1. Mar 8, 2004

### hbanana

The problem I need help with is this: In his lecture, Dr. Dyson argued that one could determine the specific gravity of a wine in the absence of a hydrometer. To demonstrate this he constructed a 100g weighted cylinder of Length (L) and cross-sectional area (A) that floated vertically. When the cylinder was placed in a glass of wine a length (L') was submerged at its equilibrium postion. He lifted the cylinder from its equilibrium position by an amount y, released it and timed its cycles. a) show that the cylinder executes simple harmonic motion when it is released. That is show the restoring force is proportional to the cylinder's displacement from its equilibrium position.
I was able to complete the following:
FB=p(V1 + V2)g
(p=density not pressure)

V2=Ay
(y=displacement)

FB=pV1g + pAyg.

I know i have to some how prove that the Force is proportional to displacement, but I'm stuck at exactly how to prove that.
This problem actually has more parts, but I can't go onto the next parts until I prove this.

2. Mar 8, 2004

### Janitor

Question

Are you allowed to simplify things by assuming that the level of the wine's surface (with respect to the glass) does not fall and rise as the cylinder bobs up and down? I.e. that the cylinder's diameter is small compared to the glass's diameter, and the cylinder's amplitude of vertical motion is not too great. (If not, then it seems to me you have to calculate the dynamics of both the cylinder and the liquid.)

Also, is your density parameter the density of the cylinder or of the wine?

Last edited: Mar 8, 2004
3. Mar 8, 2004

### hbanana

yes, we are allowed to simplify. I think my professor just wants us to prove that the specific gravity of the wine can be determined by the displacement of the cylinder. the density in my equation is the density of the fluid.

Thanks!

4. Mar 8, 2004

### Janitor

As you have already stated, motion will be simple harmonic if the restoring force is proportional to distance from equilibrium position. Symbolically,

F = -k(y-y_0)

where I have followed your lead and chosen y as my position variable, and y_0 as position at equilibrium.

In the particular case that you are dealing with, the net force on the cylinder is the difference between its weight and the weight of the wine it displaces at any given time. That is:

F = (L'-y)Apg - W

where F and y are taken as positive in the upward direction, L', cylinder cross section area A and liquid density p are as you have defined them, g is acceleration of gravity, and W is the constant weight of the cylinder (100 g times the acceleration of gravity).

I can re-write my equation as

F = -Apg{y-[L'-W/(Apg)]}

which is in the form of the first equation, with k=Apg and y_0=L'-W/(Apg).

That takes care of the first part of your question.

Last edited: Mar 9, 2004
5. Mar 8, 2004

### hbanana

thank you so much!

6. Mar 9, 2004

### Janitor

No problem, except...

I left out some g's, which I have edited into my post above. Hope this isn't too late!

The oscillating frequency (in radians per unit time, I believe) is the square root of Apg/m where m is the mass of the cylinder. So the cycle frequency is 1/(2pi) times that radian frequency. The period of oscillation is the reciprocal of the cycle frequency.

Last edited: Mar 9, 2004