The pendulum, the metre scale and the best-fit line

A stone on a string

Tie a small stone to a thread and hang it from a nail. Pull it a little and let go. It swings to and fro, and each swing takes the same time. Galileo noticed this in a church, watching a lamp. The time of a swing depends on the length of the string. Let us turn that into a measurement of $g$.

The simple pendulum

A simple pendulum is a small heavy bob on a thread that cannot stretch, hung from a rigid support. For small swings (angle less than $10^\circ$ or so) the time period is

Here $L$ is the effective length, from the point of suspension to the centre of the bob. So $L$ is the length of the thread plus the radius of the bob (plus the length of any hook). We want $g$, so $g = \dfrac{4\pi^2 L}{T^2}$.

The steps of the experiment.

1. Fix the thread firmly between two halves of a split cork in a clamp, so that the point of suspension does not move.
2. Measure the length of the thread with a metre scale. Measure the diameter of the bob with the vernier calipers and add the radius.
3. Start the swing with a small angle. Wait for a few swings. Start the stopwatch when the bob passes the mean position (lowest point) in a fixed direction, and count that pass as zero.
4. Time 20 complete oscillations. Repeat the timing and take the mean. Divide by 20 for $T$.
5. Change the length in steps of $10$ cm and repeat.

Why 20 oscillations? The stopwatch has a least count of $0.1$ s and our reaction time is about the same. For one swing of $2$ s, that is an error of $5\%$ in $T$. For 20 swings it is the same $0.1$ s in $40$ s, only $0.25\%$. Because $g$ depends on $T^2$, the gain is doubled.

The graph of $T^2$ against $L$

For one length, $g$ comes from one reading and the error is large. A graph uses all the readings together. Since $T^2 = \dfrac{4\pi^2}{g}\,L$, the graph of $T^2$ against $L$ is a straight line through the origin, and

If you measure only the length of the thread, $L_s$, the line is $T^2 = \dfrac{4\pi^2}{g}(L_s + r)$. It is still straight with the same slope, but it cuts the $T^2$ axis above zero. It cuts the length axis at $-r$, the negative of the radius of the bob. The slope, and so $g$, is the same. The graph is more forgiving than one calculation.

Move the radius from $0$ to $3$ cm. What happens to the slope? What happens to the intercept?

The least-squares line

A straight line $y = mx + c$ through $n$ points $(x_i, y_i)$ is the best when it makes the sum of the squares of the vertical gaps as small as possible. The formulas are

A line drawn by eye should look like this one. Follow these rules when you draw a graph on paper.

• Choose the scale so that the points fill most of the sheet. Use simple steps like 1, 2 or 5 squares for one unit. Do not start at zero unless the line must pass through zero.
• Label each axis with the quantity and its unit.
• Mark each point as a small cross or a dot in a circle. Do not use a big dot.
• Draw one thin straight line through the points, so that about as many are above it as below it. Do not join the points one by one.
• Find the slope from two points on the line, far apart, not from two measured points. Show the triangle on the graph.

The error in the slope. Draw the steepest and the flattest line that still pass through the cloud of points. Their slopes are $m_{max}$ and $m_{min}$. The error is $\Delta m = \dfrac{m_{max} - m_{min}}{2}$.

A worked example. A pendulum gives these readings. The length from the point of suspension to the centre of the bob is $L$, and $t$ is the time for 20 oscillations.

• $L = 0.40$ m: $t = 25.4$ s, $T^2 = 1.613\ \text{s}^2$.
• $L = 0.50$ m: $t = 28.4$ s, $T^2 = 2.016\ \text{s}^2$.
• $L = 0.60$ m: $t = 31.2$ s, $T^2 = 2.434\ \text{s}^2$.
• $L = 0.70$ m: $t = 33.7$ s, $T^2 = 2.839\ \text{s}^2$.
• $L = 0.80$ m: $t = 36.0$ s, $T^2 = 3.240\ \text{s}^2$.
• $L = 0.90$ m: $t = 38.2$ s, $T^2 = 3.648\ \text{s}^2$.
• $L = 1.00$ m: $t = 40.1$ s, $T^2 = 4.020\ \text{s}^2$.

