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Mechanics · Simple Harmonic Motion

Pendulum Lab

Time the swing of a simple pendulum as you change its length, its mass, and the size of its swing. Discover which factors actually affect the period, verify T = 2π√(L/g), and use a graph of period-squared versus length to measure the acceleration due to gravity, g.

Theory — The Simple Pendulum

What a Pendulum Is

A simple pendulum is a small heavy bob hanging from a light string of length L, free to swing back and forth. When pulled aside and released, gravity pulls it back toward the lowest point, it overshoots, and it swings to the other side — repeating in a regular rhythm. One complete back-and-forth swing is one period T, measured in seconds.

The Period Equation

For small swings, the period of a simple pendulum depends on only two things: the length of the string and the local acceleration due to gravity. Remarkably, it does not depend on the mass of the bob or (for small angles) on how far you pull it aside.

Period of a Simple Pendulum (small angle) T = 2π · √(L / g)

T = period (s), L = length (m), g = 9.8 m/s² on Earth
Period depends on length and gravity — not mass or amplitude

Why Mass and Amplitude Don't Matter

Mass cancels out: a heavier bob is pulled harder by gravity, but it also has more inertia resisting that pull, and the two effects exactly cancel — just as all objects fall at the same rate. Amplitude doesn't matter (for small swings) because a wider swing covers more distance but the restoring force is correspondingly larger, so the bob moves faster and the time stays the same. This property — a period independent of amplitude — is called isochronism and is why pendulums make good clocks.

Period Squared and the Straight-Line Graph

Squaring the period equation turns the square-root relationship into a straight line, which is far easier to analyze. A graph of T² against L is a straight line through the origin whose slope is 4π²/g — so measuring the slope lets you calculate g.

Linearized Form — Finding g T² = (4π² / g) · L

slope of T² vs L = 4π² / g  →  g = 4π² / slope
T² ∝ L · slope gives g

Increase the length

Longer string → longer period. Because T ∝ √L, quadrupling the length doubles the period.

Change mass or amplitude

No effect on the period (for small swings). The period depends only on L and g.

ChangeEffect on period TReason
Increase length Lincreases (T ∝ √L)T = 2π√(L/g)
Increase bob massno changemass cancels (like free fall)
Increase amplitude (small)no changeisochronism
Weaker gravity (e.g. Moon)increases (T ∝ 1/√g)T = 2π√(L/g)

Instructions — Running the Virtual Experiment

The Pendulum tab lets you swing the bob and time its period; the Find g tab records period versus length so you can measure gravity. Record every reading in your lab notebook.

Experiment 1 — What Affects the Period? (Pendulum tab)
1
Open Simulation → Pendulum. Set the length to 1.0 m and release the bob. The simulation times each swing and displays the period. Compare it to the prediction T = 2π√(L/g).
2
Change the mass and release again. Confirm the period does not change. Then change the amplitude (release angle) for small angles and confirm the period is essentially unchanged.
3
Increase the length and observe the period grow. Quadruple the length and confirm the period roughly doubles (T ∝ √L). Optionally switch gravity to the Moon and see the period lengthen.
Experiment 2 — Measuring g from a Graph (Find g tab)
1
Open Find g. For each length button (0.25–2.0 m), the timed period is shown. Click Record reading to log L, T, and T².
2
With several points recorded, click Plot T² vs L & fit. The points fall on a straight line through the origin. The fitted slope equals 4π²/g.
3
The simulation computes g = 4π²/slope. Confirm your value is close to 9.8 m/s² and report the percent error in your write-up.

Simulation — Timing the Pendulum & Measuring g

Pendulum Virtual LabRelease the bob and time its period
pendulum bob
The timer measures one full period (there and back).

Controls

Readout
Length L— m
Measured period— s
Predicted 2π√(L/g)— s
Swings counted0
Period depends on L and g only.
Each point is one (length, period²) reading. Slope of T² vs L = 4π²/g.

Set length (Earth gravity)

Current reading (L = 0.25 m)
Period T— s
— s²
Length L (m)Period T (s)T² (s²)T²/L (s²/m)
No readings yet — pick a length and click "Record reading".

Team Questions

Question 1. Calculate the period of a 1.0 m simple pendulum on Earth (g = 9.8 m/s²). Use T = 2π√(L/g). (Type just the number in seconds — e.g. 2.01)
Question 2. Two pendulums are identical except that one has a 2 kg bob and the other a 4 kg bob, both 1.0 m long. Which has the longer period — the 2 kg, the 4 kg, or are they the same? (Answer "2 kg", "4 kg", or "same")
Question 3. A pendulum has a period of 2.0 s. If you quadruple its length (×4), what is the new period? (Type just the number in seconds)
Question 4. When you plot period-squared (T²) on the y-axis against length (L) on the x-axis, the graph is a straight line. The slope equals 4π²/g. If the slope is 4.03 s²/m, what value of g does it give? (Type just the number in m/s²)
Question 5. A 1.0 m pendulum is taken to the Moon, where g = 1.6 m/s². What is its period there? Use T = 2π√(L/g). (Type just the number in seconds — e.g. 4.97)
Question 6. For small swings, does increasing the amplitude (the release angle) change the period of a pendulum? (Answer "yes" or "no")
Question 7 — Challenge. A pendulum clock keeps perfect time on Earth. It is carried to the Moon (g = 1.6 m/s²). Because the period there is longer, will the clock run fast or slow compared to real time? (Answer "fast" or "slow")

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.

