SHM of liquid

Classical Mechanics Level 5

A V-shaped tube attached to the ground with its vertex. It's one leg makes an angle of $30 ^\circ$ , and the other leg makes an angle of $60 ^\circ$ with the horizontal. The tube is filled with a liquid in a total length of $1\text{ m}$ . Now the liquid at one side is displaced a little.Find the Time - period of SHM.

Your answer can be represented as $\sqrt{\dfrac{a \pi^b}{c (\sqrt{d} + 1)}} \; ,$

where $a,b,c$ and $d$ are positive integers with $a,c$ coprime and $d$ square-freee.

Enter your answer as $a + b + c + d$ .

Details and Assumptions :

Neglect surface tension , viscuous force and friction.
$g = 10\text{ m/s}^2$ .

This is a part of my set: Aniket's Mechanics Challenges .

The answer is 14.

2 solutions

Abhi Kumbale
Nov 23, 2016

Same! Let me write a website-version for this.

Kelvin Hong - 3 years, 9 months ago

Kelvin Hong
Aug 20, 2017

Ready for a and b:

$a+b=1$

$a=\sqrt3 b$

So, $a=\frac{3-\sqrt3}{2}$ , $b=\frac{\sqrt3-1}{2}$ .

By calculating and arranging energy,

$K.E. = \frac{1}{2}m \dot{x}^2$

$P.E. = \frac{1}{4} mg [\sqrt3(x-b)^2+(x+a)^2]$

So , Lagrangian will be

$L = K.E.-P.E. =\frac{1}{2}m \dot{x}^2-\frac{1}{4} mg [\sqrt3(x-b)^2+(x+a)^2]$

Computing Lagrangian Equation:

$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$ , $\frac{\partial L}{\partial x}=-\frac{1}{4}mg[2\sqrt3 (x-b)+2(x+a)]$

$\frac{d}{dt} (\frac{\partial L}{\partial \dot{x}})=\frac{\partial L}{\partial x}$

Get

$a=\ddot{x} = -\frac{2\sqrt{3}+2}{4}gx$

By the knowledge of SHM, the coefficient of $x$ is $-\omega ^2$ .

So $\omega =\sqrt{\frac{\sqrt3+1}{2}g}$

$T=\frac{2\pi}{\omega}=\sqrt{\frac{4\pi^2}{5(\sqrt3+1)}}$

$a+b+c+d=4+2+5+3=\boxed{14}$