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We learnt that a spring stores and releases energy in either direction from the resting position when extended by some distance. When I tried doing this is real life by creating a very low friction surface and a spring and a mass, I noticed that the spring was far better at pulling the mass compared to pushing it. Any possible reasons or is my experimentation wrong?

Qmechanic
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Ashwin Alagiri-rajan
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    Can you explain what experiment you did and what results you saw that led you to conclude that "the spring was far better at pulling the mass compared to pushing it"? Right now, we can only guess what might have led you to that conclusion. – Tanner Swett Feb 16 '22 at 09:26
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    As written I don't think an answer is possible, and the answers below may or may not be relevant. The question is unclear. The meaning of "better" is particularly opaque. There is no way to decide if the experimentation is wrong if the experiment isn't adequately described. This question needs further details. – James K Feb 16 '22 at 10:42
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    Many springs are designed to mainly do either pulling OR pushing. The counterweight on a garage door is pull-only. It'll pull untill the windings rest on each other. The springs on a car wheel is made to push against the car weight, and pull against the weight of the wheel. If you unmount one of those without securing it, it'll push your fingers off. So, wheel springs are mounted to have their equilibrium point in a loaded state. There is a world of complexity that could have interrupted your experiment (but that is no reason to stop!) – Stian Feb 16 '22 at 10:44
  • @StianYttervik In addition, is your spring buckling when pushing? – DKNguyen Feb 16 '22 at 23:21
  • Admittedly this is not a sophisticated experiment. By better I meant that the spring readily pulled towards the mean point with a high acceleration. On the other hand when the spring was fully compressed the mass just behaved like an object in an elastic collision and the spring didn't seem to provide any restoring force. – Ashwin Alagiri-rajan Feb 17 '22 at 13:53
  • Isn't it true that most springs can be stretched - so they would pull back - far more than they can be compressed - so they would push back? Are you saying your experiment measured the result of the same degree of stretching and compression, or what? – Robbie Goodwin Feb 18 '22 at 00:26
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    Real springs do not follow Hook's law, except when they are only slightly compressed or extended. Fully compressing the spring would result in a non-Hookean response in many cases. – ZachMcDargh Feb 18 '22 at 22:05

5 Answers5

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Note that we are assuming the spring is a coil-type spring, the likes of which you will find in school labs and such.

Most springs don't follow Hooke's law when compressed to the point where the windings of the springs start touching each other. So if by pulling you mean to extend the spring and then letting go, it is expected that this works better than compressing the spring if the spring is already "tightly wound".

A good way of seeing the situation where pulling and pushing is actually symmetrical is by hanging a spring from a table with a small load applied. That way, the spring+load system settles into an equilibrium position where the spring is not tightly wound. You can then give it a small push either up or down, and you will notice that it starts to oscillate with approximately the same amplitude if your push is almost the same in the two cases.

Community
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Marius Ladegård Meyer
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    Technically they do kind of follow Hooke's Law when they're compressed so the coils touch. It just becomes Hooke's Law for a hollow metal cylinder with that wall thickness, which is a lot less willing to compress. :) – Graham Feb 16 '22 at 13:24
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    Put another way, it can be modeled as satisfying Hooke’s Law piecewise, with vastly different spring constants (torquing a wire vs. compressing the bulk material). – Chemomechanics Feb 16 '22 at 16:47
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    When you displace it just a little bit away from equilibrium, (almost) everything is a spring! – Jojo Feb 16 '22 at 20:13
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    @Joe, indeed. Taylor series FTW <3 And to think some students ask why they should bother studying the harmonic oscillator, hehe. – Marius Ladegård Meyer Feb 16 '22 at 20:15
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    Yeah I remember having this galaxy brain moment sometime in my undergrad when I realised you could approximate almost all small oscillations as simple harmonic motion by Taylor expanding the potential – Jojo Feb 16 '22 at 20:19
  • @Joe That's the same in EE when they make you learn everything about 1st and 2nd order RLC circuits like some sort of prayers. Then later on you realize that most electrical systems can be (even brutally) approximated by such circuits with proper assumptions (engineers are particularly fond of their versions of the "spherical cow" :-) – LorenzoDonati4Ukraine-OnStrike Feb 16 '22 at 22:51
  • I don't see the convincing power of the experiment suggested at the end of this answer. Even in a far from linear force field (say a hypothetical spring that suddenly becomes twice as stiff when extended less that the equilibrium position with respect to when it is extended beyond that), starting the oscillation with two opposite impulses will result in similar oscillations (same amplitude, same frequency), starting at a different phase of the oscillation. What this would exhibit is a non-sinusoidal oscillation, with different amplitudes and half-periods at both sides of the equilibrium. – Marc van Leeuwen Feb 18 '22 at 10:15
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Hooke's Law for a metal-coil spring is usually analyzed in one dimension, but a spring is a three-dimensional object. You have a couple of answers which describe a metal-coil spring compressed to the point where the coils touch, at which point it behaves like a metal block. However, a stiff metal spring may also be subject to a buckling instability, where the spring suddenly goes bloop! out to one side and is no longer any approximation of a one-dimensional object.

