One end of 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 horizental 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.

Explanation

Figure shown as below

Let small liquid column of length dx is in the left of tube at x height from horizontal
Potential energy due to dx element,
`d(PE)= dmgx" "[PE=" from mgh "]`
`=p V gx " "[therefore m= pV" and "V= A dx]`
`=p A dx gx`
`= pA g x dx`
Total potential energy of liquid in left column
`P.E= int_(0)^(h_1) Ap gx"" dx`
`=A pg int_(0)^(h_1) x dx`
`=A pg[(x^2)/(2)]_(0)^(h_1)= (A pg h_(1)^(2))/(2)`
From figure `h_(1)= l sin 45^(@)= (l)/(sqrt(2))`
`therefore P.E = (A p gl^(2))/(4)`
Total potential energy of liquid in V-tube
`P.E= (A pg l^(2))/(4)+ (Apg l^(2))/(4)`
Initial `P.E. = (A pg l^(2))/(4)" ""......."(1)`
When the suction pump is removed and due to pressure difference, let element moves towards right side by y unit.
Then the liquid column in left arm `=(l-y)`
And the liquid in right arm `=l+y`
Total potential energy `=Ap g(l-y)^(2) sin^(2) 45^(@)+ Ap g (l+y)^(2) sin^(2)45^(@)`
Final `P.E. = (A pg (l-y)^(2))/(2)+ (Ap g(l+y)^(2))/(2)" ""......"(2)`
Potential energy difference
`triangle PE =` final P.E. - initial P.E.
`=(A pg)/(2)[(l-y)^(2)+(l+y)^(2)-l^(2)]" "` [From equation (1) and (2) ]
`=(Apg )/(2) [2l^(2)+2y^(2)-l^(2)]`
`=(A pg )/(2) [l^(2)+2y^(2)]" ""........"(3)`
If change in velocity v of total liquid column
`triangle KE= (1)/(2)mv^(2)" "[" where "m= pV= pA xx 2l]`
`triangle KE = Ap lv^(2)" ""......."(4)`
From law of conservation of energy,
`triangle PE + triangle KE= 0`
`A pg (l^(2)+2y^(2))+ Apl v^(2) = 0" ""........."(5)`
Difference w.r.t time .t.
`Apg [(d)/(dt)(l^(2)+2y^(2))]+ Apl (d)/(dt)(v^2)=0`
`Apg (0+2y(dy)/(dt))+Apl (2v(dv)/(dt))=0`
`2A pg y v + Apl (2v)a=0`
Dividing by `2Apv`,
`gy+la=0" "[therefore 2Apv ne 0]`
`therefore a= -(g)/(l)y`, comparing with this to
`(d^(2)y)/(dt^(2))= -omega^(2)y`
`therefore omega = sqrt((g)/(l))`
`therefore (2pi)/(T)= sqrt((g)/(l))`
`therefore T= 2pi sqrt((l)/(g))` is time period.