Time period of oscillation of a pole floating

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SUMMARY

The discussion centers on calculating the time period of vertical oscillation for a pole floating in a liquid, with 80 cm of its length submerged. The time period is determined to be 4π/7 seconds. Key equations involve the restoring force derived from the buoyant force and the pole's density, with the equilibrium condition of 0.8 m immersion being crucial for solving the problem. Participants emphasize the necessity of knowing the full length of the rod and its density to fully resolve the oscillation dynamics.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Knowledge of buoyancy principles and Archimedes' principle
  • Familiarity with basic physics equations involving force and mass
  • Concept of equilibrium conditions in fluid mechanics
NEXT STEPS
  • Study the derivation of the time period of oscillation in SHM
  • Learn about buoyancy and its effects on floating objects
  • Explore the relationship between density, volume, and mass in fluid dynamics
  • Investigate the conditions for equilibrium in floating bodies
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Students studying physics, particularly those focusing on fluid mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to buoyancy and SHM.

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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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Anyone?
 


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.
 


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


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.
 


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


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
 


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?
 


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

ehild
 
  • #10


I got it. Thanks!
 

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