A wooden cube (density of wood $d$) of slide $l$ floats in a liquid of density $\rho$ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, its performs simple harmonics motion of period $T$, then $T$ is equal

Question

Dr. Abubacker · Accepted Answer

(b) Let at any instant, cube is at a depth $x$ from the equilibrium position then net force acting on the cube = upthrust on the portion of length $x$
$F = - \rho l^{2}xg = - \rho l^{2}gx$ …(i)
Negative sign shows that, force is opposite to $x$. Hence equation of SHM
$F = - kx$ …(ii)
Comparing Eqs. (i) and (ii)
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$k = \rho l^{2}g$
$T = 2\pi\sqrt{\frac{m}{k}}$
$= 2\pi\sqrt{\frac{l^{3}d}{\rho l^{2}g}} = 2\pi\sqrt{\frac{ld}{\rho g}}$

Q. A wooden cube (density of wood $d$) of slide $l$ floats in a liquid of density $\rho$ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, its performs simple harmonics motion of period $T$, then $T$ is equal

Explanation

Reviewed by Dr. Abubacker

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