Question 340: Waves

Question

A rectangular block of mass '$\mathrm{m}$' and crosssectional area A, floats on a liquid of density '$\rho$'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency '$n$' equal to ( $g=$ acceleration due to gravity)

Accepted Answer

The formula for the time period is given as $
\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$
The mass of displaced fluid is
$\begin{aligned}
& 	ext { Mass }=	ext { density } 	imes 	ext { volume } \
& \mathrm{m}=ho 	imes \mathrm{A} l
\end{aligned}$
At equilibrium,
Weight of the block $=$ Weight of the displaced liquid
$\begin{aligned}
& 	herefore \quad \mathrm{mg}=\mathrm{A} l ho \mathrm{g} \
& 	herefore \quad l=\frac{\mathrm{m}}{\mathrm{A} ho}
\end{aligned}$
Substituting the values in the equation
$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \
& \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{Apg}}}
\end{aligned}$
The frequency $f=\frac{1}{T}$
$	herefore \quad \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{Apg}}{\mathrm{m}}}$

Answer

$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{Apg}}{\mathrm{m}}}$

Answer

$2 \pi \sqrt{\frac{\mathrm{Apg}}{\mathrm{m}}}$

Answer

$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{m}}{\mathrm{Apg}}}$

Answer

$2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{Apg}}}$

Waves

A rectangular block of mass '\mathrm{m}' and crosssectional area A, floats on a liquid of density '\rho'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency 'n' equal to ( g= acceleration due to gravity)

Choose the correct answer: