Question

A wooden cube (density of wood ''d'') of side'' l '' floats in a liquid of density p with its upper and lower surfaces horizontal .If the cube is pushed slightly down and released ,it performs simple harmonic motion of period T. Find the expression of T dimensionally.

Answer 1

Let at any instant, cube be at a depth x from the equilibrium position, then Net force acting on the cube = Upthrust on the portion of length x Negative sign shows that force is opposite to x . Hence, equation of SHM F =- kx ......(ii)

Answer 2

Given that,

Side of cube = l

Density of wood = d

Density of liquid = ρ

Let at instant, the cube be at a depth x from the equilibrium position, then the net force acting on the cube

We need to calculate the force

force acting on the cube = upthrust on the portion of length x

$F=-\rho l^2 xg$....(I)

We know that,

The equation of simple harmonic motion

The restoring force is directly proportional to the distance.

$F\propto(-x)$....(II)

Negative sign shows the force is opposite to x.

Now, comparing equation (I) and (III)

$k=\rho l^2 g$

We need to calculate the time period

Using formula of time period

$T=2\pi\sqrt{\dfrac{m}{k}}$

$T=2\pi\sqrt{\dfrac{l^3d}{\rho l^2 g}}$

$T=2\pi\sqrt{\dfrac{ld}{\rho g}}$

Hence, The time period is $2\pi\sqrt{\dfrac{ld}{\rho g}}$

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Topic : time period

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