# Mass on elastic rope

1. Jun 18, 2014

### skrat

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?

2. Jun 18, 2014

### ehild

As you need the time at the maximum distance there is better to keep the original differential equation for z(t) , with the initial conditions z(0)=0 and dz/dt = √(2gl) at t=0.

$\ddot z+\frac {k}{m}z=g$

is a linear second order equation, with constant coefficients. The general solution is the sum of the general solution of the homogeneous part + a particular solution: $z(t)=z_h+ z_p$. You know that the general solution of the homogeneous part is $z_h=Asin(ωt)+Bcos(ωt)$and you get a particular solution by setting $\ddot z=0$: $z_p=\frac{gm}{k}$

You oly need to find ω and fit the constants to the initial conditions. Then find the time when z is maximum.

ehild

3. Jun 18, 2014

### skrat

Is this $\omega$ you mentioned the same as if the system was oscillating around equilibrium position ($\omega ^2=k/m$) or is this something else?

4. Jun 18, 2014

### ehild

Substitute the solution back into the differential equation and see.
(Yes, it is the same)

ehild

5. Jun 18, 2014

### skrat

Hmmm, ok, this should be it than:

$z(t)=Acos(ωt)+Bsin(ωt)+\frac{gm}{k}$

$z(t=0)=0=A+\frac{gm}{k}$

$\frac{dz}{dt}(t)=-Aωsin(ωt)+Bωcos(ωt)$

$\frac{dz}{dt}(t=0)=\sqrt{2gl}=Bω$ where $ω=\sqrt{\frac k m}$

So finally

$z(t)=-\frac{gm}{k}cos(\sqrt{\frac k m}t)+\sqrt{\frac m k 2gl}sin(\sqrt{\frac k m}t)+\frac{gm}{k}$

We already calculated $dz/dt$:

$\frac{dz}{dt}(t)=-Aωsin(ωt)+Bωcos(ωt)=\sqrt{\frac k m}\frac{gm}{k}sin(\sqrt{\frac k m}t)+\sqrt{\frac k m}\sqrt{\frac m k 2gl}cos(\sqrt{\frac k m}t)=0$

$\frac{gm}{k}tan(\sqrt{\frac k m}t)+\sqrt{\frac m k 2gl}=0$

$t_2=\sqrt{\frac m k}arctan(\sqrt{\frac{m2gl}{k}}\frac{k}{gm})=0.319s$

6. Jun 18, 2014

### ehild

Don't you miss a minus sign?

ehild

7. Jun 18, 2014

### skrat

Ammmm. Wow. This means that $t_2$ would be negative. It can't be negative.

8. Jun 18, 2014

### ehild

Add pi to the arctan .

ehild

9. Jun 18, 2014

### skrat

HAHA. :D Slow down.

Firstly, how do I know that I didn't make a mistake during my calculus and that adding pi will give me the right result? I mean I know now because you said so but if I would get this on exam, I would immediately rip the papers.

10. Jun 18, 2014

### ehild

The mass would go downward after reaching the end of the unstreched string till it reaches maximum depth. That will happen in the future, that means positive time not negative one. You know the tangent of the phase when that happens, but tangent is periodic function, with period pi. tan(x) =tan(x+pi) You need the smallest positive phase angle.

You did the calculus well except the sign error at the end.

And you can think also that just letting the mass drop from the end of the elastic string it starts at maximum distance from the equilibrium position and comes to a halt at maximum distance again, on the opposite side. At that position, the phase is pi.
In this situation, the mass had initial downward velocity. It will go further , so the phase will be greater than pi at the end.

ehild

11. Jun 18, 2014

### skrat

Ok, makes sense. Thanks for explanation.

$t_2=\sqrt{\frac m k}arctan(-\sqrt{\frac{m2gl}{k}}\frac{k}{gm}+\pi)=0.38s$

The total time is like I said $t_{tot}=t_1+t_2$.

Thank you very much!

12. Jun 18, 2014

### ehild

Not quite. $t_2=\sqrt{\frac m k}\left(arctan(-\sqrt{\frac{m2gl}{k}}\frac{k}{gm})+\pi\right)$

ehild

13. Jun 18, 2014

### skrat

Of course.. It's arctan.

I want to move along y axis for period pi and not along x axis like I did. Ah ...