1. The time period. $T = t/20$. For the first length, $T = \dfrac{25.4\ \text{s}}{20} = 1.27$ s, and $T^2 = 1.613\ \text{s}^2$.
2. The slope. The least-squares formulas give $m = 4.03\ \text{s}^2/\text{m}$ and $c = 0.007\ \text{s}^2$.
3. The value of g. $g = \dfrac{4\pi^2}{m}$ $= \dfrac{39.48}{4.03\ \text{s}^2/\text{m}}$ $= 9.79\ \text{m/s}^2$.

The intercept $0.007$ s² is close to zero, which says that $L$ was measured in the right way. The standard value is near $9.8\ \text{m/s}^2$.

The metre scale and the principle of moments

A uniform metre scale has its centre of gravity at the $50.0$ cm mark. If it is supported on a knife edge at that mark, it stays level. This is a good test that the scale is uniform.

The principle of moments. A body that stays in rotational equilibrium has equal clockwise and anticlockwise moments about any point:

A moment is a force times its perpendicular distance from the turning point. For weights on a scale, the same $g$ appears on both sides and cancels, so we can use masses.

Finding the mass of the scale. Hang a known mass $m$ at the $a$ cm mark. Move the knife edge until the scale balances at the mark $x$ (with $a < x < 50$). The weight of the scale acts at $50.0$ cm. About the knife edge:

For example, $m = 50.0$ g at $a = 10.0$ cm balances at $x = 39.7$ cm. Then $M = \dfrac{50.0\ \text{g} \times 29.7\ \text{cm}}{10.3\ \text{cm}}$ $= 144$ g.

The same idea finds the mass of an unknown body. Hang the body on one side and a known mass on the other, find the balance, and use the moments. Keep the knife edge at the $50.0$ cm mark, and the weight of the scale does not enter at all.

Practise with new numbers

Try it in Python

The program finds the best line through the seven readings of the worked example from the least-squares formulas. Then it finds $g$.

Common mistakes

• Measuring the length to the top of the bob. Measure to the centre of the bob.
• A big angle. The formula needs a small angle, less than $10^\circ$. For a bigger one the period is longer.
• Counting the first pass as one. Count the first pass as zero, so that 20 oscillations really means 20.
• A moving support. If the point of suspension slips, the length changes in each swing.
• Joining the points of a graph one by one. Draw one best line.
• Taking the slope from two measured points. Take two points on the line, far apart.

Remember these

• $T = 2\pi\sqrt{L/g}$, so $g = \dfrac{4\pi^2 L}{T^2}$, and the slope of $T^2$ against $L$ is $\dfrac{4\pi^2}{g}$.
• The effective length is from the point of suspension to the centre of the bob.
• Time 20 or more oscillations, so that the stopwatch error is shared among them.
• Least squares: $m = \dfrac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$.
• Moments: $m(x - a) = M(50.0 - x)$ for a scale balanced at $x$.

Check yourself

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

1. The length of a pendulum is $0.800$ m and 20 oscillations take $35.9$ s. What is g?

2. In a pendulum graph, the slope of $T^2$ against $L$ is $4.10\ \text{s}^2/\text{m}$. What is g?

3. A metre scale balances at the $50.0$ cm mark. A $60.0$ g mass is hung at the $20.0$ cm mark and an unknown mass at the $80.0$ cm mark. The scale is level. What is the unknown mass?

4. Why is a graph better than finding g from one length?

References

• NCERT, Laboratory Manual: Physics, Class XI, the experiments on the simple pendulum and on the metre scale and the principle of moments.
• OpenStax University Physics Volume 1, Chapter 15 (Oscillations), the section on the simple pendulum.