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(A) Let the first cylinder be displaced downwards by a distance $x$. Then the second cylinder will be displaced upwards by the same distance $x$ becau

Vedclass · Accepted Answer

$T = 2\pi \sqrt {\frac{{2v}}{{gs}}\,\frac{{{\rho _0}}}{{\left( {{\rho _1} + {\rho _2}} \right)}}} $

Vedclass · Answer

(A) Let the first cylinder be displaced downwards by a distance $x$. Then the second cylinder will be displaced upwards by the same distance $x$ because they are connected by a string over pulleys.The change in the buoyant force on the first cylinder is $\Delta F_1 = \rho_1 s x g$ (upward force decreases as it moves down,but here the net restoring force is considered).When the first cylinder moves down by $x$,the buoyant force on it increases by $\rho_1 s x g$ (acting upwards). When the second cylinder moves up by $x$,the buoyant force on it decreases by $\rho_2 s x g$ (acting upwards).The net restoring force on the system is $F_{net} = -(\rho_1 s x g + \rho_2 s x g) = -(\rho_1 + \rho_2) s g x$.The total mass of the system is $2m = 2\rho_0 v$.Using Newton's second law,$2m a = F_{net} \implies 2\rho_0 v a = -(\rho_1 + \rho_2) s g x$.$a = -\left[ \frac{(\rho_1 + \rho_2) s g}{2 \rho_0 v} \right] x$.Comparing this with the $SHM$ equation $a = -\omega^2 x$,we get $\omega = \sqrt{\frac{(\rho_1 + \rho_2) s g}{2 \rho_0 v}}$.The time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{2 \rho_0 v}{(\rho_1 + \rho_2) s g}}$.Since $m = \rho_0 v$,we have $T = 2\pi \sqrt{\frac{2m}{(\rho_1 + \rho_2) s g}}$. Substituting $m = \rho_0 v$,we get $T = 2\pi \sqrt{\frac{2 \rho_0 v}{(\rho_1 + \rho_2) s g}}$.

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