A hydrometer executes simple harmonic motion | vedclass

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(A) When the hydrometer is floating in equilibrium,its weight is balanced by the buoyant force.When it is pushed down by a small distance $x$,the addi

Vedclass · Accepted Answer

$T = 2 \pi \sqrt{\frac{m}{\pi r^2 \rho g}}$

Vedclass · Answer

(A) When the hydrometer is floating in equilibrium,its weight is balanced by the buoyant force.When it is pushed down by a small distance $x$,the additional buoyant force acts as the restoring force.The additional volume submerged is $V = \pi r^2 x$.The additional buoyant force is $F = \rho V g = \rho (\pi r^2 x) g$.Since this force acts in the direction opposite to the displacement,the restoring force is $F_{restoring} = -\rho \pi r^2 g x$.Using Newton's second law,$m a = -\rho \pi r^2 g x$,which gives $a = -(\frac{\rho \pi r^2 g}{m}) x$.Comparing this with the standard $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = \frac{\rho \pi r^2 g}{m}$.The time period $T$ is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{m}{\rho \pi r^2 g}}$.

$A$ hydrometer executes simple harmonic motion when it is pushed down vertically in a liquid of density $\rho$. If the mass of the hydrometer is $m$ and the radius of the hydrometer tube is $r$,then the time period of oscillation is

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