Mass on elastic rope

1. The problem statement, all variables and given/known data
On the end of ##1m## long elastic rope a ##50g## mass is hanged, therefore the rope extends for ##20cm##. Than we lift the mass to the point where rope is attached to the ceiling. Now we release the mass and let gravity do the work. How many seconds will pass until the body reaches it’s maximum distance from the ceiling?

2. Relevant equations

3. The attempt at a solution

Here is what I did:

I split the total time in two parts ##t_{tot}=t_1+t_2## where ##t_1## is time where elastic rope has no effect and ##t_2## time when taking elastic rope in account is crucial.

We can easily calculate ##t_1## from ##l=\frac 1 2 gt_1^2##.

A lot more complicated is to get ##t_2## at least the way I started…. There must be an easier way!

##mg-kz=m\ddot z##

If ##\ddot z =0## than ##k=\frac{s}{mg}##. Now let’s continue working with Newton’s equation:

##\ddot z=g-\frac k m z## note that vertical displacement is now measured from the point where rope actually has an effect. This explicitly means that the total distance of the body from the ceiling at this point is ##1m##. Meaning I am trying to find out what is happening with the body below that point.

To reduce the order of DE I used ##\dot z =v ##, which also means that ##\dot v =\frac{dv}{dz}\dot z=\frac{dv}{dz}v##. This leaves me with

##\frac{dv}{dz}v=g-\frac k m z## so

##v(z)=\sqrt{2gz-\frac k m ^2+C}##

We also know that ##v(z=0)=v_0=\sqrt{2gl}## where ##l=1m##. This exactly determines that ##C=2gl##

##v(z)=\dot z=\frac{dz}{dt}=\sqrt{2g(z+l)-\frac k m ^2}##

Now this is where all the nasty s*** begins.

##\int \frac{dz}{\sqrt{-\frac k m z^2+2gz+2gl}}=\int dt+D##

Now according to my book this can be integrated:

##-\frac{1}{\sqrt{\frac{k}{m}}}arcsin(\frac{-2\frac k m z+2g}{\sqrt{\frac{8kgl}{m}+4g^2}})=t+D##

Now if this weren’t so horrible, I would get ##z(t)## from the last equation. Than find such ##D## that ##z(t=0)=0##.

After that I would take a closer look at the equation I got for ##v(z)## and find ##z_{max}## from ##v(z_{max})=0##.

In order to find the time I am looking for, I have to than use condition ##z(t)=z_{max}## and find the right ##t_2##.

Huh. Is this even the right way to do it? Is there really no easier way? I am mistaken somewhere?

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