Take a pendulum from Lahore to the top of a mountain and it ticks at a slightly different rate. Change the bob from iron to wood and nothing changes. One formula explains both facts: T = 2π√(l/g). This note shows where it comes from and how to use it.
Chapter 14 of the PECTAA Physics 12 book is Simple Harmonic Motion. After the mass-spring system, section 14.3 treats the simple pendulum as a second practical SHM system. This note is for FSc Part 2 (2nd year, Class 12) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.
What to know
The set-up. A small, heavy, solid bob of mass m hangs from a rigid support by a light, inextensible string. Pull it aside through an angle θ to an extreme position and release it. A restoring force pulls it back, and it swings to and fro along a curved path about the mean position.
The forces. Resolve the weight mg at the extreme position:
- mg cos θ lies along the string. It is balanced by the tension, because the bob does not move along the string.
- mg sin θ acts along the path towards the mean position. This is the restoring force: F = −mg sin θ.
Newton's second law gives ma = −mg sin θ, so a = −g sin θ.
Small angles. For small θ (in radians), sin θ ≈ θ. The book's table compares θ and sin θ from 0° to 10° and shows they match closely. Also θ = S/l, where S is the arc length, and for small angles the arc S is very nearly the displacement x. So
a = −(g/l)x
With l and g fixed, g/l is constant, so a ∝ −x. The motion of a simple pendulum is simple harmonic for small angles.
Time period. Comparing a = −(g/l)x with a = −ω²x gives ω² = g/l. Using ω = 2π/T:
T = 2π√(l/g)
The time period is directly proportional to √l and inversely proportional to √g. It does not depend on the mass of the bob or on the amplitude.
Second pendulum. A pendulum whose time period is two seconds.
Measuring g. Measure the length l, time the oscillations with a stopwatch to find T, then use g = 4π²l/T².
How good is sin θ ≈ θ? A few values (rounded to four places) show the gap stays tiny:
| θ (degree) | θ (radian) | sin θ |
|---|---|---|
| 2 | 0.0349 | 0.0349 |
| 6 | 0.1047 | 0.1045 |
| 8 | 0.1396 | 0.1392 |
| 10 | 0.1745 | 0.1736 |
Even at 10° the two differ by only about half a percent. That is why the book treats a pendulum swinging through a small angle as SHM, and why the book recommends a small swing when timing a pendulum.
Key points
- Restoring force is mg sin θ, not mg cos θ.
- SHM only for small angles, about 10° or less in the book's table.
- Longer string, longer period. Stronger gravity, shorter period.
Worked examples
Example 1 (book example). Find the length of a second pendulum and its frequency where g = 9.8 m s⁻².
- T = 2 s. From T = 2π√(l/g): l = gT²/4π².
- l = (9.8 × 2²) / (4 × 3.14²) = 39.2 / 39.44 ≈ 0.994 m, about 99.4 cm.
- f = 1/T = 0.5 Hz.
Example 2. A pendulum of length 1.6 m swings where g = 9.8 m s⁻². Find T.
- T = 2π√(1.6/9.8) = 2π × 0.404 ≈ 2.54 s.
Example 3. A student's pendulum is 0.50 m long and completes 20 vibrations in 28.4 s. Find g.
- T = 28.4/20 = 1.42 s.
- g = 4π²l/T² = 4 × 9.87 × 0.50 / 2.02 ≈ 9.8 m s⁻².
Example 4. The length of a pendulum is made four times as long. What happens to T? T ∝ √l, so T doubles.
Common mistakes in the exam
- Putting mass into the formula. T does not depend on the bob's mass.
- Using large angles. Beyond small angles, sin θ ≈ θ fails and the motion is not exactly SHM.
- Using degrees in θ = S/l. The angle must be in radians.
- Calling a second pendulum's period one second. Its period is two seconds.
- Forgetting to square T when finding g or l.
Exam technique
- In the derivation, show the resolved components, then a = −g sin θ, then the small-angle step, then a = −(g/l)x.
- State clearly: "Hence the motion is SHM, and T = 2π√(l/g)."
- Time many vibrations, not one, when describing the g experiment, and divide.
- Check your board's notice for the topics in your paper.
Practice questions
- Which component of weight gives the restoring force? Answer: mg sin θ, along the path towards the mean position.
- Does T change if the bob's mass is doubled? Answer: No.
- Find T for l = 0.25 m and g = 9.8 m s⁻². Answer: T = 2π√(0.25/9.8) ≈ 1.0 s.
- Would a pendulum of fixed length have the same period on the Moon? Answer: No. g is smaller on the Moon, so T is longer.
- What happens to T if the length is halved? Answer: T falls by a factor √2, to about 0.71 of its value.
Ready to practise this? Continue in 2nd Year Physics.