Other oscillating systems

One rule for many oscillators

A pendulum is not the only thing that swings back and forth. Water sloshes in a U-tube. A buoy bobs up and down in the sea. A ruler hanging from a nail swings like a pendulum, but not quite like one.

All of them follow one rule. Find the pull back for a small displacement. If it is in proportion to the displacement and opposite to it, the motion is simple harmonic. Then

For a straight motion, the inertia factor is the mass that moves. The restoring factor is the force for each metre of displacement. Let us use this on three systems.

Liquid in a U-tube

A U-tube with a cross-section $A$ holds a liquid of density $\rho$. The whole column, from one surface round the bend to the other, has length $L$.

Push the liquid down by $y$ on the right. It rises by $y$ on the left. Now the two levels differ by $2y$. The extra column of height $2y$ pushes the liquid back:

The whole column moves, so the mass is $m = LA\rho$. Then

The density and the cross-section cancel. Only the length of the column matters. A column of water $0.60\ \text{m}$ long has $T = 2\pi\sqrt{\dfrac{0.60\ \text{m}}{19.6\ \text{m/s}^2}} = 1.10\ \text{s}$.

A floating body

A cylinder floats upright with depth $h$ under the water. Its weight equals the buoyant force, so $m = A h \rho_w$, where $A$ is its cross-section and $\rho_w$ the density of water.

Push it down a further $y$. The extra buoyant force pushes it back:

So $T = 2\pi\sqrt{\dfrac{Ah\rho_w}{A\rho_w g}} = 2\pi\sqrt{\dfrac{h}{g}}$. A block that floats with $7.0\ \text{cm}$ under water bobs with $T = 2\pi\sqrt{\dfrac{0.070\ \text{m}}{9.8\ \text{m/s}^2}} = 0.53\ \text{s}$.

In a denser liquid the same body floats higher. Its $h$ is smaller, so it bobs faster.

A body swinging about a pivot

A rigid body that swings about a pivot is a physical pendulum. Let its centre of mass be a distance $d$ from the pivot. For a small angle the torque is $\tau = -mgd\,\theta$. With $\tau = I\alpha$:

Here $I$ is the moment of inertia about the pivot. For a simple pendulum, $I = mL^2$ and $d = L$, and we get $2\pi\sqrt{L/g}$ again.

A rod about one end. A uniform rod of length $L$ has $I = \tfrac13 mL^2$ about one end, and $d = \tfrac{L}{2}$. So

A rod $1.50\ \text{m}$ long has $T = 2\pi\sqrt{\dfrac{1.0\ \text{m}}{9.8\ \text{m/s}^2}} = 2.01\ \text{s}$. It keeps time like a simple pendulum only $1.0\ \text{m}$ long. The length $\tfrac{2L}{3}$ is called the length of the equivalent simple pendulum.

Start with the hole at the end, $d = 0.50\ \text{m}$. Move it towards the middle. The period falls at first. Close to the middle it rises fast, because the pull back becomes weak. Find the hole that gives the shortest period.

Practise with new numbers

Try it in Python

The program moves the hole along a metre rule and finds the period each time.

Common mistakes

• Using $g$ in the U-tube formula. The level difference is $2y$, so it is $2g$.
• Using the height of the liquid in one arm. $L$ is the whole column, round the bend.
• Treating a rod like a bob. Its mass is spread out. Use $I$ about the pivot, not $mL^2$.
• Measuring $d$ from the end. $d$ runs from the pivot to the centre of mass.

Remember these

• Pull back in proportion to displacement means SHM: $T = 2\pi\sqrt{\text{inertia}/\text{restoring factor}}$.
• U-tube: $T = 2\pi\sqrt{L/2g}$, for any liquid.
• Floating body: $T = 2\pi\sqrt{h/g}$, with $h$ the depth under the surface.
• Physical pendulum: $T = 2\pi\sqrt{I/mgd}$. A rod about one end: $T = 2\pi\sqrt{2L/3g}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The liquid column in a U-tube is made twice as long. What happens to the period?

2. A uniform rod $0.60\ \text{m}$ long swings about one end. How long is the simple pendulum that keeps the same time?

3. A wooden block bobs in fresh water. It is moved to salty water. Does it bob faster or slower?

4. In a U-tube, what is the restoring force for each metre of displacement?