If gravity were increased, how would it change the spring constant for a given spring?

Question

I was just wondering how the value of a spring constant would change if gravity were increased. Agreeing that gravity is 9.8 m/$s^2$, if, for example, gravity was changed to 15 m/$s^2$ with nothing else to consider in altering a spring constant, how would the spring constant change? Would it increase, decrease, or stay the same with an increased value of gravity? And how does that relate to Hooke's Law?

I personally think that it would be harder to compress and easier to expand springs since gravity would more so pull down on things more, allowing for springs to be pulled down and expanded more easily.

Answer 1

The spring constant depends on the material and geometry and not a net effect of other environmental factors (ignoring temperature dependence). It measures the stiffness of the spring.

Answer 2

I was just wondering how the value of a spring constant would change if gravity were increased.

Gravity has nothing to do with the spring constant. It can only effect the net force on the spring depending on the orientation of the spring.

Agreeing that gravity is 9.8 m/𝑠2, if, for example, gravity was changed to 15 m/𝑠2 with nothing else to consider in altering a spring constant, how would the spring constant change?

It wouldn't

Would it increase, decrease, or stay the same with an increased value of gravity? And how does that relate to Hooke's Law?

Stay the same.

I personally think that it would be harder to compress and easier to expand springs since gravity would more so pull down on things more, allowing for springs to be pulled down and expanded more easily.

Yes it would be harder, but not because the spring constant changes. It would only be harder because the force of gravity would either add to the force required to extend or reduce the force requird to compress the spring because the spring is oriented vertically, but not because it would alter the spring constant. If the spring is oriented horizontally, gravity has no effect.

Hope this helps.

Answer 3

So the potential energy of a spring oriented along the $z$-axis is $$ U=\frac12 k(z-z_0)^2 $$ where $z_0$ is the position that the spring occupies at rest. Note that the spring constant $k$ has to deal with how sharp the parabola is, but the force depends on not just the spring constant but also the displacement.

When we add gravity along the $z$-axis we will have $$ U=\frac12 k(z-z_0)^2 + m g z $$ And while it looks like maybe this has a different character, constants in potential energy disappear because potential energy does not have a meaningful zero, which allows tricks like this: $$ U=\frac12 k\left(z+\frac{mg}{k}-z_0\right)^2 - \frac{m^2g^2}{2 k} + m g z_0 $$ Notice that after ignoring the constants on the right hand side, the entire effect of gravity is actually just subsumed in a redefinition of the equilibrium displacement $z_0\mapsto z_0-mg/k.$ So this adjustment $\ell=mg/k$ is everything. The spring constant $k$ doesn't change at all, the parabola still has the same shape, just shifted. This also means that changing the value of $g$ does not affect the spring constant, it just shifts the equilibrium displacement by $\ell_1-\ell_2= m(g_1 - g_2)/k$ as you remove one of these adjustments and add the other one.

So in fact to see this sort of effect, you need nonlinearities in the spring, at which point we need to introduce effective spring constants. Choose your coordinates so $z_0=0$ and assume a small quartic nonlinearity in the potential, $$ U=\frac14 \alpha z^4 + \frac12 k(z+\ell)^2 $$ Then the new equilibrium is where $U'=0$ which requires in theory solving a cubic equation, but we can assume that $\alpha$ is small enough that its new equilibrium is $z=-mg/k$ (in fact to first order in $\alpha$ the correction should then be $z=-\ell+\alpha\ell^3/k$, falling a bit higher than you would have expected) and near here the nonlinearity appears like a force $\alpha(-\ell + \delta z)^3$. Then the nonlinearity effectively appears to change the spring constant $k\mapsto k + 3\alpha \ell^2$. So increasing the gravity actually pulls us more into the non-linear regime and makes the spring stiffer, not “easier to expand” as you were apparently guessing.

But as long as Hooke's law holds good, the effect should not be easily noticeable.

Answer 4

Assuming you're using the same mass to measure $k$ (spring constant) under different gravitational accelerations, $k$ will always remains the same. With a greater acceleration due to gravity than where you originally measured the spring constant, the forces acting upon the spring will be different. However, that doesn't mean that the spring constant is really variable. As the perceived weight changes, the distance the spring compresses will change proportionately. Using Hooke's law, $W = kd$ ($W$ = Weight, $d$ = distance), we can rearrange this equation, $$\frac{W}{d} = k$$ Given that weight affects distance (compression in this case) linearly, $k$ will remain constant to your spring regardless of an alteration in gravitational acceleration. But yes, the spring will compress more under greater acceleration.

Below is an example of the proportionality between perceived weight and distance (I'm saying perceived because the same properties apply to a spring under acceleration).

Image source:

http://www.4physics.com/phy_demo/HookesLaw/HookesLawLab.html#:~:text=W%20%3D%20kx.,about%20980%20cm%2Fsec2.

Answer 5

It makes no difference on $k$ (assuming the spring is ideal).

Start with the potential energy of the spring $\frac{1}{2}kz^2$ alone and write the equation of motion \begin{align} m\ddot{z}=-kz\, , \end{align} with solution \begin{align} z(t)=A\cos(\omega t)+B\sin(\omega t)\, ,\qquad \omega^2 = k/m\, . \tag{1} \end{align}

Now add the effect of gravity to get the net potential: \begin{align} U=\frac{1}{2}kz^2+ mgz\, . \end{align} The equation of motion is now \begin{align} m\ddot{z}=-kz-mg\, . \end{align} You can then verify by direct substitution that \begin{align} z(t)=A\cos(\omega t)+B\sin(\omega t)+z_0\, ,\qquad z_0=mg/k \, ,\tag{2} \end{align} is the solution, with the same frequency $\omega$ as if there were not gravity. The solution has just been shifted by an amount $z_0$.

If gravity had an effect on the spring, it would have changed the frequency of oscillation of the system, since this frequency is always $\sqrt{k_{\small{eff}}/m}$, where $k_{\small{eff}}$ is the effective spring constant of the system.

Indeed it’s a common experiment in basic physics to compare the frequency of the mass-spring system arranged horizontally (where gravity has no effect) and vertically: to no surprise there is no change in frequency, and thus no change in the value of $k$ due to gravity.

Answer 6

The spring constant stays the same, assuming that Hooke's law can be applied for every stretching of the spring.
If a mass is attached to the spring then in the gravity field of the Moon, the Earth, or, say, Jupiter then the spring still obeys Hooke's law when stretched by the mass. The force exerted by the mass on the spring will vary in these three cases.
On the Moon, the force due to gravity pulling on the mass, and thus to the spring, is smaller than on Jupiter. On Earth the force lays somewhere between that of the Moon and Jupiter.
The spring itself, without a mass attached, will, by its own weight, be stretched the least on the Moon and the most on Jupiter. In all cases (on the Moon, Earth, and Jupiter) we can say that the relaxed state of the spring is that without a weight attached.

Of course, Hooke's law, in reality, can't be applied anymore when the spring is stretched too far. So in a very strong gravitational field, the spring constant (which isn't constant anymore) will get bigger. If the spring is hanging in a relaxed state, the constant will be bigger and be dependent on how far a mass stretches the spring in this very high gravitational field.