# Does a spring's $k$ affect the acceleration of a coupled mass?

Suppose I have a spring of spring constant $$k$$ attached to a mass $$m$$, with a constant force $$F$$ applied to the free end of the spring thus accelerating the mass. What happens to the acceleration of the mass as $$k \rightarrow\infty$$ ?

I thought it shouldn't be affected. However, as $$k\rightarrow \infty$$, the length of the spring $$x = F/k \rightarrow 0$$ and the energy stored $$\frac{1}{2}kx^2 \rightarrow 0$$. Therefore, the mass doesn't accelerate because no energy is being transferred via the spring!

I find this hard to believe and can't see where I've gone wrong in my reasoning.

• Just to clarify, we aren't looking at oscillations right? – Aaron Stevens Oct 30 '18 at 2:36
• @AaronStevens I'm looking at how it accelerates against $k$ when a constant force is applied -- if the acceleration oscillates then this is relevant to the question. – Physiks lover Oct 30 '18 at 22:10
• Well it depends. It doesn't have to oscillate, but it can. It depends on the initial conditions. But in the regime of an infinite spring constant oscillations are not going to occur. – Aaron Stevens Oct 30 '18 at 23:51

The initial acceleration is always the same, i.e. $$F/m$$.

But as $$k$$ increases, the frequency of the oscillations of the system increases ($$\omega = \sqrt{k/m}$$) and the amplitude of the oscillations decreases because the maximum displacement occurs when all the work done is stored in the spring, i.e. when $$\frac 1 2 k x^2 = Fx$$ or $$x = 2F/k$$.

So, as the spring gets stiffer, the amount of work done by the force decreases (because the spring doesn't compress so far) and everything happens faster - i.e. the rate of decrease of the acceleration from its initial value is faster.

As $$k \to \infty$$, the work done tends to zero and the time for the initial acceleration to decrease to $$0$$ also tends to zero.

• OK, so the mass doesn't move? – Physiks lover Oct 30 '18 at 2:09
• This question is not concerned with oscillations – Aaron Stevens Oct 30 '18 at 2:22
• The question is about the initial acceleration, but for $t \ge 0$ the system as described by the OP will oscillate if the force continues to be applied.. IMO it helps to understand what happens at $t = 0$ if you realize what will happen for $t \ge 0$ as well. If the force $F$ is only applied at the "instant" $t = 0$, there is no impulse and nothing moves for any value of $k$ (which is perhaps even more puzzling, unless you understand the maths of measure theory rather than just the "classical" Riemann integration taught in initial calculus courses!) – alephzero Oct 30 '18 at 8:46
• I don't see anything in the question asking about the "intial" acceleration. I have asked the OP to clarify. In any case, the $\frac12kx^2$ doesn't tell you how much work is done on the mass. – Aaron Stevens Oct 30 '18 at 10:00
• The question is about the acceleration in general, not just the initial acceleration. What happens to the steady state acceleration the oscillation is superimposed upon as $k \rightarrow \infty$ ? – Physiks lover Oct 30 '18 at 22:38

I am assuming that we are not looking at a situation like a mass oscillating on a vertical spring. It seems like we are pulling on a mass with a spring so for this answer I will assume that the constant force was built up to slowly so that no oscillations are initiated. If we wanted to include ordinations then thinking about when $$k\neq\infty$$ would need to be approached differently.

$$k$$ is qualitatively the "stiffness" of the spring. The larger $$k$$ is the stiffer the spring is. So as $$k$$ goes to $$\infty$$, you basically just have a rigid body attached to the mass. Therefore, the force you apply to the spring is $$F$$, and the force the mass experiences is also $$F$$ (assuming a massless "spring").

Another thing to notice is that $$U=\frac12kx^2$$ is not a "transfer of energy". It just tells you how much energy is stored in a spring that is stretched a distance $$x$$ from equilibrium. It doesn't tell you anything about how much energy the mass gains.

Let's be more exact here. Let's assume a massless spring that we are applying our constant force $$F$$ to. Since the total mass of the system is $$m$$, the acceleration is $$a=\frac Fm$$. This means that a force of $$F$$ also acts on the mass, and so by Newton's third law the spring actually has a force of $$F$$ acting on both sides. Notice how none of this depends on the spring constant.

Your energy argument comes from misunderstanding the equation of elastic potential energy, as explained above. We already know that the spring exerts a force $$F$$ on the mass, and so the work done by the spring on the mass over a displacement $$x$$ is just $$W=Fx$$, which also does not depend on the spring constant.

The reason $$k$$ it's irrelevant is because for a constant applied force, the spring will achieve some equilibrium applied force that is independent of the mechanism from which this force arises. This same exact analysis and reasoning can be performed if we were instead using a massless string instead of a spring, or if we glued a block to the mass in question and pulled on that instead.

• The fact that the spring energy is zero tells me that zero work can be done by the spring on the mass, otherwise you violate the conservation of energy. – Physiks lover Oct 30 '18 at 1:55
• @Physikslover It tells you that there is no elastic potential energy stored in the "spring". It doesn't say anything about the work the "spring" does on the mass. This actually would be true even if $k\neq\infty$. That equation tells you how much energy is stored in a spring due to it being stretched or compressed. It doesn't tell you how much work it performs on the mass. – Aaron Stevens Oct 30 '18 at 2:18
• @Physikslover As we pull on the mass via the spring the amount of work the spring does will increase over time, but the potential energy stored in the spring will remain constant. This shows that the potential energy stored in the spring and the work the spring force does on the mass are two different things. – Aaron Stevens Oct 30 '18 at 2:21

If k becomes infinite, it is the same as if the spring becomes a rigid rod. If a rigid rod is attached to the mass, then the acceleration will be F/m (the same as for an inextensible string).

• The question doesn't say anything about the end of the spring attached to a wall -- a constant force is applied to that end with a mass on the other end. – Physiks lover Oct 30 '18 at 22:16
• Oops. Sorry, I missed that. But here again, if k becomes infinite, it is the same as if the spring becomes a rigid rod. So the acceleration is just F/m. I'll modify my answer. – Chet Miller Oct 30 '18 at 23:35