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Say a mass of $35kg$ is placed in the middle of a trampoline and causes a vertical displacement of $22cm$. Why is the spring constant $35g/0.22$? Why are we considering the applied force rather than the force within the material itself? Why does that work? It certainly doesn't work in the case of a mass being a hung on a horizontal rope, so what logic is one meant to use in order to deduce that the spring constant is $35g/0.22$?

Here's where the question comes from, for your interest: https://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Physics/2013/Exam%20materials/WPH05_01_que_20150618.pdf

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The issue here is that for a shape like a trampoline, you are interested in the "effective force" for a certain "effective displacement". You are right that there are much larger forces present in the springs/elastic bands that hold the trampoline in place - but the experiment you did cannot tell you about those forces without some more information about the geometry.

In the case of a string under tension, you can ask two different questions: (1) how hard are we currently pulling on the end of the string, or (2) what force is needed to displace the (center of the) string sideways by a certain amount?

Again - the answers are related, but they will be quite different. If you want to know, for example, what the resonant frequency of your string would be if you attached a mass at the midpoint and tried to excite lateral motion, you need the answer to (2) above. And that's basically the problem with the trampoline - you want to know what force gives rise to a certain vertical displacement, because that's the mode of operation of the trampoline.

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  • $\begingroup$ Are you saying that during both (1) and (2) we will get a different spring constant? I know that when we consider the force within the spring we have to use the (modulus of elasticity/length) x extension of the spring. However, I have tried it out on 2 real life questions. This spring constant will simply NOT be the same when we use that equation and F=k with the weight and the vertical displacement. $\endgroup$ – Mathematician Apr 10 '17 at 4:46
  • $\begingroup$ Yes that's what I am saying - because "spring constant" depends on what you are trying to do. $\endgroup$ – Floris Apr 10 '17 at 4:52
  • $\begingroup$ The exam question that I'm doing precisely says 'find the force constant' of the trampoline. May you please tell me how one could differentiate between the two types of spring constants here? I've edited my question to include it. $\endgroup$ – Mathematician Apr 10 '17 at 4:58
  • $\begingroup$ "Force constant" is actually less ambiguous: it is the thing that relates displacement (of the trampoline) and the force it generates. It would be quite rare (and difficult) to determine the "spring constant" of something shaped like a membrane - because you are given no dimensions, that's actually impossible. What is left is the intention of the examiners. I admit it I have sympathy for your confusion. $\endgroup$ – Floris Apr 10 '17 at 5:02
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This question is not rigorous and is open to other interpretations. When asking about Spring constant, we need to know the orientation.

For example, consider an anisotropic elastic material, i.e. it stretches differently depending on the orientation. When applying a force $F$ horizontally it moves $\delta$. When applying the same force vertically $F$ it moves $2 \delta$.

Say we want to know how much it stretches when it's applied a force $F$ at 45º. Doing the maths, $F_x = F_y = F/\sqrt 2$. Then the displacement $\Delta x=\delta/\sqrt 2$ and $\Delta y=2\delta/\sqrt 2$. Considering small displacements, the displacement along the diagonal axis can be approximated to $\Delta _ {45} = \delta\sqrt{5/2}$ (Pythagoras).

But now, what is the spring constant? Depending on the orientation, it can be:

$$K_x=F/\delta$$ $$K_y=\frac{F}{2\delta}$$ $$K_{45}=\sqrt{\frac{2}{5}}\frac{F}{\delta}$$

In this trampoline question, if you have enough data about the materials and do the maths, it can be verified that the $K$ of the rope (in this case, the fabric of the material), is related to the $K$ used to estimate the vertical displacement of the trampoline.

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Provided the springs supporting the canvas and the canvas itself remain essentially elastic, and that the deformations aren't large (questionable for the case of a trampoline - but we can ignore it for now), the force-displacement relationship of a mass in the centre of the trampoline would be expected to be basically linear.

Although there isn't an actual vertical spring in the centre of a trampoline, this linear force-displacement relationship may be considered as equivalent to a spring. And that 'spring' has a spring constant which you've calculated.

Another example is an elastic beam supported at two ends. If we subject it to a point load at its mid-span, it will deflect a certain amount that relates to the flexural stiffness of the beam and its end conditions. Clearly the beam is not a vertical spring, but for all intents and purposes we can conveniently represent its mid-point deflection using a spring constant. This is the basic approach for determining, for example, the beam's dynamic response - and this is done in practice in engineering.

Hope this helps somewhat.

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  • $\begingroup$ Are you saying that during both (1) and (2) we will get a different spring constant? I know that when we consider the force within the spring we have to use the (modulus of elasticity/length) x extension of the spring. However, I have tried it out on 2 real life questions. This spring constant will simply NOT be the same when we use that equation and F=k with the weight and the vertical displacement. The point is that I don't understand why we use vertical displacement and weight with a trampoline, and apparently is works (I haven't checked it though), but it doesn't work for a rope. $\endgroup$ – Mathematician Apr 10 '17 at 4:50
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When the trampoline reaches maximum extension, the force applied by the trampoline and the gravitational force on the object are equal because the object is rest (net force must be zero; $mg = F_{trampo}$). If you consider the trampoline as a spring, then the total extension is given as $0.22 m$ and the force applied is given to be $350N$ ($mg$).

Therefore, the spring constant is given by:

$$k = \frac{F}{\Delta x} = \frac{350}{0.22}$$

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  • $\begingroup$ But why are we not considering the tension force within the trampoline and the extension of the trampoline? Why does using the weight force and the vertical displacement work in the case of a trampoline but not in the case of a horizontal rope that has been bent into a V shape by placing a mass in the middle of it? That's the pivotal question. $\endgroup$ – Mathematician Apr 10 '17 at 4:55
  • $\begingroup$ The answer to the first part is: the tension force is what provides the force to keep the object up; this is true in case of a simple spring too. The answer to the second part is: it depends on the trampoline; if you consider the trampoline to be made up of springs arranged one after another in a circular fashion. The effective spring constant would be the sum of the individual spring constant times $\frac{\Delta x}{l_o}$ for small vertical displacements. $\endgroup$ – Yashas Apr 10 '17 at 5:17
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The correct formula is:

$$F= k\Delta l$$

where $k$ is in Newtons/meter, $l$ is in meters and $F$ is the applied force.

Therefore, $$F = 35 \times 9.8 = 343N$$

The spring constant is $343/0.22$ or $1559\space Nm^{-1}$

This is the basic formula for a constant $k$ but in most springs (or materials), it is not a constant so $k$ won't be a linear curve. It depends on materials elasticity. It's more an average.

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  • $\begingroup$ I am aware the k=F/l, but this usually only works for the force WITHIN the spring. The weight of the mass is simply not the force WITHIN the spring. Even if you try this with a mass that is hung on a line and the mass causes the rope to bend into a V shape, you simply HAVE TO consider the tension force within the spring and the extension OF the spring, not the applied force and some weird vertical displacement! $\endgroup$ – Mathematician Apr 10 '17 at 4:28
  • $\begingroup$ @Mathematician Focusing on "the force WITHIN the spring" doesn't help. Draw a free body diagram for the mass. There are two forces, its weight, and the force from the trampoline. All the details of "how" the trampoline creates the vertical force on the mass are irrelevant. Of course if you redesigned the trampoline (e.g. use different physical springs to connect the cloth with the frame) you wound get a different spring constant, but the question isn't asking you about that. The "art of engineering" (and physics!) is "ignoring all the details that don't matter". $\endgroup$ – alephzero Apr 10 '17 at 10:46

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