Oscillating block in liquid

Classical Mechanics Level 3

The container shown contains a liquid of variable density which varies as $\displaystyle d = d_0 \left( 4 - \dfrac{3y}{h_0} \right) \text{ kg/m}^3$ , where $h_0$ is total height of container, $d_0$ is constant and $y$ is measured from the bottom of the container. A solid block whose density is $\dfrac{5}{2} d_0$ and mass 'm' is released from bottom of the container. Given that block will execute SHM and is time period can be written as

$\displaystyle T = 2\pi \sqrt{\dfrac{\alpha h_0}{\beta g}}$

$\text{gcd}(\alpha,\beta) = 1$ and $g$ is acceleration due to gravity. Find $\alpha + \beta$ .

Details and Assumptions

Assume block to be cubical.

The answer is 11.

1 solution

Deepanshu Gupta
Mar 30, 2015

$\displaystyle{ma={ F }_{ b }-mg\\ ma=V{ d }_{ o }(4-\cfrac { 3y }{ h } )g-mg\\ V=\cfrac { 2m }{ 5{ d }_{ 0 } } \leadsto \\ a=\cfrac { 3g }{ 5 } -\cfrac { 6gy }{ 5h } =-\cfrac { 6g }{ 5h } (y-\cfrac { 3g }{ 5 } )\\ a=\cfrac { 6g }{ 5h } (Y)\\ T=2\pi \sqrt { \cfrac { 5h }{ 6g } } }$

Did the same way!!

Ninad Akolekar - 6 years, 2 months ago

yeah! right.. did the same way

Priyesh Pandey - 6 years, 2 months ago

From where did 5th step come from? How did you eliminate 3g/5

A Former Brilliant Member - 5 years, 5 months ago

as T is proportional to (da/dx)^(-1/2) , so it doesn't change for constants like 3g/5.

Bhaskar Pandey - 3 years, 4 months ago

Oscillating block in liquid

The answer is 11.

1 solution

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