang under gravity. Given variables: Natural length of slinky - L, Number of turns per unit length - n, Spring constant - K, Mass of slinky - M

Attempt: This is NOT a homework problem, it's a problem I thought of myself and could not find any helpful articles online. Initially I assumed it to be a simple calculus in physics problem, where taking a general distance x from the point of hanging, the remaining weight of the slinky would provide the force for the extension of the upper spring, where the upper spring would have a spring constant following the equation: $$ KL = k_x l$$ Where, k is the spring constant of the upper spring. Now,equating k∆x with remaining(∆x can be gotten from n and L) weight we do get an equation however I am not getting an approach of introducing an elemental distance($dx$) which eventually can be integrated to get the length in terms of known variables.

  • $\begingroup$ Didn't you assume the spring constant constant? The spring constant does not depend on length - it is a "per length" parameter. $\endgroup$ – Steeven Sep 23 '17 at 22:05
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    $\begingroup$ Note that we are very careful to say "homework-like" problems are off-topic, not "homework" problems. We have absolutely no interest in how you came to have the problem that you do; I imagine many of the problems that I help people with come from homework. We just care that you have boiled it down to a specific conceptual struggle which we can answer in a way that might be helpful to other people than just yourself. $\endgroup$ – CR Drost Sep 23 '17 at 22:05
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    $\begingroup$ @Steeven When we analyse a certain length of the spring the spring constant multiplied with it's natural length remains constant, cases where spring's natural length does not change, we can simply take the spring constant to be a constant, but in this case or in a case where the spring is cut, the k changes according to the equation I have written. $\endgroup$ – HyperBean Sep 23 '17 at 22:14

The challenge lies in that the spring constant is an extrinsic variable: if you take two springs of spring constant $k$ and connect the two in series, then you when you place the springs under tension $T$, each spring will stretch out by length $x_i=T/k$, so in total, $x=x_1+x_2=2T/k$. Effectively, by doubling the length of the spring, you've halved the spring constant. Something that would be independent size of the system of this would be the quantity $\eta=kL$. (If you double the length of the spring, the effective spring constant halves, but the product remains the same.)

Written this way, Hooke's law looks like $x=\frac{T}{k}=\frac{LT}{\eta}$

If you denote the amount that a piece of spring (initially) between $l$ and $l+dl$ was stretched as $dx$, then $dx=\frac{T}{\eta}dl$, where $T$ is the tension in the spring at location $l$. Of course the tension in the spring will be location dependent and depends on how much weight is below this point. Anyhow, I think this should give you enough hints to help you get started solving your problem.

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  • $\begingroup$ Thanks a lot for your help, my confusion was writing the differential equation in x, but now the doubt is cleared. $\endgroup$ – HyperBean Sep 23 '17 at 22:28

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