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An object with height h, mass M, and a uniform cross-sectional area A floats up-right in a liquid with density r. (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude F is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density r of the liquid, the mass M, and the cross-sectional area A of the object. You can ignore the damping due to fluid friction.

Short Answer

Expert verified
  1. The vertical distance from the surface of the liquid to the bottom of the floated object at equilibrium is M/A.

  2. The position of the bottom of the object in the new equilibrium position is F/Ag.

  3. The period of the SHM in terms of the density of liquid and mass and the cross-sectional area of the object is T=2M/Ag.

Step by step solution

01

Calculate the vertical distance

We have, Height of the object is h and its massM and cross-sectional area isA

Formula for buoyant force on the body is,

F=Vg

Here,

F is buoyant force

P is density of fluid

V is volume of displaced fluid

g is acceleration due to gravity

Formula to calculate force is,

F=Mg

Mis mass of object.

Using above formulas we get,

Mg=VgM=V

Thus, relation for mass is,

M=V

Formula to calculate volume is,

V=Ax

Here,

Vis volume

Ais area

xis length

Again, we get,

M=Axx=M/A

Therefore, the vertical distance from the surface of the liquid to the bottom of the floated object at equilibrium is M/A.

02

Calculate the position in new equilibrium

We have, Height of the object ishand its massM and cross-sectional area isA

The density of liquid is

Formula to calculate downward force is,

F=Ax'g

x'is farther downward displacement

Rearrange forx'

x'=F/Ag

Therefore, the position of the bottom of the object in the new equilibrium position isF/Ag

03

Calculate the period of the SHM

We have, Height of the object ishand its massM and cross-sectional area isA

and the density of the liquid is

Formula to calculate spring constant is,

k=Fx

x is spring constant

Substitute Axg for F to find k,

T=2AgM=2MA蚁g

Formula for angular frequency is,

=k/M

is angular frequency

SubstituteAgfork to find

=AgM

Formula to calculate time period is,

T=2

SubstituteAg/Mforto findT

T=2AgM=2MA蚁g

Therefore, the period of the SHM in terms of the density of liquid and mass and the cross-sectional area of the object is T=2M/Ag.

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