2019 A/L Physics Past Paper · Question 34 (English Medium)

Period of oscillation of a floating test tube

A test tube is made to float upright in a fluid by placing a metal ball at the bottom of the tube as shown in the figure. Total mass of the tube and the ball is $m$ , density of the fluid is $\rho$ , and the cross-sectional area of the tube is $A$ . Effect of surface tension and the viscosity of the fluid can be neglected. If the tube is given a small vertical displacement, what is the period of oscillations of the subsequent motion of the tube?

(1) $2π\sqrt{\frac{Aρg}{m}}$
(2) $2π\sqrt{\frac{m}{Aρg}}$
(3) $2π\sqrt{\frac{2m}{Aρg}}$
(4) $2π\sqrt{\frac{m}{2Aρg}}$
(5) $2π\sqrt{\frac{mg}{A^2ρ}}$

Answer and explanation

Answer: (2) $2π\sqrt{\frac{m}{Aρg}}$

What the question asks

We want:

The period $T$ when the floating tube is pushed down a little and let go.

Give it in terms of $m$ , $A$ , $\rho$ and $g$ .

Relevant theory

If the acceleration of a moving body is always directed towards an equilibrium point, its magnitude being directly proportional to the distance from the same point, then that motion of the body is called simple harmonic motion.

If the spring obeys Hooke's law, the extension caused by a mass m attached to it is directly proportional to the tension on the spring.

When a body is completely or partially immersed in a fluid at rest, it experiences an upthrust equal to the weight of the fluid displaced by the body.

Required equations

$U = V\rho g$

$F = -kx \Rightarrow T = 2\pi\sqrt{\frac{m}{k}}$

Reasoning and calculation

The letters:

$m$ : the mass of the tube and the ball.

$A$ : the cross-section of the tube.

$\rho$ : the density of the liquid.

$x$ : how far the tube is pushed down.

At rest the push up equals the weight.

Push the tube down by $x$ . An extra volume $Ax$ of liquid is pushed aside.

So the push up grows by:

$A x\rho g$

It points up, against the push. It is proportional to $x$ .

$F = -(A\rho g)\,x$

$m a = -(A\rho g)\,x \Rightarrow a = -\frac{A\rho g}{m}x$

$T = 2\pi\sqrt{\frac{m}{A\rho g}}$

Reading the options

Look at the answers.

Inside the root, $m$ is on top and $A\rho g$ is below. Nothing else.

Only option (2) is like this.

Checks

Check the unit.

$\dfrac{[m]}{[A][\rho][g]} = \dfrac{M}{L^2 \cdot ML^{-3} \cdot LT^{-2}} = T^2$

The root of $T^2$ is a time ✓.

Option (1) has the dimension $T^{-1}$ . That is not a time.

Does it make sense?

A denser liquid gives a bigger push. So the tube bobs faster.

A wider tube gives a bigger push too. It bobs faster.

A heavier tube is harder to move. It bobs slower ✓.

Conclusion

$T = 2\pi\sqrt{\frac{m}{A\rho g}}$

The restoring force is $A\rho g x$ , so $k = A\rho g$ :

$T = 2\pi\sqrt{\dfrac{m}{A\rho g}}$

So the answer is option (2).

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