Question

A cylindrical piece of cork of base area A and height h floats in a liquid of density ρ1.

The cork is depressed slightly then released. Show that it oscillates up and down simple harmonically with a period of

$T = 2\pi \sqrt{ \frac{hρ}{gρ1} }$

where ρ is the density of cork.
(Ignore Damping due to viscosity of liquid)

​

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}}$

• Base Area of Cork = A
• Height of Cork = h
• Density of Cork = ρ
• Density of Liquid = ρ1

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

# See attached Figure for diagram

So,

We come to know that :

$\large \star {\boxed{\sf{Weight \: of \: Cork \: = \: Weight \: of \: liquid \: Displaced}}} \\ \\ \\ \small{\sf{\pink{\underline{\: \: \: \: \: \: \: \: \: \: \: \: When \: Cork \: is \: Displaced \: \: \: \: \: \: \:!\: \: \: \:}}}} \\ \\ \\ : \implies {\sf{Force \: = \: Weight \: of \: Displaced \: Water}} \\ \\ \\ : \implies {\sf{F \: = \: - \big( Volume \: \times \: Density_{liquid} \: \times \: g \big) }} \\ \\ \\ : \implies {\sf{F \: = \: - \big( Ax \: \times \: \rho 1 \: \times \: g \big) \: \: \: \: \: \: \: \: \: \: \big[ Volume \: = \: Area \: \times \: x \big]}} \\ \\ \\ : \implies {\boxed{\sf{F \: = \: -(Ax \: \times \: \rho 1 \: \times \: g)}}} -----(1)$

$\rule{200}{1}$

As we also know that :

$\large \star {\boxed{\sf{F \: = \: -kx}}} ----(2)$

For the spring motion.

Now,

Equate (1) and (2)

$:\implies {\sf{\cancel{-} k \cancel{x} \: = \: \cancel{-} A \cancel{x} \rho1 g}} \\ \\ \\ : \implies {\sf{k \: = \: A \rho 1 g}}----(3)$

$\rule{200}{1}$

As we have equation for Time period for S.H.M

$\large \star {\boxed{\sf{T \: = \: 2 \pi \: \sqrt{\dfrac{m}{k}}}}}$

Substitute value of k

$: \implies {\sf{T \: = \: 2 \pi \sqrt{\dfrac{m}{A \rho 1 g}}}}$

And take value of m as Ahρ

Substitute value of m

$: \implies {\sf{T \: = \: 2 \pi \sqrt{\dfrac{\cancel{A} h \rho}{\cancel{A} \rho 1 g}}}} \\ \\ \\ \LARGE {\boxed{\sf{T \: = \: 2 \pi \sqrt{\dfrac{h \rho }{g \rho 1 }}}}}$

$\large {\mathbb{HENCE \: \: PROVED}}$

Answer 2

Answer:

This is a question of Simple Harmonic Motion involving the concept of Fluid Properties.

Given Information:

• Area of the Base of Cork = A
• Height of the Cork = h
• Density of Liquid = ρ₁

Now we know that, in a simple harmonic motion:

→ F is directly proportional to -x

→ F = kx, where k is a constant

Also we know that,

→ F = mg

But since the force is upthrust force, we denote it as -mg.

Now, we know that:

→ Density = Mass / Volume

→ Mass = Density × Volume =  ρ₁ × V

→ Volume = Area × Displaced Length = Ax

→ Mass of the water displaced =  ρ₁Ax

Hence we get,

→ Force = -ρ₁Ax × g = -ρ₁gAx

Now we also know that,

→ F = -kx

→ k = -F/x

→ k = - ( -ρ₁gAx ) / x

→ k =  ρ₁gA

Now for an object exhibiting Simple Harmonic Motion, we know that:

→ T = 2π √ (m/k )

Mass of cork = Volume of Cork × Density of Cork

→ Mass of cork = A × h × ρ

→ T = 2π √ ( ρAh)/(ρ₁gA)

→ T = 2π √ (ρh)/(ρ₁g)