Chain on a hillock

consider any small element of mass 'dm' of length 'dx'

Let the linear mass density of the chain be ' $\lambda$ '

Then the forces acting on this small component is dT ( change in tension from right of this component to left) and

dm(g sin (x)) along the hillock

or we have $dm\quad g\quad sin\theta \quad -\quad dT\quad =\quad dm\quad a\quad \\ \\$

Now being an inextensible string, also due to lack of friction there is no chance of slacking,

so acceleration of all components have same magnitude but different directions.

Now

being a free chain, its tension at the ends are 0 , (qualititavely it can also be explained as arising from the fact that net tension should cancel out as an internal force so $\int { dT }$ should be 0 when over whole chain it is integrated)

also

$gsin\theta \quad dm\quad =\quad gsin\theta \quad \lambda \quad dl\quad =\quad g\lambda \quad (dl\quad sin\theta )\\ \\ =g\lambda dy$

thus we have

$dm\quad g\quad sin\theta \quad -\quad dT\quad =\quad dm\quad a\quad \\ \\$

= $g\lambda dy-dT=adm$

Now when i slightly displace the chain,, the lengths of chain on either hill change by l+dl and l-dl, and height to h+dh and h-dh,

Then integrating expression for part 1 under limits

we get

$g\lambda dy-dT=adm\\ \\ g\lambda (h2-h1)=ma(downward\quad along\quad the\quad hill)\quad for\quad h1=h+dh\quad and\quad h2=h-dh\\ \\ we\quad have\quad -2g\lambda dh=ma\quad or\quad \\ -\frac { g\lambda 2 }{ m } dh=a=-2\frac { g\lambda }{ m } sin\Phi \quad dl=\quad -2\frac { g }{ l } dl=\ddot { l } (represents\quad acceleration\quad along\quad length)\\ \\$

$so\quad 2\pi \sqrt { \frac { l }{ 2g } } =\quad 1.40496$

Now isnt that a beautiful result , it is equivalent to a pendulum with half its length, please correct me if I am wrong