A piece of wood has length,breadth and height | vedclass

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(D) The time period of $S.H.M.$ for small vertical oscillations of a floating body is given by $T = 2 \pi \sqrt{\frac{h'}{g}}$,where $h'$ is the depth

Vedclass · Accepted Answer

$2 \pi \sqrt{\frac{ad}{g}}$

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(D) The time period of $S.H.M.$ for small vertical oscillations of a floating body is given by $T = 2 \pi \sqrt{\frac{h'}{g}}$,where $h'$ is the depth of the object submerged in the liquid.For a floating body,the weight of the object equals the weight of the displaced liquid.Let $V$ be the volume of the wood,$V = a 	imes b 	imes c$.Mass of wood $= V 	imes d 	imes ho_w = (abc) 	imes d 	imes ho_w$ (where $ho_w$ is the density of water).Volume of displaced water $V_{disp} = b 	imes c 	imes h'$.Weight of displaced water $= (bc h') 	imes ho_w 	imes g$.Equating the two: $(abc) 	imes d 	imes ho_w 	imes g = (bc h') 	imes ho_w 	imes g$.Solving for $h'$,we get $h' = ad$.Substituting this into the time period formula: $T = 2 \pi \sqrt{\frac{ad}{g}}$.

$A$ piece of wood has length,breadth and height,'$a$','$b$' and '$c$' respectively. Its relative density is '$d$'. It is floating in water such that the side '$a$' is vertical. It is pushed down a little and released. The time period of $S.H.M.$ executed by it is ($g=$ acceleration due to gravity).

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