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Question-36855




Question Number 36855 by Tinkutara last updated on 06/Jun/18
Commented by Tinkutara last updated on 06/Jun/18
Answered by ajfour last updated on 11/Jun/18
(τ_A /τ_B ) = (6/( (√(11)))) .
$$\frac{\tau_{{A}} }{\tau_{{B}} }\:=\:\frac{\mathrm{6}}{\:\sqrt{\mathrm{11}}}\:. \\ $$
Commented by Tinkutara last updated on 11/Jun/18
How?
Commented by ajfour last updated on 11/Jun/18
Commented by ajfour last updated on 11/Jun/18
(τ_A /τ_B )=(((D/(v+u))+(D/(v−u)))/((2D)/( (√(v^2 −u^2 )))))   with v=1.2u=((6u)/5)  v+u=((11u)/5)  ;  v−u=(u/5) ;   (√(v^2 −u^2 )) = ((√(11))/5)  (τ_A /τ_B ) = (((1/(11))+1)/(2/( (√(11))))) = (6/( (√(11)))) .
$$\frac{\tau_{{A}} }{\tau_{{B}} }=\frac{\frac{{D}}{{v}+{u}}+\frac{{D}}{{v}−{u}}}{\frac{\mathrm{2}{D}}{\:\sqrt{{v}^{\mathrm{2}} −{u}^{\mathrm{2}} }}}\: \\ $$$${with}\:{v}=\mathrm{1}.\mathrm{2}{u}=\frac{\mathrm{6}{u}}{\mathrm{5}} \\ $$$${v}+{u}=\frac{\mathrm{11}{u}}{\mathrm{5}}\:\:;\:\:{v}−{u}=\frac{{u}}{\mathrm{5}}\:;\: \\ $$$$\sqrt{{v}^{\mathrm{2}} −{u}^{\mathrm{2}} }\:=\:\frac{\sqrt{\mathrm{11}}}{\mathrm{5}} \\ $$$$\frac{\tau_{{A}} }{\tau_{{B}} }\:=\:\frac{\frac{\mathrm{1}}{\mathrm{11}}+\mathrm{1}}{\frac{\mathrm{2}}{\:\sqrt{\mathrm{11}}}}\:=\:\frac{\mathrm{6}}{\:\sqrt{\mathrm{11}}}\:. \\ $$
Commented by Tinkutara last updated on 14/Jun/18
Thanks Sir!

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