Simple Harmonic Motion
14.0 Simple Harmonic Oscillation of a Fluid Column
14.0 Simple Harmonic Oscillation of a Fluid Column
1. Fluid in Inclined Column
Initially the liquid in both the sides will be at the same height as shown in fig SHM 28 (a).
Now after displacing the liquid by a distance $x$, as shown in fig SHM 28 (b).
Due to the difference in the level of liquid in column, the restoring force arises and is given as,
$$\begin{equation} \begin{aligned} F = - \rho gA\left( {{h_1} + {h_2}} \right)\quad ..\left( i \right) \\ {h_1} = xsin{\theta _1}\quad ..\left( {ii} \right) \\ {h_2} = xsin{\theta _2}\quad ..\left( {iii} \right) \\\end{aligned} \end{equation} $$
Putting equation $(ii)$ & $(iii)$ in $(i)$ we get,
$$\begin{equation} \begin{aligned} F = - \rho gA\left( {sin{\theta _1} + sin{\theta _2}} \right)x \\ ma = - \rho gA\left( {sin{\theta _1} + sin{\theta _2}} \right)x \\ a = \frac{{ - \rho gA\left( {sin{\theta _1} + sin{\theta _2}} \right)x}}{m}\quad ..\left( {iv} \right) \\ a = - {\omega ^2}x\left( {SHMacceleration} \right)\quad ..\left( v \right) \\\end{aligned} \end{equation} $$
From equation $(iv)$ & $(v)$ we get,
$$\begin{equation} \begin{aligned} \omega = \sqrt {\frac{{\rho gA\left( {sin{\theta _1} + sin{\theta _2}} \right)}}{m}} \\ \therefore T = 2\pi \sqrt {\frac{m}{{\rho gA\left( {sin{\theta _1} + sin{\theta _2}} \right)}}} \\\end{aligned} \end{equation} $$
2. Fluid in Vertical Column
By considering the equation given below, we can find out the time period of a vertical liquid column.
$$T = 2\pi \sqrt {\frac{m}{{\rho gA\left( {sin{\theta _1} + sin{\theta _2}} \right)}}} $$
For vertical liquid column,
$$\begin{equation} \begin{aligned} {\theta _1} = 90^\circ \\ {\theta _2} = 90^\circ \\ m = 2\left( {\rho Al} \right) \\\end{aligned} \end{equation} $$
So, the equation is reduces to, $$T = 2\pi \sqrt {\frac{l}{g}} $$
