Grade by Grade
C. Energy and Momentum · Work, Energy and Simple Harmonic Motion

The Simple Pendulum

C3.2 C2.3About 70 minutes
Learning goals
  • I can explain why a pendulum swinging through small angles moves in simple harmonic motion.
  • I can derive and use T = 2π√(L/g).
  • I can explain why a pendulum's period does not depend on its mass or (for small swings) its amplitude.
  • I can use a pendulum to measure the gravitational field strength g.

Words to know

simple pendulum
A small, heavy bob on a light, inextensible string that swings about a fixed pivot.
small-angle approximation
For small angles in radians, sin θ ≈ θ.
restoring force (pendulum)
The component of gravity along the arc, −mg sin θ, which pulls the bob back toward the lowest point.
seconds pendulum
A pendulum with a period of 2.0 s (one second per one-way swing), about 0.99 m long on Earth.

Clocks that swing

For a long time, pendulum clocks were among the most accurate clocks in the world. The Dutch scientist Christiaan Huygens invented the pendulum clock in 1656, and the clockmaker Salomon Coster built the first ones to his design. They worked because a pendulum swinging through small arcs takes almost exactly the same time for every swing, big or small. In this lesson you will see why a pendulum is (nearly) a simple harmonic oscillator, derive its period, and use it to measure g at home.

The restoring force on a pendulum

A simple pendulum is a small, heavy bob of mass m on a light string of length L. When the string makes an angle θ with the vertical, two forces act on the bob: the string's tension F_T and gravity mg.

Split gravity into two components:

  • mg cos θ along the string (balanced by part of the tension; together they keep the bob on its circular path),
  • mg sin θ along the arc, pointing back toward the lowest point.

So the restoring force along the arc is F = −mg sin θ.

This is not quite proportional to the displacement, because of the sine. But for small angles, with θ in radians, sin θ ≈ θ, and the arc length from the bottom is x = Lθ. Then

F ≈ −mg(x/L) = −(mg/L)x

That is Hooke's-law form with an effective spring constant k = mg/L. A pendulum swinging through small angles is a simple harmonic oscillator.

arc length x = LθθLmgF_Tmg cos θmg sin θRestoring force along the arc:F = −mg sin θSmall angles: sin θ ≈ θ = x/LF ≈ −(mg/L)xLike a spring with k = mg/L:T = 2π√(L/g)
Forces on a pendulum bob. The component mg sin θ along the arc pulls the bob back toward the bottom. For small angles it is close to −(mg/L)x, which makes the motion simple harmonic.

The period of a pendulum

Put k = mg/L into T = 2π√(m/k):

T = 2π√(L/g)

  • The mass cancels: a heavy bob and a light bob on the same length of string have the same period.
  • The period depends only on the length and on g. Four times the length gives twice the period.
  • For small amplitudes the period does not depend on the amplitude. The approximation sin θ ≈ θ is very good for small angles: for amplitudes up to about 15°, T = 2π√(L/g) is accurate to better than 0.5%. Larger swings take slightly longer.
  • Because T depends on g, timing a pendulum is a simple way to measure g: g = 4π²L/T².

Energy view: a pendulum is also an energy converter, as in Conservation of Mechanical Energy: Tracks, Pendulums and Loops: Eg at the ends of the swing (h = L(1 − cos θ)) becomes Ek at the bottom.

L (m)T (s)0.511.522.51230.25 m: 1.0 s1.0 m: 2.0 s2.25 m: 3.0 sT = 2π√(L/g)4 × the length→ 2 × the period
Period against length for a pendulum on Earth. The curve is a square-root graph: making the pendulum four times as long only doubles its period.
Worked example

Example 1: A playground swing

Treat a swing with 1.5 m chains as a simple pendulum. Find its period for small swings.

  1. T = 2π√(L/g) = 2π√(1.5 ÷ 9.8) = 2π√0.1531 = 2π(0.3912).
  2. T = 2.5 s (2.46 s).

About 2.5 s per complete back-and-forth swing. To push in rhythm, a parent pushes about once every 2.5 s.

Worked example

Example 2: The seconds pendulum

A traditional grandfather clock has a pendulum that takes 1.0 s for each one-way swing, so its period is 2.0 s. How long is it?

  1. Solve T = 2π√(L/g) for L: L = gT²/(4π²).
  2. L = (9.8)(2.0)² ÷ 39.48 = 39.2 ÷ 39.48 = 0.99 m.

About 0.99 m, which is why grandfather clocks are so tall.

Worked example

Example 3: Measuring g

A student uses a pendulum 0.800 m long and times 20 complete swings in 35.9 s. Find g.

  1. T = 35.9 ÷ 20 = 1.795 s.
  2. g = 4π²L/T² = 4π²(0.800) ÷ (1.795)² = 31.58 ÷ 3.222 = 9.80 N/kg.

g ≈ 9.80 N/kg. Timing 20 swings makes the reaction-time error about 20 times smaller per swing.

