# Maximum height mass attached to a vertical spring with conservation of energy

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?

• As highest point I mean the point with lowest absolute value of $z$ of course, which is the highest point from the earth! – Euler_Salter Nov 24 '16 at 1:59
• "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? – Gert Nov 24 '16 at 2:09
• 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? – Euler_Salter Nov 24 '16 at 2:17
• "Is it even possible?" Let me have a look. It might be an odd way of doing it. – Gert Nov 24 '16 at 2:30
• If you get an answer that is "obvious", why do you think you did something wrong? – sammy gerbil Nov 24 '16 at 3:45

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.
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 $z_u=z_d$.