AU -tube of uniform bore is placed with arms vertical. The length of the liquid in the two arms of the U-tube is L. The time period T of the oscillation of the liquid column when it is displaced by y is

Question

AU -tube of uniform bore is placed with arms vertical. The length of the liquid in the two arms of the U-tube is L. The time period T of the oscillation of the liquid column when it is displaced by y is (A) T=2 π √(L/g) (B) T=2 π √(2 L/g) (C) T=2 π √(L/2 g) (D) T=2 π √(y/g)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a3891e47d0a9286dd26d4868a7037868-.png" /> (d2 y/d t2)=-(F/m)=-((2 y A ρ g/L A ρ)) ⇒ (d2 y/d t2)=-((2 g/L)) y or ω2=(2 g/L) ⇒ T=2 π √(L/2 g)

Q. AU -tube of uniform bore is placed with arms vertical. The length of the liquid in the two arms of the U-tube is L. The time period T of the oscillation of the liquid column when it is displaced by y is

$T = 2 π \frac{L}{g}$

$T = 2 π \frac{2 L}{g}$

$T = 2 π \frac{L}{2 g}$

$T = 2 π \frac{y}{g}$

$\frac{d ^{2} y}{d t ^{2}} = - \frac{F}{m} = - (\frac{2 y A ρ g}{L A ρ})$
$\Rightarrow \frac{d ^{2} y}{d t ^{2}} = - (\frac{2 g}{L}) y$
or $ω^{2} = \frac{2 g}{L} \Rightarrow T = 2 π \frac{L}{2 g}$

Q. AU -tube of uniform bore is placed with arms vertical. The length of the liquid in the two arms of the U-tube is L. The time period T of the oscillation of the liquid column when it is displaced by y is

T=2πgL​​

T=2πg2L​​

T=2π2gL​​

T=2πgy​​

dt2d2y​=−mF​=−(LAρ2yAρg​) ⇒dt2d2y​=−(L2g​)y or ω2=L2g​⇒T=2π2gL​​

$T = 2 π \frac{L}{g}$

$T = 2 π \frac{2 L}{g}$

$T = 2 π \frac{L}{2 g}$

$T = 2 π \frac{y}{g}$

$\frac{d ^{2} y}{d t ^{2}} = - \frac{F}{m} = - (\frac{2 y A ρ g}{L A ρ})$
$\Rightarrow \frac{d ^{2} y}{d t ^{2}} = - (\frac{2 g}{L}) y$
or $ω^{2} = \frac{2 g}{L} \Rightarrow T = 2 π \frac{L}{2 g}$