I was going through Newton's laws of motion and came across the following problem:

A spring of mass $M$ is pulled such that at a given instant, the velocity of both ends is of magnitude $v$ and in opposite directions with respect to each other. Find the kinetic energy of spring.

  • I thought that as both ends are given equal velocities it will remain at rest and hence kinetic energy should be $0$. But the answer was incorrect.

  • The second thought that came was that the relative velocity of one end w.r.t another is $2v$. Therefore answer should be $\dfrac{1}{2}m(2v)^2$. But this approach was also wrong.

Please explain the resolution in detail and also let me know where did I go wrong.


closed as off-topic by Dvij Mankad, John Rennie, Emilio Pisanty, sammy gerbil, Kyle Kanos Jul 30 '18 at 10:15

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Although this is really a ''too homework-like'' question and it doesn't ask about a very specific basic concept (and I have even voted to close this question for these reasons), I would like to answer this as the solution is fairly interesting. It also invokes some concepts about springs that are otherwise not-so-discussed, and might not be obvious to one attempting the question afresh.

So, let's dive in!

Both the ends of the spring are given some speed $v$ in opposite directions as compared to each other. I presume these directions are along the length of the spring. Now, this means, due to symmetry, that the mid-point of the spring is at rest (but not the whole spring). In other words, both the halves of the spring are stretching out with their end-points moving at speeds $v$ with the mid-point of the spring being at rest.

Let's say that the natural length of the spring is $l$ with its mid-point at the origin. Now, if at an instant, the total elongation in the length of the spring is $\Delta l$, then the point on the spring which was initially at a coordinate $x$ would now lie at $x+x\dfrac{\Delta l}{l}$. This is a consequence of the fact that the strain in an ideal spring is assumed to be uniform. Thus, at any instant, the displacements $\Delta x_1$ and $\Delta x_2$ of two of the points on spring from their natural positions $x_1$ and $x_2$ are related by $$\dfrac{\Delta x_1}{\Delta x_2}=\dfrac{x_1}{x_2}$$ Thus, the velocity of a point on spring that was at $x_1$ in the natural position is given by $$u=v\dfrac{2x}{l}=2v\dfrac{x}{l}$$ Therefore, the kinetic energy can be obtained as $$K=\displaystyle\int dK=2\displaystyle\int_{0}^{\frac{l}{2}} \dfrac{1}{2}\Bigg(\dfrac{M}{l}\Bigg)\Bigg(2v\dfrac{x}{l}\Bigg)^2dx$$Here, $2$ ahead of the integral is representative of the exploitation of symmetry. I am integrating from $0$ to $\dfrac{l}{2}$ instead of from $-\dfrac{l}{2}$ to $\dfrac{l}{2}$ and am accounting for the error through the factor of $2$. If one worries that the linear mass density shouldn't remain $\dfrac{M}{l}$ when the spring is stretched then she would be right but one should also notice that the $\dfrac{M}{l}$ in this calculation refers to the fact that the linear mass density was this much in the natural state of the spring---that density is relevant for the integration here because the marker $x$ is marking the natural-state positions of the points on spring (that is why it runs from $-\frac{l}{2}$ to $\frac{l}{2}$).

With these clarifications, I proceed further with evaluating the integral as follows $$K=\dfrac{4Mv^2}{l^3}\displaystyle\int^{\frac{l}{2}}_{0}x^2 dx$$ Or, $$K=\dfrac{4Mv^2}{l^3}\Bigg[\dfrac{x^3}{3}\Bigg]^{\frac{l}{2}}_{0}=\dfrac{4Mv^2}{l^3}\Bigg[\dfrac{l^3}{24}\Bigg]=\dfrac{1}{6}Mv^2$$

  • $\begingroup$ But I am unable to understand it! 😅 $\endgroup$ – Lol Olo Jul 29 '18 at 7:27
  • $\begingroup$ This kind of calculation is the way to arrive at the usual prescription of including $\frac{1}{3}M_\text{spring}$ in the "mass" of a mass-on-spring oscillator. So seeing $\frac{1}{6}$ for the coefficient in the both-ends-free case is reassuring. $\endgroup$ – dmckee Jul 29 '18 at 17:52
  • $\begingroup$ It is inconsistent to vote to close a question (which implies that you think it should not be answered on this site) and to answer it. $\endgroup$ – sammy gerbil Jul 30 '18 at 8:28
  • $\begingroup$ Please don't answer "do my homework for me" type questions. This isn't a homework help site and answering them only encourages more of these types of questions that we don't want. $\endgroup$ – Kyle Kanos Jul 30 '18 at 10:15
  • $\begingroup$ @sammygerbil I was not entirely decided as to whether this is such a type of question that should not be answered at all or not. Thus, I both voted for it to close (thus, discouraging the behavior of asking homework question) and answered it (thus, acknowledging the fair enough interesting character of the question). It is an inconsistent behavior---reflexive of the competing modes of thinking. $\endgroup$ – Dvij Mankad Jul 30 '18 at 10:39

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