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If your mass is attached to the bottom of an un-stretched vertically hanging spring and released, the spring is initially exerting no force, and the mass will accelerate downward due to gravity. If you choose the zero point for gravitational energy at the equilibrium position where mg – kx = 0, it will start with gravitational energy mgx, and pass ...

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If the mass is released it will also have kinetic energy. You must include the kinetic energy in your expression.

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Now if the mass is released, the tension will be the restoring force of the spring and since the mass is neither going up nor going down But at that point the velocity (kinetic energy) is non-zero. So the claim that it neither goes up or down is false. In your energy conservation statement, you have zero kinetic energy and that happens only at maximum ...

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Add some kind of damping element. This is regularly done when you require a spring to isolate vibrations, i.e. in a car suspension: The damping element slows down the relaxation of the spring and at the same time dissipates energy to reduce oscillations.

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insert it in oil cool it down with liquid nitrogen

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If the inital force $Kx$ from the spring on block A is less than or equal to the maximum static friction between the two blocks then friction from block B will exert a force $F$ to the right on block A that is sufficient to prevent relative motion between the blocks. Note that no relative motion does not means that block A is at equilibrium - if block B ...

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It's easier to think this situation in terms of centre of mass frame. Let $F_2>F_1$. Hence $F_{net}=F_2-F_1$ $$F_{net}=(m_1+m_2) a_{CM}$$ $$a_{CM}=\frac{F_2-F_1}{m_1+m_2}$$ Where $a_{CM}$ is acceleration of center of mass of our system. Since whole system moves with some constant acceleration, our individual blocks $m_1,\,\,m_2$ experience the pseudo ...

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Please see the lecture on http://www.unizor.com under Physics 4 Teens > Waves > Transverse Waves > Musical Strings 1. It models a musical string (like the one on a violin) with a model you describe. It explains why oscillations with small vertical deviation are close to harmonic. The angular frequency $\omega$ of these oscillations is \omega^2 = \...

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If you release the mass from $\Delta y = 0$ and do not stop it at its equilibrium point $\Delta y = \frac {mg} k$ then it accelerates until it reaches its equilibrium point and then decelerates until it is momentarily stationary at $\Delta y = \frac {2mg} k$ (because it has inertia it cannot stop at its equilibrium point, even though the net force on it here ...

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The interatomic displacements are very small, since an elastically deformed body consists of zillions of atoms. Yet, for sufficiently big deformation Hooke's law breaks: first the deformation becomes non-linear, then plastic (i.e., the object does not return to its initial state anymore), and eventually it breaks. Wikipedia figure of deformation stages

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As far as I know as per Coulomb's law or Newton's law, attraction force or repulsion force should decrease as the distance increases between two particles(charged). This is true; for example, because of electron distribution fluctuations, even neutral atoms are attracted to each other as $\sim r^{-6}$, where $r$ is the separation distance. However, they're ...

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Newton's and Coulomb's laws apply to point particles interacting by gravitational and electrostatic forces only. That's far from the case in a solid, or even a diatomic molecule for which there are many interacting particles that are subject to Coulomb's law (Newton is negligible) and the laws of quantum mechanics. The situation of a diatomic molecule is ...

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Overall you have made the problem more difficult by making the statement Let the height at the bottom of oscillation = 0. Choosing the datum as the equilibrium position is the more usual way. In the way you have formulated the question the diagram below shows: the unextended spring (natural length), the system (spring & mass) at equilibrium, the system ...

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Your solution is correct for $|\omega| < 10$. When $\omega=10$ the spring exerts an inwards force of $100x$ N on the mass at extension $x$ metres, but the centripetal force required to keep the mass moving in a circle with radius $2+x$ metres is $200 + 100x$ N, which is always greater than $100x$ N. So when $\omega=10$ the spring is not strong enough to ...

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If the mass was displaced a further distance $dx$ so the extension is $x+\delta x$ then the restoring force is $F = k(x+\delta x) - mg = kx + k \delta x -kx = k\delta x$ towards the equilibrium position. The restoring force is proportional to the distance from the equilibrium position and that's the condition for SHM. In SHM the extremes of the oscillation ...

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The first thing to note is what k is. In a linear equation like this, k is what is known as a “constant of proportionality”. Funnily enough (although I hope this doesn’t confuse you), the letter k is often used as a general constant of proportionality, but in this equation k is the symbol given to the “Spring constant”. It is a property of the spring. If we ...

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The force during extension of the spring is not constant. $F = kx$ means that the force at that very extension or compression $x$ from the equilibrium position is $F$. The work done is given. You need to determine the spring constant $k$ first by integrating the spring force between the limits $0$ to $x_1$ and equating it to the work done. Then when you have ...

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