A small amount of mercury is filled into massless cylindrical test tube with an even bottom and length $l=0,200m$. The test tube is put into a large swimming pool filled with water.

a) How much mercury can be filled into the test tube before it sinks? What is the height it will reach in the tube?

b) If less than the maximum amount of mercury is put into the test tube, pushing the tube into the water and then letting go creates a vertical oscillation. What is the frequency if 75% of the maximum amount (from part a) is put into the test tube? (Friction and surface tension can be ignored)

This is a practice question from an old German physics book, which unfortunately has no solutions. I hope my translation makes sense.

I used Archimedes principle to solve the first one.

$\rho_{water}\cdot V_{water}\cdot g=\rho_{mercury}\cdot V_{mercury}\cdot g$

Cancelling $g$ leaves:

$\rho_{water}\cdot V_{water}=\rho_{mercury}\cdot V_{mercury}$

Rewriting $V$:

$\rho_{water}\cdot l \cdot \pi r^2=\rho_{mercury}\cdot h \cdot\pi r^2$

$h=\frac{l\cdot \rho_{water}}{\rho_{mercury}}$

If I plug in the densities and $l=0.200m$, I get $h=0.015m$.

So I can fill the test tube to $0.015m$ with mercury before it starts to sink.

Is that correct?

I have no idea how to do the second part. Where should I start? Thanks.


Over the course of your physics career you'll come across many questions asking you to work out the frequency of some oscillating system, and (unless you have an unusually sadistic professor) the system in question is always some varient on a simple harmonic oscillator.

In this case your floating tube looks like the tube on the left:

Floating tube

Now suppose we push the tube down by a distance $x$ as shown on the right. If the area of the tube is $A$, then we displace a volume of water $V = Ax$, and the weight of water displaced is just:

$$ W = V\rho_w g = A\rho_w gx $$

Archimedes principle tells us the the upwards force $F$ is equal to the weight of water displaced, and Newton's second law tells us that $F = ma$, where $m$ is the mass of the tube + mercury and $a$ is the upwards acceleration $d^2x/dt^2$. So we get:

$$ m \frac{d^2x}{dt^2} = -A\rho_w gx $$


$$ \frac{d^2x}{dt^2} = -\frac{A\rho_w g}{m} x $$

which is just the equation for simple harmonic motion. You just need to express $m$ as the height of mercury (which will make the $A$ factor cancel out) then put in 75% of the height you calculated in part one. Then solve the equation, or do what I do and just Google for the solution :-)

  • $\begingroup$ Thank you very much. This was such a great answer! Have a nice day. $\endgroup$ – qmd Jan 14 '15 at 18:59
  • 3
    $\begingroup$ @Rzeta: you're welcome. Note that we don't usually answer homework like questions. I answered only because you, and every other student of physics reading this, are going to come across thinly disguised questions about simple harmonic oscillators dozens of times. It's important that you learn to recognise them. $\endgroup$ – John Rennie Jan 15 '15 at 6:59

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