My question is that if such a spring has mass, then to produce tension it, will the forces that are to be applied on the opposite sides of it be equal , without causing any acceleration in the spring as a whole.

I think that the answer should be that such forces would not be equal, because if such a string is tied to a wall on one end (A) and is pulled from the other (B), then it will have different values of tension at different points, the highest being at A and lowest being at B. Using that analogy we may be requiring different forces at different endpoints of the string.

The above reasoning also implies that you could apply a greater force from either end and the lower force will simply apply at the other end.

In opinion of friends at my locality, the forces should be equal just like in the case of massless string.


Many high school kids relate to this , so this is a legit question.

  • $\begingroup$ You are right, they may have different values because of finite velocity of "transporting change" along the spring. Such transport of mechanical change is usually called sound :) so it point you how fast equilibrium of forces is obtained. Sound speed in metals is usually an order or two bigger than in air. $\endgroup$ – kakaz Dec 10 '17 at 20:00
  • $\begingroup$ Is the force applied suddenly, or is this a static situation? $\endgroup$ – Chet Miller Dec 11 '17 at 1:37
  • $\begingroup$ I think I got your point but, could you give me some references for what you just said. Thanks @kakaz $\endgroup$ – delstin jujawill Dec 11 '17 at 4:10
  • $\begingroup$ the forces are applied simultaneously and yes the spring is at rest. @ChesterMiller $\endgroup$ – delstin jujawill Dec 11 '17 at 4:12
  • $\begingroup$ It is seldom analysed system, but discussed usually in various relativistic articles and books, because relativists analyses finite speed of deformation propagation. In normal life usually it is not important. $\endgroup$ – kakaz Dec 11 '17 at 5:14

If k is the spring constant for the overall spring under static conditions, then for a differential section of the spring $d s$ along its length, the more general tensile force equation is $$F=kL\frac{\partial u}{\partial s}\tag{1}$$where L is the total length of the spring, u is the axial displacement from an initial location of the unstretched spring, and s is the unstretched axial distance coordinate; the local axial strain in the spring is $$\epsilon=\frac{\partial u}{\partial s}$$When $\Delta s = L$, we have the tension $F=ku_L$, where $u_L$ is the displacement of the right end of the spring relative to the left end; this is consistent with the overall force equation when the spring is deformed statically. When the spring is being deformed dynamically, both the tension F and the displacement u are functions of time and position along the spring: $F = F(t,s)$ and $u = u(t,s)$.

If m is the total mass of the spring, then the amount of mass dm between axial locations s and s + ds is given by: $$dm=\frac{m}{L}ds\tag{2}$$

If we carry out a differential force balance on this parcel of spring mass between s and s + ds, we have: $$\frac{m}{L}ds\frac{\partial^2 u}{\partial t^2}=F(s+ds)-F(s)$$This yields: $$\frac{m}{L}\frac{\partial^2u}{\partial t^2}=\frac{\partial F}{\partial s}\tag{3}$$If we combine Eqns. 1 and 3, we obtain: $$\frac{1}{c^2}\frac{\partial ^2u}{\partial t^2}=\frac{\partial ^2u}{\partial s^2}\tag{4}$$where c is the "wave velocity" along the spring, given by:$$c=\sqrt{\frac{k}{m}}L\tag{5}$$ One would solve this wave equation, subject to the boundary conditions on forces and the initial conditions of zero displacements and velocities, to obtain the displacement and tension as a function of position and time.


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