Finding a formula for displacement of a mass on a spring using v.

[hamishmidd]

Homework Statement

A mass m is hung from a spring with spring constant k. The mass is kicked upwards such
that it has a speed of v when the mass is at the equilibrium position. What is the maximal displacement of the mass from the equilibrium position as the mass subsequently
oscillates?

Relevant Equations

Ek=1/2mv^2, U=1/2kx^2, kx=mg (at equilibrium position)

I have tried to answer this using the relevant equations I am provided on my formula sheet, however I get stuck pretty close to the end. I start with 1/2mv^2=1/2kx^2 at the equilibrium position, and kx=mg, x=mg/k. This gets me to v^2=mg^2/k, but I don't know where to go from there. The potential answers are:
(A) x = v*sqrt(m/k) (B) x =v^2/2g (C) x =sqrt(2mv/k) (D) x = vt +1/2gt^2 (E) None of the above

 

[kuruman]

A vertical spring-mass system oscillates about its equilibrium position exactly like a horizontal spring-mass system. The only difference is that the vertical spring is at equilibrium when the spring is stretched by $\Delta x=mg/k$ whilst the horizontal spring is not stretched at equilibrium.

Answer this question as if you had a horizontal spring. Note that you are asked to find the maximal displacement from the equilibrium position. What is another name for it?

 

[haruspex]

I start with 1/2mv^2=1/2kx^2 at the equilibrium position,

Further to @kuruman's advice, I'll point out that there is no such standard equation.
There's $1/2mv_{max}^2=1/2kx_{max}^2$, and there's $1/2mv^2(t)+1/2kx^2(t)=E$, where x is displacement from equilibrium.
At equilibrium, $x=0, v=v_{max}$.