# Homework Help: Period of a bobbing cork

1. Jul 1, 2009

### alexander_i

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

A cork floats in water. The cork is a cylinder with radius 1 cm and height 3.4 cm. The density of the cork is 0.55 g/cc. Calculate the period of oscillation if the cork is pushed down a little and released.

2. Relevant equations

I think I need help with the restoring force... please.

3. The attempt at a solution

-1st I found the mass of the cork - M=D*V =
D = (.55g/cm3)*(1kg/1000g)*(100cm/m)3 = 550kg/m3
V = pi*r2*h = pi*(.01m)2*.034m = 1.068E-5 m3

M = 550kg/m3*1.068E-5m3 = 5.874E-3 kg

-Then setting up my differential:
Fnet = Frestore + Fgravity

Fr = -Dh2o*Vh2o : (Density of water * volume of water displaced)

Fg = mg

ma = mg - Dh2o*Vh2o

volume is dependent on y, or the height, so V = pi*r2*y

-rearranging the equation my'' + Dh2o*pi*r2*y = mg

divide by m --> y'' + Dh2o*pi*r2*y/m = g

and setting y=ert

r1=+isqrt(Dh2o*pi*r2*/m)
r2=-isqrt(Dh2o*pi*r2*/m)

I don't need the particular solution because we need to calculate the period, and

y(t) = Acos{sqrt(Dh2o*pi*r2*/m)t}
+ Bsin{sqrt(Dh2o*pi*r2*/m)t}

the period should be 2*pi/(sqrt(Dh2o*pi*r2*/m) right?

I got .859s but this is not correct. If anyone has some advice, I would be much obliged.

2. Jul 1, 2009

### turin

I would first suggest to redefine y so that y=0 when the cork is at equilibrium. Then, you a have a VERY popular (in physics) type of 2nd order diff. eq., from which you can directly read off the frequency. Actually, it is not really necessary to redefine y, but I think it might make the equation easier for you to recognize.

BTW, y=e^rt is certainly not correct.

3. Jul 1, 2009

### RoyalCat

Try phrasing the restoring force as a force of the form $$\vec F(y) = -C\vec y$$, and think about what that says about the system.
A good way to start is to look at what the net force on the cork is at equilibrium, and what it is when you've displaced it by a height of $$y$$ into the water (Remember that the force $$mg$$ doesn't change).

A good mental analogy to make in this case would be comparing it to a vertical spring, I assume you've already dealt with that problem, try and remember how you dealt with the effect of gravity there, it's very similar here.