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There are various methods to find anything we like for a mass attached to a vertical spring, however I wanted to use the conservation of energy.

Suppose $m$ hangs from the spring of normal length $l$ and we have that the equilibrium point is at distance from the top of $z_0$, which I've found to be $z_0 = \frac{kl+mg}{k}$. Then I want to displace the mass downwards by an amount $z_d$.

I would like to find the highest point the mass will be lifted by the spring thanks to this displacement.

I've tried this way: I know that at $z = z_0+z_d$, the Kinetic Energy $K = 0$. And I also know that $K=0$ when the mass will be at the highest point.

Hence I take $V(z) = -mgz + \frac{1}{2}k(z-l)^2$ and substitute $z= z_o+z_d$ to get $V(z_0+z_d) = -mg(z_o+z_d)+\frac{1}{2}k((z_o+z_d)-l)^2$.

Then I take the same expression and substitute $z=z_{max}$ and I get $V(z_{max})=-mgz_{max}+\frac{1}{2}k(z_{max}-l)^2$.

Hence I put these two quantities equal to each other. However all I get is that $z_{max} = z_0+z_d$, which is obvious. What did I do wrong?

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  • $\begingroup$ As highest point I mean the point with lowest absolute value of $z$ of course, which is the highest point from the earth! $\endgroup$ – Euler_Salter Nov 24 '16 at 1:59
  • $\begingroup$ "Then I want to displace the mass downwards by an amount $z_d$. I would like to find the highest point the mass will be lifted by the spring thanks to this displacement." You're aware the mass will enter into a harmonic oscillation, right? $\endgroup$ – Gert Nov 24 '16 at 2:09
  • $\begingroup$ Yeah, were the motion would be $z(t) = A\cos{wt}+B\sin{wt}+\frac{mg}{k}$ with $A,B$ constants and $w = \sqrt{\frac{k}{m}}$ (I think, haven't checked). But I would like to solve it without using the equations of motion etc, if possible. I'd like to use the conservation of energy only! Is it even possible? $\endgroup$ – Euler_Salter Nov 24 '16 at 2:17
  • $\begingroup$ "Is it even possible?" Let me have a look. It might be an odd way of doing it. $\endgroup$ – Gert Nov 24 '16 at 2:30
  • $\begingroup$ If you get an answer that is "obvious", why do you think you did something wrong? $\endgroup$ – sammy gerbil Nov 24 '16 at 3:45
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Introducing the natural unextended length of the spring $l$ only complicates the analysis.

The static equilibrium condition for the mass, with down as positive, is $mg - kz_o =0$ where $z_o$ is the extension of the spring.

Assume the zero of potential energy (both spring and gravitational) to be at the level of the bottom of the unextended spring.
You took the zero of gravitational potential energy at the top of the spring so necessitating the introduction of the unextended length of the spring.

Now consider the mass, the spring and the Earth as the system.
If the spring is then extended a further distance $z_d$ below the static equilibrium position then the energy stored in the system is $\frac 12 k (z_o+z_d)^2 - mg(z_o+z_d)$
Now assume that the mass is $z_u$ above the equilibrium position then the energy stored in the system is $\frac 12 k (z_o-z_u)^2 - mg(z_o-z_u)$

Equating these two energies and using the static equilibrium condition you can show that $z_u=z_d$.

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