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2

Hooke’s Law is a first-order linear approximation and is only valid over a small range of deformation. You cannot use it to conclude anything about the forces between atoms or molecules outside of this range - except that, as your example shows, these forces cannot be linear over their whole range.


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I don't think this will reduce to a simple spring by letting $\eta$ go to zero. The dashpot allows the spring to relax over time to its zero stress state. Reducing $\eta$ to zero means the spring will relax instantaneously. That is, the dashpot end of the spring is free to move. I think what you want is to let $\eta$ go to infinity which, in effect, connects ...


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The springs are massless, so the tension at any point in either spring is the same value, $F$. For the two individual springs, with extensions $x_1$ and $x_2$, we have# $F = k_1x_1=k_2x_2$ and when considered as a single spring they have an effective spring constant $k$ such that $F = k(x_1+x_2) \\\displaystyle \Rightarrow F = k\left(\frac F {k_1} + \frac F {...


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When the springs (assumed to be massless) are hung upside down, they will have a zero extension. So in fact $$mg = k_1x+k_2x$$ when the mass is added to the system. You wrote that the resultant force $$F_r = F_1+F_2$$ and concluded that this should be $2F$ as if the force in both springs were equal. So what you should have done is write the resultant ...


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ok so the first condition for the series spring system is : The spring force in the entire system is the same i.e the tension in the springs is the same therefore $k_1x_1=k_2x_2$ $k_e(x_1+x_2)=k_1x_1$ where $k_e$ is the equivalent spring constant so if the system is vertical let gravitation force, mg, be $F$ hence $x_1=F/k_1$ ,$x_2=F/k_2$ and $x_1+x_2=F/k_e$ ...


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Chemical bonds apply at the atomic level and the potential tying atoms and molecules up to liquids and solids cannot be approximated by the potential of a spring, it is too complicated for that. The atoms and molecules are in orbitals, which allow spaces where positive and negative electromagnetic forces allow for bonding,( see for links this answer of mine ...


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You were correctly looking at the general form of the response of a harmonic oscillator, where you get the $\frac{1}{\omega_n^2-\omega^2}$ proportionality term between the displacement and the force (I omit some constants here). From here, if $\omega_n\ll\omega$, you get a minus sign. So the force and displacement are just like in the 1st figure, at a 180 ...


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It's not conservation of energy, it's the minimization of energy. The potential energy with gravity "on" is: $$U(x)=\frac 1 2 kx^2+mgx $$ which is minimized via: $$\frac{dU}{dx} = kx + mg =0$$ or $$ x=-\frac{mg} k$$ Thus, the energy stored in the spring is: $$ E_0=\frac 1 2 kx^2=\frac 1 2 \frac{(mg)^2}k$$ while the gravitational potential energy is:...


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You are confusing two moments in time here. One is the instant the spring is maximally compressed, the other is when one of the blocks is (almost) at rest relative to the rail (ground). These are not occuring at the same time, but it is difficult to see because you need to pay precise attention to two quantities at the same time (spring compression and block ...


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