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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» <html><body><p></p>Solution :Figure shown as below <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P2_C14_E04_040_S01.png" width="80%"/> <br/> Let <a href="https://interviewquestions.tuteehub.com/tag/small-1212368" style="font-weight:bold;" target="_blank" title="Click to know more about SMALL">SMALL</a> liquid column of length dx is in the left of tube at x height from horizontal <br/> Potential energy due to dx element, <br/> `d(PE)= dmgx""[PE=" from mgh "]` <br/> `=p V gx ""[therefore m= pV" and "V= A dx]` <br/> `=p A dx gx` <br/> `= pA g x dx` <br/> <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> potential energy of liquid in left column <br/> `P.E= int_(0)^(h_1) Ap gx"" dx` <br/> `=A pg int_(0)^(h_1) x dx` <br/> `=A pg[(x^2)/(2)]_(0)^(h_1)= (A pg h_(1)^(2))/(2)` <br/> From figure `h_(1)= l sin 45^(@)= (l)/(<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(2))` <br/> `therefore P.E = (A p gl^(2))/(4)` <br/> Total potential energy of liquid in V-tube <br/> `P.E= (A pg l^(2))/(4)+ (Apg l^(2))/(4)` <br/> Initial `P.E. = (A pg l^(2))/(4)"""......."(1)` <br/> When the <a href="https://interviewquestions.tuteehub.com/tag/suction-1233369" style="font-weight:bold;" target="_blank" title="Click to know more about SUCTION">SUCTION</a> pump is removed and due to pressure difference, let element moves towards right side by y unit. <br/> Then the liquid column in left arm `=(l-y)` <br/> And the liquid in right arm `=l+y` <br/> Total potential energy `=Ap g(l-y)^(2) sin^(2) 45^(@)+ Ap g (l+y)^(2) sin^(2)45^(@)` <br/> Final `P.E. = (A pg (l-y)^(2))/(2)+ (Ap g(l+y)^(2))/(2)"""......"(2)` <br/> Potential energy difference <br/> `triangle PE =` final P.E. - initial P.E. <br/> `=(A pg)/(2)[(l-y)^(2)+(l+y)^(2)-l^(2)]""` [From equation (1) and (2) ] <br/> `=(Apg )/(2) [2l^(2)+2y^(2)-l^(2)]` <br/> `=(A pg )/(2) [l^(2)+2y^(2)]"""........"(3)` <br/> If change in velocity v of total liquid column <br/> `triangle KE= (1)/(2)mv^(2)""[" where "m= pV= pA xx 2l]` <br/> `triangle KE = Ap lv^(2)"""......."(4)` <br/> From law of conservation of energy, <br/> `triangle PE + triangle KE= 0` <br/> `A pg (l^(2)+2y^(2))+ <a href="https://interviewquestions.tuteehub.com/tag/apl-363630" style="font-weight:bold;" target="_blank" title="Click to know more about APL">APL</a> v^(2) = 0"""........."(5)` <br/> Difference w.r.t time .t. <br/> `Apg [(d)/(dt)(l^(2)+2y^(2))]+ Apl (d)/(dt)(v^2)=0` <br/> `Apg (0+2y(dy)/(dt))+Apl (2v(dv)/(dt))=0` <br/> `2A pg y v + Apl (2v)a=0` <br/> Dividing by `2Apv`, <br/> `gy+la=0""[therefore 2Apv ne 0]` <br/> `therefore a= -(g)/(l)y`, comparing with this to <br/> `(d^(2)y)/(dt^(2))= -omega^(2)y` <br/> `therefore omega = sqrt((g)/(l))` <br/> `therefore (2pi)/(T)= sqrt((g)/(l))` <br/> `therefore T= 2pi sqrt((l)/(g))` is time period.</body></html>


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