Angular frequency of mercury in U pipe

[songoku]

Homework Statement

A mercury is filled to a U-pipe so that the total length of the mercury is h (no figure given on the question). If the mercury column of one leg of the pipe is pushed then released, the mercury will oscillate with small amplitude. The angular frequency is
$$a. \sqrt{\frac{2g}{h}}$$

$$b. \sqrt{\frac{g}{2h}}$$

$$c. \sqrt{\frac{g}{h}}$$

Homework Equations

Maybe a = -ω²x

The Attempt at a Solution

Because there is no figure given, I think h is the height of the mercury column of both legs of the U-pipe, measured from the same point. But I don't know how to find the ω...

 

[Curious3141]

Homework Statement

A mercury is filled to a U-pipe so that the total length of the mercury is h (no figure given on the question). If the mercury column of one leg of the pipe is pushed then released, the mercury will oscillate with small amplitude. The angular frequency is
$$a. \sqrt{\frac{2g}{h}}$$

$$b. \sqrt{\frac{g}{2h}}$$

$$c. \sqrt{\frac{g}{h}}$$

Homework Equations

Maybe a = -ω²x

The Attempt at a Solution

Because there is no figure given, I think h is the height of the mercury column of both legs of the U-pipe, measured from the same point. But I don't know how to find the ω...

No, h is the *total length* of the mercury in the entire tube (not the height of each leg). The distinction is important.

Start by figuring out the net pressure at a point in the mercury at the bottom of the tube (which approximately corresponds to the centre of mass of the mercury) when the column is depressed by x on one side. Your expression will involve the density of mercury, which you can call ρ. Remember that the column is depressed on one side, but raised on the other by the same amount x. Then figure out the force on the mercury column (which can be assumed to be acting at the centre of mass). For this, you'll need to include the cross-sectional area of the column, call this A.

Then use F = ma to work out the acceleration of the mercury column. Get m in terms of h, A and ρ. Now you'll have an expression for a in terms of x, ρ and g. Finally, compare with a = -ω²x to figure out what ω should be.

 

[songoku]

No, h is the *total length* of the mercury in the entire tube (not the height of each leg). The distinction is important.

Start by figuring out the net pressure at a point in the mercury at the bottom of the tube (which approximately corresponds to the centre of mass of the mercury) when the column is depressed by x on one side. Your expression will involve the density of mercury, which you can call ρ. Remember that the column is depressed on one side, but raised on the other by the same amount x. Then figure out the force on the mercury column (which can be assumed to be acting at the centre of mass). For this, you'll need to include the cross-sectional area of the column, call this A.

Then use F = ma to work out the acceleration of the mercury column. Get m in terms of h, A and ρ. Now you'll have an expression for a in terms of x, ρ and g. Finally, compare with a = -ω²x to figure out what ω should be.

OK let me try

The net pressure = P = ρ.g.2x

The force on mercury column = P.A = ρ.g.2x.A

F = m.a
ρ.g.2x.A = ρ.A.h.a
a = 2gx / h

By comparing, ω = √(2g/h) --> finished ! But am I correct?

 

[Curious3141]

OK let me try

The net pressure = P = ρ.g.2x

The force on mercury column = P.A = ρ.g.2x.A

F = m.a
ρ.g.2x.A = ρ.A.h.a
a = 2gx / h

By comparing, ω = √(2g/h) --> finished ! But am I correct?

Yes, well done.

 

[songoku]

Yes, well done.

Thanks