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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. |
Answer» Solution :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. |
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