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The force of gravity will simply move the equilibrium point, being a costant force. So if $l_0$ is the initial lenght of the spring, the applying the orizontal component of the force $Mg\sin \theta$, you'll get the new lenght in equilibrium that is (Hooke's Law) $$ \vec{F} = -k\vec{(x - x_0)} \Rightarrow x = F/k + x_0$$ $$x_1 = Mg\sin \theta / k + x_0$$ ...


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Yes, your circuit on the right models a chain of springs intersperced with masses. Think about it. It should be obvious from frequency analisys alone that the left circuit can't be right. Springs with masses are going to low pass filter any force or motion input. The left circuit clearly high pass filters, not low pass.


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From the free body diagram you must have $$ \begin{align} m_1 \ddot{x}_1 &= F_1 - F_2 \\ m_2 \ddot{x}_2 &= F_2 - F_3 \\ m_2 \ddot{x}_3 &= F_3 \end {align} $$ with the spring forces defined as $$ \begin{align} F_1 & = -k_1 x_1 \\ F_2 & = -k_2 (x_2-x_1) \\ F_3 & = -k_3 (x_3-x_1) \end{align} $$ The above is combined as $$ ...


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You have a mass moving under the influence of a simple massless spring. The equations of motion are $F = -k x = m a$. These can be solved using direct integration. $$ \begin{aligned} a =\frac{{\rm d}v}{{\rm d}t} & = -\frac{k}{m} x \\ \int \frac{{\rm d}v}{{\rm d}t} \, {\rm d}x & = - \int \frac{k}{m} x \,{\rm d} x \\ \int v \, {\rm d}v ...


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There are a few problems with your derivation. Your main mistake is using formulae that assume the acceleration of the spring is constant (it is not). In one dimensional systems with nonuniform acceleration, we have $$ v(t) = v_0 + \int_{t_0}^t a(t)dt . $$ Furthermore, the $v^2 = v_0^2 + 2a \Delta x$ no longer works for nonuniform accelerations either. ...


2

One can't discuss the difference between factors of $1/3$ and $1/6$ without mathematics. The difference between these two numbers – and generally, any fact about any numbers – is all about mathematics. If all values of $x$ between $0$ and $A$ were equally likely, the average value of $kx^2/2$ would be $kA^2/6$ as you say because the average value of $X^2$ ...



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