First slide

Simple hormonic motion

Question

A brass cube of side a and density $σ$ is floating in mercury of density $ρ$ . If the cube is displaced a bit vertically, it executes S.H.M. Its time period will be

Moderate

A

$2 π \sqrt{\frac{σa}{ρg}}$
B

$2 π \sqrt{\frac{ρa}{σg}}$
C

$2 π \sqrt{\frac{ρg}{σa}}$
D

$2 π \sqrt{\frac{σg}{ρa}}$

Solution

As a is the side of cube $σ$ is its density.
Mass of cube is $a^{2}$ $σ$ its weight = $a^{3} σg$
Let h be the height of cube immersed in liquid of density $ρ$ in equilibrium then, $F = a^{2} hρg = Mg = a^{3} σg$
If it is pushed down by y then the buoyant force
$F^{'} = a^{2} (h + y) ρg$
Restoring force is $∆ F = F^{'} - F = a^{2} (h + y) σg - a^{2} hσg$
= $a^{2} yρg$
Restoring acceleration = $\frac{∆ F}{M} = - \frac{a^{2} yρg}{M} = - \frac{a^{2} ρg}{a^{2} σ} y$
Motion is S.H.M.
$\Rightarrow T = 2 π \sqrt{\frac{a^{3} σ}{a^{2} ρg}} = 2 π \sqrt{\frac{aσ}{ρg}}$

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