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It is difficult to give a definitive answer. The spring constant refers to an ideal spring which obeys Hooke's Law (extension proportional to load). Real springs and especially elastic materials such as bungee cord do not necessarily follow this law. Usually they do initially for 'smallish' loads, but they depart from it gradually at large loads. What ...

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While I see the 'use' of a Lagrange multiplier in this question, I don't really see the 'need' for it, so let's make it a little simpler. The Lagrangian for this system is found from: $$T = \frac{1}{2}[M(r_1\dot\theta)^2 + m_2(\dot{r_2}^2 +(r_2\dot\theta)^2)]$$ $$V = -(Mr_1+mr_2)g\cos\theta + \frac{1}{2}k(r_2-l_0)^2$$ $$L = ... 0 Try to consider the changes in KE, and how it alternates with EPE and GPE. You should also try and think about what happens to the velocity and acceleration of the mass when it reaches equilibrium, and when it is at maximum amplitude above and below the equilibrium point (consider the energies involved in those 3 specific points). This should give you some ... 2 So given a spring with spring constant k can one predict what dissipative force the spring will exert when extended? The answer is "No" because they depend on different things. The stiffness depends on the elasticity of the bonds between the atoms/molecules which make up the spring and the damping depends on the permanent distortion of the bonds which are ... 1 Spring stiffness is not responsible for energy loss. Consider a spring with stiffness k but no damping. The work done in compressing the spring by a distance x will be stored as the potential energy(PE) of the spring. For a spring compressed by a distance x, the PE is given by \frac12 k x^2. This energy is not lost and can be used by letting the ... 4 In relativity we only use the rest mass, also known as the invariant mass, of an object. In days past the concept of a relativistic mass was used, but this is now strongly deprecated as it has caused endless confusion. For example an obvious question is whether the increase of relativistic mass with speed can cause an object to become a black hole (tl;dr it ... 0 Yes it will do. You can take the potential energy to be zero when the spring is neither compressed or stretched. In special relativity the total invariant mass of the system would then include a contribution from the potential energy / c^2. The concept of mass in general relativity is quite subtle, but for weak gravitational fields, the Newtonian limit for ... 2 I think you are asking why this argument remains valid when the mass is oscillating, and does not only apply when it is static. I think the answer to this is that the forces in the springs depend only on their extensions (F=kx), not how quickly the extension is changing (F=kx+bx'+cx"). So at any given extension x the tension is the same regardless of the ... 3 Because the springs are considered massless. So if there were a force difference between the ends, you would get infinite acceleration. And this is valid not only for the ends, but for any two points, if there is no mass between them. The issue is somehow explained in this post: Is the tension in both ends the same (on a massed string)? And indeed, if the ... 0 As Farcher points out, the system is subject to a constant unbalanced force, so the resulting motion is constant acceleration. It is never in equilibrium : it is neither static nor moving with constant speed. Your eqn 1 is correct : K1x1=K2x2. However, the tension in the springs is constant, so x1 and x2 are constants, and when you differentiate they ... 0 Your approach looks correct. I didn't check your arithmetic, but I suspect that you got the right answer and that the issue here is that your expectations of Interactive Physics are not realistic. Interactive Physics is a simplified educational version of the Computer Aided Engineering program called Adams. This program is intended to produce results ... 0 Draw an free body diagram for each of the mass, apply N2L, eliminate the acceleration a to find the tension T. 2 I've simulated the case The two springs had the same initial length, and the block in the picture is in equilibrium. See how it is deviated towards the spring with bigger stiffness to decrease the resultant moment, and stop the rotation of the block. 0 The thing that the springs must have in common is their length x. This comes from the mass which is attached to the springs, having different lengths does not make sense in this setup. From this you can compute the forces. Say the spring constants are k_1 and k_2. Then the net force exerted is k_1 x + k_2 x. In equilibrium, this matches the ... 1 Your casual observation is astute; there is a relationship between the coils of a spring and sinusoids:$$e^{ix} = \cos x + i \sin x 

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A typical wound-into-a-cylinder spring is a helical in shape. And yes that means that subjected to tension and viewed from the side the transverse displacement will be sinusoidal to the same precision with which it approximates a helix.

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No, more force is required because of the higher spring constant. If the spring with the higher spring constant happens to have more mass, then inertia will come into play, but consider the following you can have two springs with different spring constant and same mass the expansion of a spring can be done arbitrarily slowly so that inertia (which opposes ...

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The idea of a massless spring is sometimes useful as is a spring whose mass is much, much less than the other masses within the system under consideration. In this case a massless spring implies that there will be an infinite acceleration when the stretching force is released so better to consider a spring with some mass. First assume that there are no ...

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