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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.

Answers (1)

Let initial height = h₀ of liquid in both columns; due to pressure difference, if the liquid in arm A is pressed by x, then liquid in arm B will rise by x.

Let us consider dx as an element of height, then,

Mass $dm = A.dx.\rho,$ where, Area of a cross-section of tube = A

PE of left dm element columns = (dm) gh

& PE of dm in left columns = $dm = A.dx.\rho gx$

Thus, total PE in left column = $\int_{0}^{h}A\rho gx dx=A\rho g\left [ \frac{x^{2}}{2} \right ]^{h_{0}}_{0}$

$= A\rho g\frac{ h_{1}^{2}}{2}$

From the figure

$\sin 45^{\circ} = \frac{h_{1}}{l}$

$h_{1} = h_{2} = l \sin 45^{\circ} =\frac{1}{\sqrt{2}}$

$h_{1}^{2} = h_{2}^{2} =\frac{ l^{2}}{2}$

Thus, PE in left column = PE in right column = $A\rho g \frac{ l^{2}}{4}$

Thus, total potential energy = $A\rho g \frac{ l^{2}}{2}$

Let the element move towards right by y units due to the pressure difference, then,

Liquid column in left arm = (l – y)

Liquid column in right arm = (l + y)

PE in left arm = $A\rho g (l - y)^{2} \sin^{2} 45^{\circ}$

PE in right arm = $A\rho g (l + y)^{2} \sin^{2} 45^{\circ}$

Total $PE = A\rho g \left ( \frac{1}{2} \right )^{2} [(l-y)^{2} + (l + y)^{2}]$

Final $PE =\frac{ A\rho g}{\sqrt{2}} [l^{2} + y^{2} - 2ly + l^{2} + y^{2} + 2ly]$

$=\frac{ A\rho g}{2} (2l^{2}+2y^{2})$

Change PE = Final PE – Initial PE

$=\frac{ A\rho g}{2} (2l^{2}+2y^{2})-A\rho g \frac{l^{2}}{2}$

$=\frac{ A\rho g}{2} (2l^{2}+2y^{2})$

If there is change in velocity(v) of liquid column,

$\Delta KE = \frac{1}{2}mv^{2}$

$m = (A.2l) \rho$

Thus, $\Delta KE = A\rho lv^{2}$

Now, change in total energy = $\frac{A\rho g}{2 }(l^{2} + 2y^{2}) + - A\rho lv^{2}$

Total change in energy,

$\Delta PE + \Delta KE = 0$

Thus, $\frac{A\rho g}{2 }(l^{2} + 2y^{2}) + - A\rho lv^{2} =0$

$\frac{A\rho}{2} [g(l^{2} + 2y^{2}) + 2lv^{2}] = 0$

$\frac{A\rho}{2}$ is not equal to 0, thus,

$[g(l^{2} + 2y^{2}) + 2lv^{2}] = 0$

Now, let us differentiate w.r.t. t,

$g \left [0 + 2 \times 2\times2 \frac{dy}{dt} \right ] + 2l.2v.\frac{dv}{dt} = 0$

$\frac{d^{2}y}{dt^{2}} + \frac{g}{l} .y= 0$

Since 4v is not equal to 0,

$\frac{d^{2}y}{dt^{2}} + \omega ^{2}y= 0$

Thus, $\omega ^{2}=\frac{g}{l}$

$\frac{2\pi }{T}=\sqrt{ \frac{g}{l}}$

Thus,

$T=2\pi\sqrt{ \frac{l}{g}}$

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