Consider the liquid in the length dx. It’s mass is Ardx at a height x.

PE = Aρdx gx

The PE of the left column

similarly, P.E. of the right column=

h_{1} = h_{2} = l sin 45^{o} where l is the length of the liquid in one arm of the tube.

Total P.E. = Aρgh^{2} = Aρgl^{2} sin^{2} 45^{o} = Aρgl^{2}/2

If the chamge in liquid level along the tube in left side in y, then length of the liquid in left side is l-y and in the right side is l+y.

Total P.E. Aρg(l - y)^{2} sin^{2} 45^{o} + Aρg(l + y)^{2} sin^{2} 45^{o}

Change in PE = (PE)_{f} - (PE)_{1}

_{}

_{Change in total energy = 0}

Change in total energy = 0

Differentiating both sides w.r.t time

Aρg[0 + 2y(dy/dt)] + 2Aρlyy = 0

2Aρgy + 2Aρly = 0

ly + gy = 0

y + (g/l)y = 0

*ω*^{2} = g/l

*ω = *√g/l

T = 2π√l/g