Question

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.

Answer 1

Consider the diagram shown below

Let us consider an infinitesimal liquid column of length dx at a height x from the horizontal line.

If ρ = density of the liquid

A = cross-sectional area of V-tube

Pe of element dx will be given as PE = dmgx = (Aρdx)gx  ......[∵ dm = pV = ρAdx]

Where Aρdx = dm = mass of element dx

∴ Total PE of the left column = `int_0^(h_1) Aρgxdx`

= `Aρg int_0^(h_1) x  dx`

= `Aρg |x^2/2|_0^(h_1)`

= `Aρg  h_1^2/2`

But, `h_1 = l sin 45^circ`

∴ PE = `(Aρg)/2 l^2 sin^2 45^circ`  ......(i)

In a similar way,

PE of right column = `(Aρg)/2 l^2 sin^2 45^circ`  ......(ii)

Total PE = `(Aρg)/2 l^2 sin^2 45^circ + (Aρg)/2 l^2 sin^2 45^circ`

= `2 xx 1/2 Aρgl^2 (1/sqrt(2))^2`

= `(Aρgl^2)/2`  ......(iii)

If due to the pressure difference created y element of the left side moves on the right side, then liquid present in the left arm = l – y

Liquid present in the right arm = l + y

∴ Total PE = `Aρg(l - y)^2 sin62 45^circ + Aρg (l + y)^2 sin^2 45^circ`

Changes in PE = `(PE)_"final" - (PE)_"initial"`

or `ΔPE = (Aρg)/2 [(l - y)^2 + (l + y)^2 - l^2]`

= `(Aρg)/2 [l^2 + y^2 - 2ly + l^2 + y^2 + 2ly - l^2]`

= `(Aρg)/2 [2(l^2 + y^2)]`

= `Aρg(l^2 + y^2)`  ........(iv)

If v is the change in velocity of the total liquid column, then the change in KE

`ΔKE = 1/2mv^2`

But `m = Aρ(2l)`

∴ `ΔKE = 1/2 Aρ2lv^2 = Aρlv^2`  ......(v)

From equations (iv) and (v),

  `ΔPE + ΔKE = Aρg(l^2 + y^2) + Aρlv^2`  ......(vi)

The system is conservative.

∴ Change in total energy = 0

From equation (vi),

`Aρg(l^2 + y^2) + Aρlv^2` = 0

Differentiating both sides with respect to time (t), we get

`Aρg [0 + 2y (dy)/(dt)] + Aρl(2v) (dv)/(dt)` = 0

But, `(dy)/(dt) = v` and `(dv)/(dt) = a`  ......[Acceleration]

⇒ `Aρg (2yv) + Aρl(2v)a` = 0

⇒ `(gy + la) 2Aρv` = 0

2Aρv = constant and 2Aρv ≠ 0

∴ `la + gy` = 0

`a + (g/l)y` = 0

or `(d^2y)/(dt^2) + (g/l)y` = 0

This is the standard differential equation for SHM of the form

`(d^2y)/(dt^2) + ω^2y` = 0

∴ `ω = sqrt(g/l)`

∴ T = `(2pi)/ω = 2pi sqrt(l/g)`