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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° 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.

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Consider the liquid in the length dx. It’s mass is Ardx at a height x.

PE = Aρdx gx

The PE of the left column

similarly, P.E. of the right column=

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

Total P.E. = Aρgh2 = Aρgl2 sin2 45o = Aρgl2/2

If the chamge in liquid level along the tube in left side in y, then length of the liquid in left side is l-y and in the right side is l+y.

Total P.E. Aρg(l - y)2 sin2 45o + Aρg(l + y)2 sin2 45o

Change in PE = (PE)f - (PE)1 

Change in total energy = 0

Change in total energy = 0

Differentiating both sides w.r.t time

Aρg[0 + 2y(dy/dt)] + 2Aρlyy = 0

2Aρgy + 2Aρly = 0

ly + gy = 0

y + (g/l)y = 0

ω2 = g/l

ω = √g/l

T = 2π√l/g

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