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My question would be, where is the reaction force corresponding to the spring force (Fe) in this game? It's the force that the block exerts on the spring. To see this, refer to the free body diagrams of the block and spring below showing the action-reaction pairs. Per Newton'w 3rd law, the force the spring exerts on the block, $F_e$, is equal and opposite ...

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Hooke’s Law $F=kx$, where $F$ is the restoring force, $k$ Is the spring constant, and $x$ is the displacement, is defined for static equilibrium. For the special case of a massless spring, though, the relation can be used in dynamic situations as well and can therefore be applied to the analysis of simple harmonic motion, for example. The reason is that the ...

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This is the FBD From here you can obtain the equation of motion, you should get second order differential equation plus first order differential equation. Edit $$m_1\,\ddot x_2=F_\sigma$$ and put dummy mass between the spring and the damper $$m_d\,\ddot x_1=F_k-F_\sigma$$ with $~m_d=0~$ and $$F_k-F_\sigma=0\quad,F_\sigma=\sigma\,(\dot x_1-\dot x_2)\quad, ... 2 We can use symmetry to gain insight here. The first case involves a spring with spring constant k and a mass m at either end. We could imagine the spring being fixed at the motionless middle, corresponding to the center of mass. The relevant spring on either side is now half as long, so its spring constant is 2k. The oscillating mass on either side is ... 1 to obtain the eigen frequencies ~\omega_i~ you start with the equations of motion case I$$m\,\ddot x_1+k\,(x_1-x_2)=0\\ m\,\ddot x_2-k\,(x_1-x_2)=0$$or$$ \begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{bmatrix}+ \underbrace{\frac{k}{m}\,\begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix}}_{\mathbf{K}}\,\begin{bmatrix} ...

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Since pendulum undergoes rotation around the point at which is suspended, it is more convenient to talk in terms of "Moment which the force exerts (as opposed to talking in terms of force alone)" and in terms of "Moment of inertia (instead of mass, which we consider for linear motion)". But the above relation would still hold, i.e. $\... 1 In physics two bodies collide if the velocity (or momentum) of one body is changed due to other, now this does not mean that they need to be in contact. Perfectly elastic collision is a collision in which there is 100% gain of kinetic energy which is converted into potential energy during collision. While perfectly inelastic collision means that there is no ... 1 Larger resistance doesn't mean less current in this case. When magnetic field on a conductor changes, a "current" is induced in the coil. Notice that it is current not voltage. The voltage comes from the current running through some impedance. When you are increasing the resistance, the current is not changing much but the total joule loss is ... 1 start with the position vector to the mass m $$\mathbf R=\left[ \begin {array}{c} x+ \left( r+l_{{0}} \right) \sin \left( \theta \right) \\ \left( r+l_{{0}} \right) \cos \left( \theta \right) \end {array} \right]$$ from here you obtain the velocity$\mathbf v~$hence the kinetic energy $$T=\frac 12 M\dot x^2+\frac 12 m \mathbf v\cdot \mathbf v$$ and the ... 1 The simplest way to keep track of this is to write$\ell=\ell_0+\epsilon r$and use$\epsilon \theta$rather than$\theta$. Expanding your Lagrangian in powers of$\epsilon$, you can then use$\epsilon$as a counter to keep track of "smallness". The anharmonic terms are those with 3 or more powers of$\epsilon\$ in the Lagrangian, or two or more ...

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