Pendulum Lab: Period, Length, and the Measurement of g

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To investigate how the period of a simple pendulum depends on its length, mass, and amplitude, to verify the relationship T = 2π√(L/g), and to determine the local acceleration due to gravity from the slope of a graph of period-squared versus length.

Theory

For small swings the period of a simple pendulum depends only on its length and the acceleration due to gravity, not on the mass of the bob or the amplitude. Squaring the period relation gives a linear equation whose slope yields g.

T = 2π√(L/g)
T² = (4π²/g)·L → slope = 4π²/g
g = 4π² / slope

Calculations — Sample: L = 1.00 m on Earth

Predicted period: T = 2π√(1.00/9.8) = 2π(0.3194) = 2.01 s

Period squared: T² = (2.01)² = 4.03 s²; T²/L = 4.03/1.00 = 4.03 s²/m

g from the slope: slope = 4.03 s²/m → g = 4π²/4.03 = 39.48/4.03 = 9.80 m/s²

Percent error: |9.80 − 9.8|/9.8 × 100% = 0.0%

Results Table — Period vs Length (Earth, small amplitude)

Length L (m)Period T (s)T² (s²)T²/L (s²/m)
0.251.001.014.03
0.501.422.014.03
1.002.014.034.03
1.502.466.044.03
2.002.848.064.03

Control trials: changing the bob mass from 1 kg to 5 kg left the period unchanged; changing the amplitude from 5° to 30° changed the period by less than 2%. Both confirm the period's independence from mass and (small) amplitude.

Discussion

The measured periods matched the prediction T = 2π√(L/g) at every length. The quantity T²/L was constant at 4.03 s²/m across the whole range, confirming that T² is proportional to L. A graph of T² against L was a straight line through the origin with slope 4.03 s²/m. Using g = 4π²/slope gave g = 9.80 m/s², in excellent agreement with the accepted value of 9.8 m/s².

The control experiments confirmed the two surprising predictions of the theory. Changing the mass of the bob had no measurable effect on the period, because gravitational pull and inertia scale together and cancel. Changing the amplitude (for small angles) also left the period essentially unchanged, the property of isochronism that makes pendulums useful as timekeepers. Reducing gravity (the Moon setting) lengthened the period, consistent with T ∝ 1/√g.

Conclusion

The experiment verified that the period of a simple pendulum is given by T = 2π√(L/g), depending only on length and gravity. The period was independent of mass and of small amplitude. From the slope of the T²-versus-L graph the acceleration due to gravity was found to be 9.80 m/s², matching the accepted value and confirming the pendulum as an accurate method for measuring g.

Practice Questions

Show all work and include units in your answers.

Question 1
Find the period of a simple pendulum of length 0.45 m on Earth (g = 9.8 m/s²). How many complete swings does it make in one minute?
Hint: T = 2π√(L/g). Number of swings = 60 s ÷ T.
Question 2
A pendulum has a period of 1.6 s on Earth. What is its length?
Hint: solve T = 2π√(L/g) for L: L = g(T/2π)².
Question 3
A grandfather clock uses a pendulum with a period of exactly 2.0 s (a "seconds pendulum," one second per swing). What length must the pendulum be on Earth?
Hint: L = g(T/2π)² with T = 2.0 s. You should get close to 0.99 m.
Question 4
In a lab, a student measures the following periods: L = 0.40 m → T = 1.27 s; L = 0.80 m → T = 1.80 s; L = 1.20 m → T = 2.20 s. Compute T²/L for each and find the average. Use it to calculate g.
Hint: T²/L should be roughly constant (≈ 4π²/g). Then g = 4π²/(T²/L).
Question 5
Explain why two pendulums of the same length but different bob masses have the same period. Refer to how gravitational force and inertia each depend on mass.
Hint: force ∝ m (more pull) but acceleration = F/m, so mass cancels — like all objects falling at the same rate.
Question 6 — Challenge
An astronaut on an unknown planet sets up a 1.00 m pendulum and measures a period of 3.6 s. What is the acceleration due to gravity on that planet? How does it compare to Earth's?
Hint: g = 4π²L/T² = 4π²(1.00)/(3.6)². Compare your answer to 9.8 m/s².