Frequency of a mass bobbing up and down in water

In summary, the mass is initially at rest and then is pushed down by a force. The new position is then solved for with a differential equation.
  • #1
nfcfox
36
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!
 
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  • #2
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.
 
  • #3
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.
 

1. What factors affect the frequency of a mass bobbing up and down in water?

The frequency of a mass bobbing up and down in water is affected by the length of the string or object attached to the mass, the mass of the object, and the tension in the string or object. These factors determine the period of the motion, which is inversely proportional to the frequency.

2. How is frequency related to the depth of the water?

The depth of the water does not have a direct effect on the frequency of a mass bobbing up and down. However, as the depth increases, the amplitude of the motion decreases due to the increased resistance of the water. This can indirectly affect the frequency by changing the period of the motion.

3. Does the density of the object affect its frequency in water?

Yes, the density of the object does affect its frequency in water. Objects with higher densities will have a lower frequency because they experience greater resistance from the water. Objects with lower densities will have a higher frequency due to less resistance.

4. Can the frequency of a mass bobbing up and down in water be affected by external forces?

Yes, external forces such as wind or currents can affect the frequency of a mass bobbing up and down in water. These forces can change the amplitude and period of the motion, resulting in a change in frequency.

5. How is the frequency of a mass bobbing up and down in water calculated?

The frequency of a mass bobbing up and down in water can be calculated using the formula: f = 1/T, where f is the frequency in Hertz (Hz) and T is the period of the motion in seconds. The period can be determined by measuring the time it takes for the mass to complete one full cycle of bobbing up and down in water.

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