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} }T=2πgρ1hρ
where ρ is the density of cork.
(Ignore Damping due to viscosity of liquid)
Answers
AnSwEr:
\small{\underline{\blue{\sf{Given :}}}}
Given:
Base Area of Cork = A
Height of Cork = h
Density of Cork = ρ
Density of Liquid = ρ1
\small{\underline{\green{\sf{Solution :}}}}
Solution:
# See attached Figure for diagram
So,
We come to know that :
\begin{lgathered}\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)\end{lgathered}
⋆
WeightofCork=WeightofliquidDisplaced
WhenCorkisDisplaced!
:⟹Force=WeightofDisplacedWater
:⟹F=−(Volume×Density
liquid
×g)
:⟹F=−(Ax×ρ1×g)[Volume=Area×x]
:⟹
F=−(Ax×ρ1×g)
−−−−−(1)
As we also know that :
\large \star {\boxed{\sf{F \: = \: -kx}}} ----(2)⋆
F=−kx
−−−−(2)
For the spring motion.
Now,
Equate (1) and (2)
\begin{lgathered}:\implies {\sf{\cancel{-} k \cancel{x} \: = \: \cancel{-} A \cancel{x} \rho1 g}} \\ \\ \\ : \implies {\sf{k \: = \: A \rho 1 g}}----(3)\end{lgathered}
:⟹
−
k
x
=
−
A
x
ρ1g
:⟹k=Aρ1g−−−−(3)
As we have equation for Time period for S.H.M
\large \star {\boxed{\sf{T \: = \: 2 \pi \: \sqrt{\dfrac{m}{k}}}}}⋆
T=2π
k
m
Substitute value of k
: \implies {\sf{T \: = \: 2 \pi \sqrt{\dfrac{m}{A \rho 1 g}}}}:⟹T=2π
Aρ1g
m
And take value of m as Ahρ
Substitute value of m
\begin{lgathered}: \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 }}}}}\end{lgathered}
:⟹T=2π
A
ρ1g
A
hρ
T=2π
gρ1
hρ
\large {\mathbb{HENCE \: \: PROVED}}
HENCEPROVED
if this attachment will help you
mark as brainlest answer
