One end of a $V$-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of $45^{\circ}$ each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in $V$-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

Options:

None

Solution:
Let consider the diagram shown below
Let the liquid column in both columns are at $h_0$ height initial. Now due to pressure difference the liquid column $A$ arm pressed by $x$ and in arm $B$ lifts by $x$. So diffrence in vertical height between two levels $=2 x$.
Consider at an infinitesimal liquid column of length $d x$ at a height $x$ from the horizontal line.
If $\rho=$ density of the liquid
$A=$ cross-sectional area of $V$-tube
PE of element $d x$ will be given as
$$
\mathrm{PE}=(d m) g x=(A \rho d x) g x \quad[\because d m=\rho V=\rho A d x]
$$
where $A \rho d x=d m=$ element mass
$\therefore$ Total PE of the left column
$$
\begin{aligned}
\text { P.E. } &=\int_0^{h_1} A \rho g x d x \\
&=A \rho g \int_0^{h_1} x d x \\
&=A \rho g\left[\frac{x^2}{2}\right]_0^{h_1}=A \rho g \frac{h_1^2}{2}
\end{aligned}
$$
From figure, $h_1=l \sin 45^{\circ}$
$$
\mathrm{PE}=\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ}
$$
In a similar way,
PE of right column $=\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ}$
Total PE $=\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ}+\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ}$
$\therefore$ Total potential energy
$$
=2 \times \frac{1}{2} A_{\rho g} l^2\left(\frac{1}{\sqrt{2}}\right)^2=\frac{A \rho g l^2}{2}
$$
Due to pressure difference created $y$ element of left side moves on the right side,
then liquid present in the left arm $=l-y$
and liquid present in the right $\mathrm{arm}=l+y$
P.E. of liquid column in left arm
$$
=A \rho g(l-y)^2 \sin ^2 45^{\circ}
$$
P.E. of liquid column in right arm
$$
\begin{aligned}
&=A \rho g(l+y)^2 \sin ^2 45^{\circ} \\
&\therefore \text { Then Total Potential Energy } \\
&=A \rho g(l+y)^2 \sin ^2 45^{\circ}+A \rho g(l-y)^2 \sin ^2 45^{\circ}
\end{aligned}
$$
Changes in $\mathrm{PE}=(\mathrm{PE})_{\text {final }}-(\mathrm{PE})_{\text {initial }}$ or
$$
\begin{aligned}
\triangle P E &=\frac{A \rho g}{2}\left[(l-y)^2+(l+y)^2-l^2\right] \\
&=\frac{A \rho g}{2}\left[l^2+y^2-2 l y+l^2+y^2+2 l y-l^2\right] \\
&=\frac{A \rho g}{2}\left[2\left(l^2+y^2\right)\right]=A \rho g\left(l^2+y^2\right)
\end{aligned}
$$
If change in K.E. (v) of the total liquid column, then change in $\mathrm{KE}$
$$
\Delta \mathrm{KE}=\frac{1}{2} m v^2
$$
So, $m=A \rho(2 l)$
$$
\therefore \Delta \mathrm{KE}=\frac{1}{2}(A \rho 2 l)^2=A \rho l v^2
$$
From Eqs. (iv) and (v),
Chance in total energy,
$$
\Delta \mathrm{PE}+\Delta \mathrm{KE}=A \rho g\left(l^2+y^2\right)+A \rho l v^2
$$
System being conservative.
$\therefore$ Change in total energy $=0$
From Eq. (vi),
$$
A \rho g\left(l^2+y^2\right)+A \rho l v^2=0
$$
Differentiating both sides with respect to time $(t)$, then,
$$
A \rho g\left[0+2 y \frac{d y}{d t}\right]+A \rho l(2 v) \frac{d v}{d t}=0
$$
$$
\begin{aligned}
&\text { But, } \frac{d y}{d t}=v \text { and } \frac{d v}{d t}=a \quad \text { [acceleration] } \\
&\Rightarrow A \rho g(2 y v)+A \rho l(2 v) a=0 \\
&\Rightarrow(g y+l a) 2 A \rho v=0 \\
&2 A \rho v=\text { constant and } 2 A \rho v \neq 0 \\
&\therefore l a+g y=0 \\
&a+\left(\frac{g}{l}\right) y=0 \\
&\text { or } \frac{d^2 y}{d t^2}+\left(\frac{g}{l}\right) y=0
\end{aligned}
$$
It is equation of SHM oscillator and the standard differential equation for SHM of the form is
$$
\begin{aligned}
&\frac{d^2 y}{d t^2}+\omega^2 y=0 \\
&\omega=\sqrt{\frac{g}{l}} \\
&\text { So, } \frac{2 \pi}{T}=\sqrt{\frac{g}{l}} \quad\left(\because \omega=\frac{2 \pi}{T}\right) \\
&T=2 \pi \sqrt{\frac{l}{g}}
\end{aligned}
$$

Topic: Physics, Oscillations