Time period of oscillation of a pole floating

Click For Summary

Homework Help Overview

The problem involves a pole floating in a liquid, with a specific portion of its length immersed. The original poster seeks to determine the time period of vertical oscillation after displacing the pole and releasing it.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the need for additional information, such as the full length of the rod and its cross-sectional area, to proceed with the solution. There are considerations about the restoring forces acting on the pole when displaced, including the effects of buoyancy and weight.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some have offered insights into the restoring forces and the conditions for equilibrium, while others express the need for more information to clarify the situation.

Contextual Notes

There is a noted lack of information regarding the full length of the rod, which is seen as critical for solving the problem. The discussion also highlights the importance of understanding the equilibrium condition of the pole in the liquid.

zorro
Messages
1,378
Reaction score
0
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.
 
Physics news on Phys.org


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!
 

Similar threads

Replies
4
Views
6K
Replies
7
Views
4K
  • · Replies 26 ·
Replies
26
Views
6K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
4K
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K