Pendulums, Springs and Liquid Columns

1One formula, many oscillators

A pendulum, a loaded spring and the water sloshing in a U-shaped tube look nothing alike. Yet each has a restoring force proportional to displacement, so each obeys $a=-{\omega }^{2}xa\; =\; -\backslash omega^2\; x$ and has a period $T=\frac{2\pi }{\omega }T\; =\; \backslash frac\{2\backslash pi\}\{\backslash omega\}$. The work in this lesson is always the same three steps: (1) displace the system by a small amount and find the resultant force; (2) show that it has the form $F=-(\text{constant})xF\; =\; -(\backslash text\{constant\})\backslash ,x$; (3) divide the constant by the mass to get ${\omega }^2\backslash omega^2$. The result tells you what the period depends on, and what it does not.

2The mass–spring system

For a mass $mm$ on a horizontal spring of stiffness $kk$, Lesson 1 gave $a=-\frac{k}{m}xa\; =\; -\backslash frac\{k\}\{m\}x$. So ${\omega }^2=\frac{k}{m}\backslash omega^2\; =\; \backslash frac\{k\}\{m\}$ and

A heavier mass gives a longer period; a stiffer spring gives a shorter one. The amplitude and $gg$ do not appear.

A vertical spring

Hang the mass from a vertical spring. At equilibrium the spring has stretched by $ee$, and the spring force balances the weight: $ke=mgke\; =\; mg$. Now pull the mass down a further distance $yy$. Taking downwards as positive, the resultant force is

The weight is cancelled by the extra tension at equilibrium, so the vertical spring has exactly the same period, $T=2\pi \sqrt{\frac{m}{k}}T\; =\; 2\backslash pi\backslash sqrt\{\backslash frac\{m\}\{k\}\}$. Since $\frac{m}{k}=\frac{e}{g}\backslash frac\{m\}\{k\}\; =\; \backslash frac\{e\}\{g\}$, you can also write $T=2\pi \sqrt{\frac{e}{g}}T\; =\; 2\backslash pi\backslash sqrt\{\backslash frac\{e\}\{g\}\}$: the period depends only on how far the mass stretched the spring.

Springs in parallel and in series

Two springs side by side, sharing the load (parallel), act as one spring of stiffness $k=k_1+k_{2}k\; =\; k\_1\; +\; k\_2$. Two springs joined end to end (series) each carry the whole load, so their extensions add, and $\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}\backslash frac\{1\}\{k\}\; =\; \backslash frac\{1\}\{k\_1\}\; +\; \backslash frac\{1\}\{k\_2\}$. Use the combined $kk$ in $T=2\pi \sqrt{\frac{m}{k}}T\; =\; 2\backslash pi\backslash sqrt\{\backslash frac\{m\}\{k\}\}$.

3The simple pendulum

A simple pendulum is a small, heavy bob on a light, inextensible string of length $LL$ (measured from the pivot to the centre of the bob). Displace it through an angle $\theta \backslash theta$. The tension acts along the string and does no work along the path. The weight has a component $mg\sin \theta mg\backslash sin\backslash theta$ along the arc, directed back towards the lowest point.

Measure the displacement along the arc: $s=L\theta s\; =\; L\backslash theta$, with $\theta \backslash theta$ in radians. Newton's second law along the arc gives

This is not yet SHM, because $\sin \theta \backslash sin\backslash theta$ is not proportional to $\theta \backslash theta$. But for small angles $\sin \theta \approx \theta \backslash sin\backslash theta\; \backslash approx\; \backslash theta$, and then

So the formula is good to well under 1% for swings up to about 10°. The period depends on the length and on $gg$, but not on the mass of the bob (the mass cancels, because the weight and the inertia are both proportional to $mm$) and, for small swings, not on the amplitude. Measuring $TT$ and $LL$ is therefore a way to measure $gg$.

4A liquid in a U-tube

Blow gently into one arm of a U-tube containing water and then stop: the water level swings up and down several times before settling. Let the liquid have density $\rho \backslash rho$, the tube a uniform cross-section $AA$, and the whole liquid column a length $ll$ measured along the tube. Push one surface down by $yy$; the other rises by $yy$, so one side now stands $2y2y$ higher than the other.

The unbalanced column of height $2y2y$ has weight $\rho A(2y)g\backslash rho\; A(2y)g$, and it pushes the whole liquid back towards equilibrium. The whole liquid, of mass $\rho Al\backslash rho\; Al$, moves together, so

The density and the cross-section cancel: water and paraffin of the same column length oscillate with the same period (if friction is small).

5Experiment 1.2: measuring g with a simple pendulum

Aim: to measure the acceleration of free fall from the period of a simple pendulum.

Apparatus: a small dense bob on a thin thread, a clamp stand with two small wooden blocks (to grip the thread so the pivot is sharp), a metre rule, a stopwatch, a fiducial marker (a pin or a line on card) placed behind the lowest point.

Method:

1. Clamp the thread between the blocks and measure $LL$ from the bottom of the blocks to the centre of the bob.
2. Displace the bob through no more than about 10° and release it. Start timing as the bob passes the fiducial marker, and time 20 complete oscillations. Repeat and take the mean.
3. Calculate $TT$ and $T^{2}T^2$. Repeat for five lengths from 0.40 m to 1.20 m.
4. Plot $T^{2}T^2$ against $LL$. Since $T^2=\frac{4{\pi }^2}{g}LT^2\; =\; \backslash frac\{4\backslash pi^2\}\{g\}L$, the graph should be a straight line through the origin with gradient $\frac{4{\pi }^2}{g}\backslash frac\{4\backslash pi^2\}\{g\}$.

Illustrative results from a school laboratory:

The best-fit line has gradient $4.04{\mathrm{s}}^2{\mathrm{m}}^{-1}4.04\backslash ,\backslash mathrm\{s^2\backslash ,m^\{-1\}\}$, so $g=\frac{4{\pi }^2}{4.04}=9.78\mathrm{m}{\mathrm{s}}^{-2}g\; =\; \backslash frac\{4\backslash pi^2\}\{4.04\}\; =\; 9.78\backslash ,\backslash mathrm\{m\backslash ,s^\{-2\}\}$. A value a little below 9.81 is expected in Rwanda: $gg$ is slightly smaller near the equator and at high altitude, as Unit 5 explains.