Time period of oscillation of a pole floating

In summary: When displaced by a small distance x, restoring force, ma=Vρg=Axρgw2=Aρg/m=Aρg/ALσ, where L is the full length of the rod and σ is its density. I can't proceed any further without knowing the values.
  • #1
zorro
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Time period of oscillation of a pole floating...

Homework Statement



A pole is floating in a liquid with 80cm of its length immersed. It is pushed down a certain distance and released. What is its time period of vertical oscillation?


The Attempt at a Solution



I think the information given is insufficient. They should have given full length of the rod or area of its cross section.

When displaced by a small distance x, restoring force, ma=Vρg=Axρg
w2=Aρg/m=Aρg/ALσ, where L is the full length of the rod and σ is its density. I can't proceed any further without knowing the values.

The answer given is 4π/7 s.
 
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  • #2


Anyone?
 
  • #3


Abdul Quadeer said:
When displaced by a small distance x, restoring force, ma=Vρg=Axρg

The restoring force will not be equal in both the directions. Once excess buoyant force and once excess weight would act.
 
  • #4


yeah, atleast full length must be given. Other 80cm has no meaning
 
  • #5


ashishsinghal said:
The restoring force will not be equal in both the directions. Once excess buoyant force and once excess weight would act.

There is a net restoring force only in the upward direction.
 
  • #6


During the oscillation when the float will be above its equilibrium position, then?
It would be nice if you post a diagram.
 
  • #7


Abdul Quadeer said:
When displaced by a small distance x, restoring force, ma=Vρg=Axρg
w2=Aρg/m=Aρg/ALσ, where L is the full length of the rod and σ is its density. I can't proceed any further without knowing the values.

Use the condition that 0.8 m is immersed in equilibrium.

ehild
 
  • #8


ashishsinghal said:
During the oscillation when the float will be above its equilibrium position, then?

I understand what you mean now. When the pole is pushed downward, there is a net restoring force acting in the upward direction and that is sufficient to write the equation for S.H.M. When it is above the equilibrium position, there is a different restoring force but its magnitude is same as the previous one.

ehild said:
Use the condition that 0.8 m is immersed in equilibrium.
ehild

where?
 
  • #9


You get information about the length and density of the rod from the equilibrium condition.

ehild
 
  • #10


I got it. Thanks!
 

Related to Time period of oscillation of a pole floating

1. What factors affect the time period of oscillation of a pole floating?

The time period of oscillation of a pole floating is affected by several factors, including the length and mass of the pole, the density and viscosity of the fluid it is floating in, and the amplitude of the oscillations.

2. How can the time period of oscillation of a pole floating be calculated?

The time period of oscillation of a pole floating can be calculated using the formula T = 2π√(I/mgd), where T is the time period, I is the moment of inertia of the pole, m is its mass, g is the acceleration due to gravity, and d is the displacement of the pole.

3. Is the time period of oscillation of a pole floating affected by the shape of the pole?

Yes, the shape of the pole can affect its time period of oscillation. A thinner and longer pole will have a longer time period compared to a shorter and thicker pole, all other factors remaining constant.

4. How does the time period of oscillation of a pole floating change with depth in the fluid?

The time period of oscillation of a pole floating decreases with increasing depth in the fluid. This is because the buoyant force acting on the pole decreases as it sinks deeper, leading to a decrease in the effective mass and therefore a shorter time period.

5. Can the time period of oscillation of a pole floating be changed by adjusting the fluid density or viscosity?

Yes, the time period of oscillation of a pole floating can be changed by adjusting the fluid density or viscosity. A denser or more viscous fluid will result in a longer time period, while a less dense or less viscous fluid will result in a shorter time period.

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