Spring force equations when there are two separated springs supporting a mass

[aspodkfpo]

Homework Statement:

n/a

Relevant Equations:

n/a

There is a trampoline drawn here and a graph of the spring force vs height.
I don't see why the spring force is decreasing at a decreasing rate with respect to height above trampoline.
F= kx = k * h/sin(theta), letting theta be between the horizontal and the spring.

 

[haruspex]

spring force is decreasing at an increasing rate with respect to height above trampoline.

Is it? Looks to me like the force is zero at all heights above the trampoline rest position and decreases at a decreasing rate as the trampoline rises from its lowest position to its rest position.

 

[aspodkfpo]

Is it? Looks to me like the force is zero at all heights above the trampoline rest position and decreases at a decreasing rate as the trampoline rises from its lowest position to its rest position.

Decreasing rate oops. How would you reason for a decreasing rate?

I think they meant for the spring force to be the symmetrical about the y-axis, hence the (both directions) bracket.

 

[haruspex]

How would you reason for a decreasing rate?

Write the force as a function of h and the constant width of the trampoline, not referring to theta.

I think they meant for the spring force to be the symmetrical about the y-axis, hence the (both directions) bracket.

No, they mean that the graph to the left of the y-axis does not depend on whether the trampoline is going up or down, i.e. no hysteresis.

 

[haruspex]

Following your questions on other problems in the Olympiad paper, I looked through some more of it and discovered that your question in this thread also comes from there, as question 13. This sheds a different light on the sketched curve.

The illustration under Marker's Comments does seem to show hysteresis. It ought to have arrows on the curves to make it clearer, but it seems to show different paths for ascent and descent. The force would be weaker as the springs contract, so the upper part (shown as a straight line) is descent.
This interpretation is supported by the comments in d ii), but I am rather confused. First, you were asked to sketch the curve, and the given solution has no hysteresis, then d ii) reads as though you are given the graph with the hysteresis in the answer booklet.

Anyway, this is not right. As I posted, we need to write the force as a function of h and constants. If the width of the trampoline is 2w then the extension of the springs is $\sqrt{h^2+w^2}-w$, giving a vertical force component of $kh(1-(1+\frac{h^2}{w^2})^{-\frac 12})$ per spring.
For small h this approximates a total force of magnitude $k\frac{h^3}{w^2}$. This is what explains the curve under "solution".
So the straight line portion in the diagram under Marker's Comments is spurious, whatever it was intended to illustrate. The correct picture would be two curves, each roughly cubic, but forming a hysteresis loop.

Edit:
Having looked at the answer booklet it is a little clearer - I was reading two consecutive questions as one.
Also, I see that the first graph was of force in the springs, which would be quadratic approaching the origin, whereas the second graph is force on the jumper, which is cubic as I wrote above.
The correct form of the second graph would be two cubics meeting at the origin. At the other end, they would be joined by a more-or-less straight vertical line.

 

[aspodkfpo]

Following your questions on other problems in the Olympiad paper, I looked through some more of it and discovered that your question in this thread also comes from there, as question 13. This sheds a different light on the sketched curve.

The illustration under Marker's Comments does seem to show hysteresis. It ought to have arrows on the curves to make it clearer, but it seems to show different paths for ascent and descent. The force would be weaker as the springs contract, so the upper part (shown as a straight line) is descent.
This interpretation is supported by the comments in d ii), but I am rather confused. First, you were asked to sketch the curve, and the given solution has no hysteresis, then d ii) reads as though you are given the graph with the hysteresis in the answer booklet.

Anyway, this is not right. As I posted, we need to write the force as a function of h and constants. If the width of the trampoline is 2w then the extension of the springs is $\sqrt{h^2+w^2}-1$, giving a vertical force component of $kh(1-(1+\frac{h^2}{w^2})^{-\frac 12})$ per spring.
For small h this approximates a total force of magnitude $k\frac{h^3}{w^2}$. This is what explains the curve under "solution".
So the straight line portion in the diagram under Marker's Comments is spurious, whatever it was intended to illustrate. The correct picture would be two curves, each roughly cubic, but forming a hysteresis loop.

Can you clarify what this means, "Looks to me like the force is zero at all heights above the trampoline rest position"?

