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The equations of motion are two coupled linear second order differential equations. Since the system is linear, any linear combination of solutions is still a solution, hence the set of solutions is a linear space. It can be proved that the solutions of a system of $n$-th order linear differential equations is a linear space of dimension $n$. This means ...


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It means does the solution of equation (5) satisfy eq (6)? You can check it by putting the solution in equation. As for your next question,it has just shifted the equilibrium point of the system around which your particle oscillated. You can gain more if you can solve the differential equation.


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Mathematically this is a differential equation. But, a solution is any function that you write down and replace $y(t)$ with $guess(t)$ and $y(t)_{tt}$ with $guess(t)_{tt}$, and the equality will still hold. The implication of including gravity is mathematically making it $F(y,y_{tt})=const \neq 0$ instead of having a homogeneous and boring $F(y,y_{tt})=0$. ...


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$Mg$ is a constant so only changes the equilibrium position, not the angular frequency. Equation 6 can be rearranged to give $$\frac{d^2y}{dy^2}=-\frac{k}{M}(y+L_0+\frac{Mg}{k})$$ This makes the new equilibrium position $L_0+\frac{Mg}{k}$ .


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The general trick is to convert the one second-order differential into two first-order ones. This is done by introducing an extra variable to carry around: \begin{align} \frac{\mathrm dv}{\mathrm dt}&=f(x,\,t)\\ \frac{\mathrm dx}{\mathrm dt}&=v \end{align} for whatever function $f(\cdot)$ you need. This can easily be applied to numerical integration ...


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Yes, assuming that gravity changes slowly compared to the damping of the system spring - weight. The effect of a sudden change of gravity would be equivalent to replace the weight by a heavier one for the same spring length. It would oscillate before rest at the new equilibrium position, with a bigger spring length. For a constant fluctuation gravity the ...


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You can do experiments (or thought experiments) to get a feel about it, using a bungee cord and a bag with weights, which you can switch (like small bottles of water). To see a constant gravity force (like in real world), hang a bag with weights such that the cord is stretched a little, but not to its full potential. To simulate a gravity force which ...


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Yes. The spring will stretch further if the gravity increases. But remember the the acceleration due to gravity has a direction, and of course, the spring will increase its length only if the acceleration of gravity has the right direction.


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Both approaches are actually the same, if you do them correctly. I will address your second case first. You are correct to use conservation of energy and say that the potential energy stored in the spring at the lowest point is equal to the sum of the kinetic energy and the potential energy due to gravity at the equilibrium point. So you were correct with ...


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