One end of a V-tube containing mercury is connected to a suction pump

1.	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 45º 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.
Answer» Let the liquid in the length dx is at a height x. Its mass = Aρdx PE = (Aρdx)dx The PE of the left column = \(\int^{h_1}_0\) gxdx =A \({[\frac{x^2}{2}]}^{h_1}_0\) = Aρg\(\frac{l^2sin^2\,45^o}{2}\) h₁ = h₂ = l sin 45º where l = length in one arm of the tube. Total PE = Aρgl²sin²45º = \(\frac{Aρgl^2}{2}\) If the change in liquid level in the tube in left side is y, Then length of liquid on left side = l − y And right side = l + y Total PE = Aρg(l − y)² sin²45º + Aρg(l + y)² sin²45º Change in PE = (PE)_F − (PE)_i =\(\frac{Aρg}{2}\) [(l − y)² + (l + y)² − l² ] = \(\frac{Aρg}{2}\)[l² + y² - 2ly + l² + y² + 2ly - l² ] = \(\frac{Aρg}{2}\)[y² + l²] Change in KE = \(\frac{1}{2}\)Aρ2ly² Change in total energy = 0 ∆(PE) + ∆(KE) = 0 \(\frac{Aρg}{2}\) [2y² + l²] + Aρly² = 0 Differentiating both wrt time Aρg[0 + yẏ] + 2aρlẏÿ = 0 2Aρg + 2Aρglÿ = 0 lÿ + gy = 0 Ÿ + \(\frac{g}{l}\)y = 0 Or ÿ = − \(\frac{g}{l}\)y ∴ w² = \(\frac{g}{l}\) or w = \(\sqrt{\frac{g}{l}}\) or T = \(2\pi\sqrt{\frac{l}{g}}\)

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 45º 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.

Answer»

Let the liquid in the length dx is at a height x.

Its mass = Aρdx

PE = (Aρdx)dx

The PE of the left column = \(\int^{h_1}_0\) gxdx

=A \({[\frac{x^2}{2}]}^{h_1}_0\) = Aρg\(\frac{l^2sin^2\,45^o}{2}\)

h₁ = h₂ = l sin 45º where l = length in one arm of the tube.

Total PE = Aρgl²sin²45º = \(\frac{Aρgl^2}{2}\)

If the change in liquid level in the tube in left side is y,

Then length of liquid on left side = l − y

And right side = l + y

Total PE = Aρg(l − y)² sin²45º + Aρg(l + y)² sin²45º

Change in PE = (PE)_F − (PE)_i =\(\frac{Aρg}{2}\) [(l − y)² + (l + y)² − l² ]

= \(\frac{Aρg}{2}\)[l² + y² - 2ly + l² + y² + 2ly - l² ] = \(\frac{Aρg}{2}\)[y² + l²]

Change in KE = \(\frac{1}{2}\)Aρ2ly²

Change in total energy = 0

∆(PE) + ∆(KE) = 0

\(\frac{Aρg}{2}\) [2y² + l²] + Aρly² = 0

Differentiating both wrt time

Aρg[0 + yẏ] + 2aρlẏÿ = 0

2Aρg + 2Aρglÿ = 0

lÿ + gy = 0

Ÿ + \(\frac{g}{l}\)y = 0

Or ÿ = − \(\frac{g}{l}\)y

∴ w² = \(\frac{g}{l}\) or w = \(\sqrt{\frac{g}{l}}\) or T = \(2\pi\sqrt{\frac{l}{g}}\)