Frequency of a mass bobbing up and down in water

Click For Summary
SUMMARY

The discussion centers on the analysis of a mass bobbing in water, modeled as simple harmonic motion (SHM). The key equations involve the net force on the mass, which is influenced by gravity and buoyancy, as described by Archimede's Principle. The derived angular frequency, ω, is expressed as ω=sqrt(-g/X_new+(A*rho_w*g)/m), where A is the cross-sectional area, ρ_w is the water density, and m is the mass. The equilibrium position, X_0, must be determined to accurately describe the system's behavior.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM) principles
  • Familiarity with Archimede's Principle and buoyancy
  • Knowledge of differential equations and their applications in physics
  • Basic concepts of forces, including gravitational and buoyant forces
NEXT STEPS
  • Study the derivation of the frequency of a mass-spring system
  • Learn about the applications of Archimede's Principle in fluid mechanics
  • Explore the relationship between restoring forces and SHM
  • Investigate the effects of varying cross-sectional area on buoyancy and motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and fluid dynamics, as well as educators seeking to clarify concepts related to simple harmonic motion and buoyancy.

nfcfox
Messages
36
Reaction score
0

Homework Statement


This was a test question I had today but basically, initially the mass is at rest as the buoyant force opposes the force of gravity. Then we push it down X meters and let it go. This can be described by SHM. We are also given the density of water, a cross sectional area of the mass, and the actual mass, M.

Homework Equations


A=area (not amplitude)
X_0= equilibrium position
X_new= furthest value it can be bobbed up or down
Archimede's Principle

The Attempt at a Solution


Net force=mg-A*(X_0+X_new)*rho_w*g note that this is zero at the equilibrium point where X_new equals 0.
ma=mg-A*(X_0+X_new)*rho_w*g
a=g-(A*(X_0+X_new)*rho_w*g)/m

The general form of a SHM equation is x(t)=Acos(ωt) which can be solve to x''(t)=-ω^2Acos(ωt)

I assumed ωt to be 0 so as we are using the amplitude. We know the amplitude to be X_0+X_new however I chose X_0 to be zero for simplicity (I know this isn't correct as the volume would be zero but this was just the position). Thus, the amplitude is now X_new. I then set the acceleration equal to x''(t) to solve for ω:

g-(A*X_new*rho_w*g)/m=-ω^2X_new

ω=sqrt(-g/X_new+(A*rho_w*g)/m)

If anyone can confirm this, that'd be awesome!
 
Last edited:
Physics news on Phys.org
nfcfox said:

Homework Statement


This was a test question I had today but basically, initially the mass is at rest as the buoyant force opposes the force of gravity. Then we push it down X meters and let it go. This can be described by SHM. We are also given the density of water, a cross sectional area of the mass, and the actual mass, M.

Homework Equations


A=area (not amplitude)
X_0= equilibrium position
X_new= furthest value it can be bobbed up or down
Archimede's Principle

The Attempt at a Solution


Net force=mg-A*(X_0+X_new)*rho_w*g note that this is zero at the equilibrium point where X_new equals 0.
ma=mg-A*(X_0+X_new)*rho_w*g
a=g-(A*(X_0+X_new)*rho_w*g)/m

The general form of a SHM equation is x(t)=Acos(ωt) which can be solve to x''(t)=-ω^2Acos(ωt)

I assumed ωt to be 0 so as we are using the amplitude. We know the amplitude to be X_0+X_new however I chose X_0 to be zero for simplicity (I know this isn't correct as the volume would be zero but this was just the position). Thus, the amplitude is now X_new.

The differential equation contains the constant term mg. The solution for X_new is SHM, but X_0 is not zero. It is the depth in equilibrium. You have to find it with the condition a=0, X_new=0.
 
SHM occurs when the restoring force obeys Hookes law, f = -kx. If you can show that the restoring force for the bob has this form and find k, then you can invoke the standard expression for the frequency of a mass-spring system.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
17
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 15 ·
Replies
15
Views
6K
Replies
7
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K