Worked example

Example 4: A pendulum on the Moon

On the Moon, g = 1.62 N/kg. What is the period of the 1.5 m swing from Example 1 there?

  1. T = 2π√(1.5 ÷ 1.62) = 2π√0.9259 = 2π(0.9623) = 6.0 s.
  2. Check by ratio: T_Moon ÷ T_Earth = √(9.8 ÷ 1.62) = 2.46, and 2.46 × 2.46 s = 6.0 s. ✓

About 6.0 s, so a swing would feel slow and floaty on the Moon.

Just for fun

A grandfather clock is really a physics experiment that has been running in someone's living room for decades: its pendulum is about one metre long precisely so that each one-way swing takes one second.

Hands-on activity

Measure g with a pendulum

You need: a piece of string about 1.2 m long, a heavy small object such as a large metal nut or a few washers, tape, a tape measure, a protractor, and a phone stopwatch.

  1. Tie the bob to the string and tape the top firmly to the top of a door frame or the edge of a high shelf (a table edge is only high enough for the shorter lengths). Measure L from the pivot to the centre of the bob.
  2. Pull the bob aside to about 10° and release it from rest (do not push). Time 20 complete swings and divide by 20 to get T.
  3. Repeat for four or five lengths (for example 1.00, 0.80, 0.60, 0.40 and 0.20 m).
  4. For each length calculate g = 4π²L/T². Or, better: plot T² against L. The graph should be a straight line through the origin with slope 4π²/g, so g = 4π² ÷ slope.
  5. Test the claims: does changing the bob's mass change T? Does releasing from 5° instead of 15° change T?

What to look for: your values of g should fall between about 9.6 and 10.0 N/kg. Mass makes no measurable difference; a very large release angle (40° or more) makes T noticeably longer. Common errors: measuring L to the top of the bob, and miscounting swings (count "zero" when you start the timer).

Your turn

1

Find the period of a 0.50 m pendulum.

2

What length of pendulum has a period of 1.0 s?

3

The bob of a pendulum is replaced by one with twice the mass. What happens to the period?

4

A pendulum is made four times as long. What happens to its period?

5

On Mars, g ≈ 3.7 N/kg. Find the period of a 1.0 m pendulum there.

6

Why must the swing be small for T = 2π√(L/g) to apply?

7

Find the frequency of a 2.0 m pendulum.

More practice: warm-up

1

Write the formula for the period of a simple pendulum.

2

Which two quantities does a pendulum's period depend on?

3

Which force provides a pendulum's restoring force?

4

What is the effective spring constant of a pendulum?

5

Write the formula for g from a pendulum's length and period.

More practice: on your own

1

A 1.20 m pendulum makes 10 swings in 22.0 s. Find g.

2

A pendulum clock runs slow (its pendulum takes too long per swing). Should the pendulum be lengthened or shortened? Explain.

3

A child on a 2.5 m swing is pushed once per complete swing. About how many pushes per minute is that?

4

What length gives a period of 3.0 s?

5

Air resistance slowly reduces a pendulum's amplitude from 10° to 5°. What happens to its period?

Challenge

1

Derive T = 2π√(L/g) starting from the restoring force on a pendulum bob.

2

A pendulum clock keeps perfect time where g = 9.81 N/kg. It is moved to a place where g = 9.78 N/kg. Does it run fast or slow, and by about how many seconds per day?

3

A 1.0 m pendulum hangs in an elevator that accelerates upward at 2.0 m/s². Find its period, and compare with its period when the elevator is at rest.

Check your understanding

Quiz

1.

The period of a simple pendulum swinging through small angles depends on:

2.

The period of a 2.0 m pendulum on Earth is closest to:

3.

Why is a pendulum only approximately a simple harmonic oscillator?

4.

To halve the period of a pendulum, its length must be:

5.

A pendulum's period is measured as 2.00 s and its length as 0.993 m. The value of g is:

6.

Two pendulums have the same length. One bob is iron, the other is wood of the same size. Ignoring air resistance, their periods are:

For parents and teachers

Big idea (C3.2, C2.3): for small swings, a pendulum's restoring force is nearly proportional to its displacement, so it moves in simple harmonic motion with period T = 2π√(L/g). The period does not depend on the mass or (for small swings) the amplitude.

How to help: do the measure-g activity together. Students are often surprised that changing the bob's mass does not change the period; let them test it.

Common difficulties:

  • Measuring L to the top of the bob instead of to its centre.
  • Timing one swing instead of many, or counting a one-way swing as a full period.
  • Using the formula for large swings (over about 15°).
  • Calculator in degree mode when using sin θ ≈ θ (θ must be in radians).