I have one spring $k$ with one end fixed and a mass $m$ on the other.
Is there any difference to its period when the system is vertical (g) versus when it lays on a table without friction?
The equilibrium changes, obviously.
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1$\begingroup$ Could you add the Newton's equations to your question? Perhaps the answer will even be obvious from the form of the equations. $\endgroup$– Roger V.Commented Jul 6, 2021 at 14:00
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The answer is no, the period doesn't change. Let's look at it more formally. The equation of motion of horizontal pendulum is $$m\ddot{x}=-kx$$ The equation of motion for vertical pendulum is $$m\ddot{x}=-kx+mg$$ If we use substitution like $y=x-mg/k$, then the second equation can be rewritten as $$m\ddot{y}=-ky$$ Here we used the fact that $mg/k$ doesn't depend on $t$, thus $\ddot{x}=\ddot{y}$ As the equations for horizontal pendulum in terms of $x$ and vertical pendulum in terms of $y$ is the same, we get the same solution and thus the same period for both cases. The only difference is that in the vertical case the point around which the mass oscillates is displaced down by an amount of $mg/k$.
