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-1

The answer is no. The net force at the maximum elongation points has the same magnitude. This is because the rest point of the spring is modified by the gravitational weight of the mass. The mass oscillates around this new rest point, and at the points of maximum amplitude the net force is the same. One can tell that the gravitational force can be ...

2

In accordance with Hooke's law the force is linear with distance. Incorporating gravity only means that the equillibirum position of the spring has changed, the "zero" around which it oscillates. The gravitational pull is already compensated by the spring. Thus the magnitude of the force is euqal at $-A$ and $+A$. Edit: When the gravitational pull on the ...

5

The equilibrium position in this case is not where the spring is not stretched, it is actually stretched by a $\Delta x$ amount with $F_{spring}(0) = k\Delta x$. So the spring force on point A is a little smaller than in point -A, since $F_{spring}(A) = -k(A-\Delta x)$ and $F_{spring}(-A) = k(A+\Delta x)$ so it compensates the "extra" force. You have ...

0

I agree with your explanations. Just an afterthought, have you considered how much extra strength it requires to balance without a cane. Walking with a cane makes balancing easier taking stress away from the legs which would otherwise compensate for missteps.

1

First off, let's answer the question: what part of that equation would change? $f\left(\epsilon\right)$ is always the same if you're in equilibrium. Scattering won't change that, because scattering alone won't move you out of equilibrium. If anything, scattering has the opposite effect. What can change is $Z\left(\epsilon\right)$. How it would change ...

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