 

[haruspex]

Can you clarify what this means, "Looks to me like the force is zero at all heights above the trampoline rest position"?

The y-axis is the trampoline rest position, i.e. the trampoline is flat and the jumper somewhere in the air above.
For the force on the jumper, second graph, it is obviously zero to the right of the y axis.
For the force in the springs, first graph, in a real trampoline the springs would still be exerting a force. They are never relaxed.

 

[aspodkfpo]

The y-axis is the trampoline rest position, i.e. the trampoline is flat and the jumper somewhere in the air above.
For the force on the jumper, second graph, it is obviously zero to the right of the y axis.
For the force in the springs, first graph, in a real trampoline the springs would still be exerting a force. They are never relaxed.

Why is there hysteresis?

 

[haruspex]

Why is there hysteresis?

Because there are always losses in springs. The energy you get out is less than the energy you put in. If you push on a spring to some point, reaching a force F, as you release it the force it exerts back on you drops below F immmediately.

 

[aspodkfpo]

Following your questions on other problems in the Olympiad paper, I looked through some more of it and discovered that your question in this thread also comes from there, as question 13. This sheds a different light on the sketched curve.

The illustration under Marker's Comments does seem to show hysteresis. It ought to have arrows on the curves to make it clearer, but it seems to show different paths for ascent and descent. The force would be weaker as the springs contract, so the upper part (shown as a straight line) is descent.
This interpretation is supported by the comments in d ii), but I am rather confused. First, you were asked to sketch the curve, and the given solution has no hysteresis, then d ii) reads as though you are given the graph with the hysteresis in the answer booklet.

Anyway, this is not right. As I posted, we need to write the force as a function of h and constants. If the width of the trampoline is 2w then the extension of the springs is $\sqrt{h^2+w^2}-1$, giving a vertical force component of $kh(1-(1+\frac{h^2}{w^2})^{-\frac 12})$ per spring.
For small h this approximates a total force of magnitude $k\frac{h^3}{w^2}$. This is what explains the curve under "solution".
So the straight line portion in the diagram under Marker's Comments is spurious, whatever it was intended to illustrate. The correct picture would be two curves, each roughly cubic, but forming a hysteresis loop.

Edit:
Having looked at the answer booklet it is a little clearer - I was reading two consecutive questions as one.
Also, I see that the first graph was of force in the springs, which would be quadratic approaching the origin, whereas the second graph is force on the jumper, which is cubic as I wrote above.
The correct form of the second graph would be two cubics meeting at the origin. At the other end, they would be joined by a more-or-less straight vertical line.

How did you get through this algebra?
"
$\sqrt{h^2+w^2}-1$, giving a vertical force component of $kh(1-(1+\frac{h^2}{w^2})^{-\frac 12})$ per spring.
For small h this approximates a total force of magnitude $k\frac{h^3}{w^2}$. "

 

[haruspex]

How did you get through this algebra?
"
$\sqrt{h^2+w^2}-1$, giving a vertical force component of $kh(1-(1+\frac{h^2}{w^2})^{-\frac 12})$ per spring.
For small h this approximates a total force of magnitude $k\frac{h^3}{w^2}$. "

When the centre of the trampoline is h below the unstretched position, each half forms a right angled triangle. The hypotenuse is $\sqrt{w^2+h^2}$, so the extension is $\sqrt{w^2+h^2}-w=w(\sqrt{1+\frac{h^2}{w^2}}-1)$ (had a typo previously), the force in each half is $kw(\sqrt{1+\frac{h^2}{w^2}}-1)$ and the vertical component of each is $kw\frac{h}{\sqrt{w^2+h^2}}(\sqrt{1+\frac{h^2}{w^2}}-1)=kh(1-(1+\frac{h^2}{w^2})^{-\frac 12})$.
For small h/w we can use the binomial expansion $(1+\frac{h^2}{w^2})^{-\frac 12}=1-\frac 12\frac{h^2}{w^2}+\frac 38 \frac{h^2}{w^2}...$ which I will truncate at $1-\frac 12\frac{h^2}{w^2}$.
That gives the total vertical force as $\frac{kh^3}{w^2}$.