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This probably isn't going to be in time for your lab tomorrow, but I'll attempt to shed some light on this: In this sort of fluid-structure interaction with an oscillating/vibrating body, there are generally three fluid effects that need to be considered: added mass, added stiffness and added damping. Added mass is basically the fluid's mass adding to that ...


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If the two devices are sharing the load, they will each show approximately half the weight. But if you hang one below the other, then each feels the full load. Think about it this way: the one nearest the suitcase "feels" the suitcase. It doesn't care whether you are holding it with your hand, attached to a spring, or whether it is dangling from the side of ...


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After the elongation of x units in the spring, the spring develops energy in the form of spring compression or elongation energy(the work done in elongation is converted into spring energy), spring tries to get into its mean position. Now, if the system is isolated the spring will perform SHM otherwise friction and other forces will make it stop its motion. ...


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Since you have "done the work" (as much in the comments as in the original question...) but are left with a small uncertainty about the concept of what force applies where, I feel I can now write an actual answer. The following diagram explains it (gravity operates from right to left in my picture so the text aligns better): The spring compresses by ...


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A simple model for a coil spring would be that, when the spring is subjected to a force, the entire coil is subjected to a torsion $\tau$. This torque causes the coil the twist by an angle, which can be approximated with, $$ \theta = \frac{l\tau}{IG}, $$ where $\theta$ is the angle of twist in radians, $l$ the length of the coil (not to be confused with ...


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In the standard approximation it does not depend on the length but on the number of active windings. As you can find e.g. here, the spring constant $k$ is $$k=\frac{G d^4}{8 n D^3}$$ whre $d$ is the wire diameter, $D$ the coil diameter, $n$ the number of active coils, and $G$ the shear modulus. So if you cut it in half $n \rightarrow n/2$ the spring ...


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You can understand in a simple way the factor $1/3$ which gives you the approximate solution in the low frequency regime (more on this later) in the following way. Start by writing the kinetic energy of your system as: $$K = \frac{1}{2} m \dot{\delta}(\ell)^2 + \int_0^\ell \frac{1}{2} \lambda \dot{\delta}(u)^2 du$$ where $\delta(u)$ is the displacement of ...


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A simple figure of the system at hand is given below. In writing the kinetic energy we use the velocities of particles. In particular, we square the velocities. And velocity is nothing but the derivative of the displacement. Here we have two particles where we showed their displacements by $x_1$ and $x_2$. That means kinetic energy for the first particle ...


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No, it can't. The unstretched spring extension is the equilibrium solution to the unforced equation $$m\ddot x+\gamma\dot x=-k(x-x_0),$$ whereas the stretched extension is the equilibrium solution to the forced equation $$m\ddot x+\gamma\dot x=-k(x-x_0)-mg.$$ The equilibrium positions are completely insensitive to the damping constant $\gamma$, since you ...


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Your assumption that the force at A is equal to $mg$ is not correct. It might have been correct if the entire system was in equilibrium. Then you would have been justified in using the principle of equilibrium. However, this is a dynamic system, and you have to apply Newton's 2nd Law instead. To solve this problem, let us define three variables, $x$, $y$ ...


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In wristwatches, they tend to use a spiral wound spring called a mainspring - this allows you to have significant (but decreasing) torque over a large number of turns (lot of stored energy in a small volume). It is not constant force, torque or power although the design can strive to have a plateau where torque is "more constant". From the above linked wiki ...


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The result is not physical, so there must be a mistake! The potential energy is not correct. The negative sign in the gravity term suggests that the lowest potential is at theta=0, whereas this is actually highest potential. I guess with this change you'll have a transition from positive to negative with oscillation to falling over being the two cases.


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Two equations that may be helpful to you, depending on exactly what you're trying to do with your spring: Hooke's Law, which is written in equation form as $F=-k\Delta x$ states that the restoring force that returns a spring to equilibrium is proportional to its distance stretched, where $k$ is the spring constant. The spring constant is a measure of how ...


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What you observe as mechanical deformation of a steel spring is an actual displacement (motion) of the atoms constituting the spring. In places, atoms will be slightly closer to their neighbors (compression) and in some other places actually futher apart (tension). The combination of compression on one side and compression on the other side of a beam or a ...


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First, let's review the basic ideas of simple harmonic motion (I'm assuming an early university level). Starting with Newton's equation: $$F=ma$$ and using Hooke's law $$ma=-kx$$ then recognizing that acceleration is the second derivative of position x $$mx''= -kx$$ We know that simple harmonic motion is sinusoidal, so we substitute $x=\sin(\omega t)$ ...



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