The harmonic oscillator, which is how a graduate student would refer to a Hooke's Law device, pops up in all kinds of places across physics, not just in coil springs. The leaf spring is more obviously designed to be compressed, and appears in old vehicle suspensions. Modern vehicle suspensions generally use big, beefy coil springs, but I'm pretty sure even those are installed so that the chassis compresses the spring onto the axle. Every time you drive a car over a bump, a spring pushes you upwards, rather than pulling.

rob
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    Coil springs in car suspension are of course constrained to prevent buckling, often with the shock absorber (damping element) in the middle of the coil and providing some of the constraint. Leaf springs are still found on the back of some heavier vehicles - my van has them for example. That's not for reasons relating to their spring behaviour, but because the assembly is lower and fits below the load space – Chris H Feb 16 '22 at 09:38
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    The important function of the coils on your car is to push the wheel back onto the surface after the bump, though. Unsprung wheels have terrible traction on uneven roads, it spends most of the time without contact. – Stian Feb 16 '22 at 10:47
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    I look forward to the journal article that describes the "bloop!" effect. :) – Barmar Feb 16 '22 at 14:58
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When you use the spring to push an object, and the loops in the spring come closer together and touch, the Hooke's law relation $$F=-kx$$ no longer has such a relationship (especially if the loops are in direct contact) and $k$ gets larger increasing $F$ in such a situation.

Of course when pulling an object (and the loops in the spring are not so close and Hooke's law works well), the potential energy stored in the spring can go into moving the object.

joseph h
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  • If the loops are in direct contact, doesn't Hooke's Law still apply, but with a somewhat larger value of $k$? (As long as the spring material is only subject to elastic deformation.) – Andrew Morton Feb 16 '22 at 12:10
  • This and the other answers should clarify that the spring doesn’t follow Hooke’s law with the same $k$. Bottomed-out springs still follow Hooke’s law (up to the yield point), just with a different $k$, as Andrew notes. – Chemomechanics Feb 16 '22 at 16:51
  • @Chemomechanics That's true and I've edited the answer. – joseph h Feb 16 '22 at 19:57
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A spring in equilibrium. You can pull it out by the number of winds times the circumference of one winding. How for you can compress it? The number of windings times the distance between two windings. Well, actually one less.

So, say the winding is $2\pi R$ in circumference, and there are $n$ windings. Then you can pull it out $2n\pi R$ meter. If the distance, in equilibrium, between two windings is $d$, then you can compress it $(n-1)d$ meter.

When these are equal? Well, then

$$2n\pi R=(n-1)d$$,

so (setting $n-1=n$)

$$2\pi R=d$$

So for a spring with $\frac{d}{R}=2\pi$ you have (more or less) linear behavior in both directions.

MatterGauge
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Without getting into Hooke's law or any such considerations, one way to think about the problem qualitatively is to consider the extreme scenario where the "spring" isn't one at all, i.e., where $k=0$. In this case, we basically have a string, not a spring. Strings can pull, but they can't push at all. What a non-zero $k$ does is simply allow for some pushing while at the same time not reducing the pulling "capability". For there to be a symmetry of sorts between pushing and pulling, one would need $k\rightarrow \infty$.

Tfovid
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    A spring with non-zero small $k$ will not behave similarly to a string: it will (almost) not resist to elongation. How do you make a transition to $k=0$? – svavil Feb 16 '22 at 11:15
  • k=0 would be represented by a slack string, which can neither push nor pull - as long as the string remains slack, no matter how far the ends are separated, the force applied at either end is zero. k=0 does not represent a spring that can pull but not push, it represents a spring that can't apply any forces at all. – Nuclear Hoagie Feb 16 '22 at 13:45
  • @NuclearHoagie I disagree that $k=0$ is synonymous to a slack string. Hooke's law doesn't account for the full picture, or else how would one model the tension on an extended string from which a mass hangs (for example)? – Tfovid Feb 16 '22 at 21:42
  • @Tfovid k=0 implies F=0 regardless of displacement. If the string is not slack, it exerts a non-zero tension force by definition, meaning k is not zero. You could model a string as a one-way spring that only exerts force when it extends but not compresses. But a "spring" with k=0 can never exert any force at all, and can neither push nor pull. You could model a string as a type of spring, with non-zero k for extension, and k=0 for compression. – Nuclear Hoagie Feb 17 '22 at 14:23
  • @NuclearHoagie Re: "If the string is not slack, it exerts a non-zero tension force by definition, meaning k is not zero". $k$ doesn't make sense unless it is multiplied by displacement, whereas the tension on its own doesn't have anything to do with displacement as it is translation invariant. – Tfovid Feb 18 '22 at 11:57
  • @Tfovid A real string stretches a bit, with tension starting at zero when it's slack, and then increasing once the string is stretched to its nominal length. Any non-ideal string is somewhat elastic, with tension increasing to the breaking point as you continue to stretch the string. My overall point is that discussing k for a string isn't a good analogy in the first place, and if you do want to do that, it certainly can't have k=0 as that implies an inability to exert any force at all. – Nuclear Hoagie Feb 18 '22 at 14:27