Forces applied on both sides of a spring to stretch it? 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 spring 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 spring.
P.S
Many high school kids relate to this, so this is a legit question.
 A: 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.
A: Suppose the spring is lying unstretched on a horizontal surface (so we can ignore gravity) and one end is fastened to a wall. To stretch the spring you apply a pulling force at the other end. Initially the pulling force will exceed the resistive force of the wall- that's because the centre of gravity of the spring moves away from the wall as it stretches, so the pulling force has to accelerate the mass of the spring. However, a point of equilibrium will be reached at which the pulling force and the resisting force balance out.
If you pull the string very rapidly, then its behaviour before it settles into equilibrium can be much more complicated, with various transient vibrations, but the end state will be the same.
