# Frequency of a mass bobbing up and down in water

1. Nov 16, 2016

### nfcfox

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
A=area (not amplitude)
X_0= equilibrium position
X_new= furthest value it can be bobbed up or down
Archimede's Principle
3. 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: Nov 16, 2016
2. Nov 16, 2016

### ehild

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.

3. Nov 17, 2016

### Staff: Mentor

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.